Introduction to Physics
This is my guide on the basics of physics.
Table of Contents
Measurement and Uncertainty
The beginning of physics started with measuring quantities. It soon became apparent that not all data meant the same. It was important to have reliable measurements. However, that was often a problem. Measuring things was often a random affair. The devices used were not all made the same either. Many measuring devices were poorly made and inaccurate at first. So measurements will never be perfect. They will not because humans are error prone and the devices we use are as well.
Uncertainty
Every device, like a ruler, has a built in amount of uncertainty. It is vital to state this when publishing measurements. It also gives the uncertainty as a percentage. This can help make it clear. It is the ratio of the uncertainty to the measured value, multiplied by 100.
If we measure a piece of metal to be 6.5 cm long, we can assume it has an uncertainty of .1 cm. To get the percentage of uncertainty we divide:
\[ \frac{0.1}{6.5} = .015 = 1% \]
Significant Figures
This is the number of reliably known digits in a number. Non-zero digits are usually significant. Zeroes vary according to the situation. In basic lab work, you never want to state more precision than you have. This is just inaccuracy. If your measurements have 3 significant digits, then your calculations should have only 3 significant digits. Many people do this, don’t be one of them, please. If you are dealing with numbers that contain different amounts of significant digits, then always use the least amount. Consider the number 12 and 23.4. I use these two numbers in a calculation. The result should have 2 significant figures in it.
Scientific Notation
When you are dealing with numbers that contain many digits it is convenient to use scientific notation. A number like 100,00,000 can be written as \( 1.0 * 10^7 \). This is just a standard way to display numerical values. Another example is the number .00000383. I would write it as \( 3.83 * 10^-6 \).
Derived Quantities
Derived quantities come from the known base quantities. Base quantities are length, time, mass, electric current, temperature, substances, and luminosity. So a derived quantity uses these to form a different unit. Speed uses time and length units, for example.
Conversion of Units
Anything that we can measure comprises, a number, and the associated unit. Without the unit, the number means nothing. Sometimes there are multiple units involved. When this happens, we must convert to something meaningful to us. Lets convert 15 inches to centimeters. 1 in = 2.54 cm.
\[ 15 in * \frac{2.54 cm}{in} = 38.1 = 38 cm \]
You see that the inches cancelled out. This is the trick to knowing if you set it upright. The correct units will cancel out and it will leave you with the one unit that you want.
Example 1
We consider the age of the universe to be 14 billion years old. Let’s write this in scientific notation.
\[ 14,000,000,000 = 1.4 * 10^10 \]
Example 2
How many significant figures do the following numbers have? 214, 81.6, and .0086?
214 = 3 significant digits, 81.6 = 3 significant digits, and .0086 has 2 significant digits
Example 3
Write out the following numbers in scientific notation. 21.8, .0068
21.8 = 2.18 * 10^1 and .0068 = 6.8 8 10^-3
Write out the following number. \( 8.69 * 10^4 \)
\[ 8.6900 \]
This section was about the basics of measurements and uncertainty. When measuring, you must always take into account the uncertainty of your tool. As you state your calculation, give the correct number of significant digits.
Vectors
Quantities like mass and temperature are described with a number called a scalar. Other quantities like displacement, velocity, or force have a direction associated with them and are called vectors. When looking at a vector on paper, a vector is an arrow with the length of the arrow representing the number and direction of the vector.
A vector can be described with a number and an angle, like \( A=19 \angle 25 \deg \). When doing basic math operations, you should use the component form to write vectors. So, if a line were placed with the tail of the vector at the origin of the coordinate system, x and y portions can be calculated like this:
\( y = 19 \sin 25^{\circ} = 8.02 \)
\( x = 19 \cos 25^{\circ} = 17.21 \)
This is how you calculate vectors.
If you have the x and y vectors, you can calculate the length of the corresponding line by:
\( \sqrt{x^{2} + y{2}} \)
\( \sqrt{8.02^{2} + 17.21^{2}} \)
\( \sqrt{64.32 + 296.18} \)
\( \sqrt{360.5} = 19 \)
To find the original angle we use:
\( \theta = \tan^{-1} \frac{y}{x} \)
So:
\( \theta = \tan^{-1} \frac{8.02}{17.21} \)
\( \theta = 25^{\circ} \)
Example 1
Write out the components of the vector \( c = 47 \angle 193^{\circ} \)
The x component is:
\( x = 47 \cos 193^{\circ} = -45.8 \)
The y component is:
\( y = 47 \sin 193^{\circ} = -10.6 \)
The use of unit vectors simplifies the mathematical operations on vectors. In two dimensions, unit vectors are vectors of unit value and in the +x and +y directions.
So the first vector would be written as \( 18.4i + 13.8j \) and the second vector is
\( -45.8i - 10.6j \).
You can add vectors by adding the individual components. If you have:
\( (18.4i + 13.8j) + (-45.8i - 10.6j) \)
You just add the individual components together. The combined vector becomes:
\( -27.4i + 3.2j \)
Subtracting vectors can also be done. It is a bit quirky, however. You will use the opposite sign of the second vector.
\( (18.4i + 13.8j) - (-45.8i - 10.6j) \)
This gives the result of:
\( 64.2i + 24.4j \)
We should also know how to work backward when given the components. When we do this properly it will give us the length of the vector and its angle.
Let us use the last vector:
\( 64.2i + 24.4j \)
\( V = \sqrt{64.2^{2} + 24.4^{2}} = 68.7 \)
\( \theta = \tan^{-1}\frac{24.4}{64.2} = 20.8^{\circ} \)
When put in vector form:
\( V = 68.7 \angle 20.8^{\circ} \)
Example 2
Add the vectors:
\(A=13 \angle 50^{\circ}\)
\(B=15 \angle -60^{\circ} \)
\(C=17 \angle 20^{\circ} \)
The first step is to put each vector in its component form.
\( A = \sin 50 * 13 = 10 = y \)
\( A = \cos 50 * 13 = 8.4 = x \)
\( A = 8.4i + 10.0j \)
\( B = \sin(-60) * 15 = -13 = y \)
\( B = \cos(-60) * 15 = 7.5 = x \)
\( B = 7.5i -13.0j \)
\( C = \sin 120 * 17 = -8.5 = y \)
\( C = \cos 120 * 17 = -14.7 = x \)
\( C = -14.7i -8.5 j \)
Sum all the i’s and all the j’s. You get:
\( 1.2i -11.5j \)
That is your x and y dimensions
Now use the Pythagorean theorem with those values
\( \sqrt{11.5^{2} + 1.2^{2}} = 11.6 \)
That gives you the length of the new vector
Now, let’s find the new angle
We use the inverse tangent to do this
\( \tan^{-1} \frac{-11.5}{1.2} = -84^{\circ} \)
So the new vector is:
\( 11.6 \angle -84^{\circ} \)
Use as many diagrams as possible when doing problems like this. Diagrams help keep your thought processes straight and help prevent silly errors.
Example 3
Find the resultant of the two forces \(800(N) \angle 47^{\circ}\) and \(600(N) \angle 140^{\circ}\)
I will write out all the calculation steps first, before I actually compute them
\(800 \cos 47^{\circ} = 545.6 \)
\(800 \sin 47^{\circ} = 585.1 \)
\(600 \cos 140^{\circ} = -460.0 \)
\(600 \sin 140^{\circ} = 385.7 \)
Vector 1 is: \(545.6i + 585.1j\)
Vector 2 is: \(-460.0i + 385.7j\)
Add the vectors together
We get:
\(85.6i + 970.8j\)
Use Pythagorean theorem
\(\sqrt{85.6^{2} + 970.8^{2}}\)
The length of our new vector is:
\(= 974.6 \)
The new angle of our vector is:
\(\tan^{-1}\frac{970.8}{85.6} = 85^{\circ} \)
So:
\(974.6(N) \angle 85^{\circ}\)
Scalars
The dot product of two vectors produces a scalar. Scalars, as you know, do not have a direction associated with them. There are two ways to compute a dot product.
\(A*B = AB \cos \theta\)
And
\(A*B = a_xb_x + a_yb_y\)
Example 4
Form the dot product of \(A = 23 \angle 37^{\circ}\) and \(B = 14 \angle -35^{\circ}\)
Use the first definition
\(A*B = A*B \cos \theta = 23*14 \cos 72^{\circ} = 100\)
Now use the second definition
This requires the components of both vectors
We learned how to do this earlier, it is the same process
We get:
\(18.4i + 13.8j\) and \(11.5i + (-8.0)j\)
\(A*B=100\)
Cross Product
The cross product of A*B also produces a vector.
\(A*B \sin \theta\)
Example 5
Form the cross product of \(A = 23 \angle 37^{\circ}\) and \(B = 47 \angle 193^{\circ}\)
This is A*B* sine of the difference in the angles.
The difference in angles is 156 degrees.
\(A*B = 23*47*\sin 156^{\circ} = 440\)
We can also use determinants to get the same result
Calculate the components of each vector:
Cross multiply. This is why it is called the cross product
\(A*B = 18.4(-10.6) - 13.8(-45.8) = 440\)
If it is a simple problem then use the first definition any time you can. Otherwise, use the second definition.
Kinematics
The motion of a particle is described by giving position, velocity, and acceleration, usually as a function of time. We start with one dimension. Distance means total distance traveled, while displacement is the actual difference between endpoint and the beginning.
\[Velocity = \frac{displacement}{time}\]
and:
\[Acceleration = \frac{v2-v1}{time}\]
This velocity calculation fits our present definition and is equivalent to drawing a straight line between points on the smooth curve of x versus t. Calculating velocity this way presents a problem, because, depending on the interval, we will get different answers for the velocity. To find the velocity at t=1, we found x at t=1 and x at t=2 and performed the velocity calculation. This is not a good approximation to the velocity at t=1 because the average is between t=1 and t=2. For better approximations, shorten the time interval centered about t=1. Getting it as small as you can will give better results. So, the slope of the straight line tangent to the curve at x=1 represents the velocity at x=1.
Instantaneous Velocity and Acceleration
A more versatile definition of velocity is \(\frac{\Delta x}{\Delta t}\), where the interval \(\Delta t\) is very small and centered about the time where the velocity is desired. This approach leads to a general method for obtaining an expression for velocity that can be evaluated at any point rather than going through the numeric calculation whenever we want a velocity. This definition is stated as the instantaneous velocity.
\[v = \lim\limits_{\Delta t \to \0} = \frac{dx}{dt}\]
This definition of the derivative is the slope of the tangent to the curve evaluated at the point in question. If we want to find the velocity of any particle traveling according to a polynomial relationship between x and t, all we need is a general technique for finding the slope of the tangent to the polynomial at any point.
The instantaneous acceleration is defined as the value of \(\Delta v / \Delta t\) as \(\Delta t\) approaches zero.
\[a = \lim\limits_{\Delta t \to \0} = \frac{dy}{dt}\]
The instantaneous acceleration is the slope of the tangent to the curve of v versus t. The general expression for the slope of any polynomial of the form \(x = ct^{n}\). The expression for the slope at any point is \(cnt^{n-1}\). Stating this in calculus terms, for any function \(x = ct^{n}\), the derivative of the function is \(cnt^{n-1}\). This can be verified in the case of the parabola by taking successively smaller intervals of \(\Delta t\) and \(\Delta x\) at any point t to verify that the slope at any point is 2t.
Kinematic Equations of Motion
The kinematic equations of motion are derived under the assumption of constant acceleration. While this may seem at first to be a restriction, there are a large number of problems where the acceleration is constant. The simplest and most obvious are falling body problems, or problems involving bodies falling on or near the surface of the earth where the acceleration due to gravity is a constant.
Starting with the assumption of constant acceleration, we can write:
\[a = \frac{v_f - v_o}{t}\]
This can be rearranged to this form:
\[ta = v_f - v_o \longrightarrow ta + v_o = v_f \longrightarrow v_f = v_o + ta \]
We define the average velocity as:
\[v_{avg} = \frac{v+ v_o}{2}\]
We define displacement as:
\[x = x_o + v_{avg}t\]
We can now substitute \(v_{avg}\) with the previous equation.
\[x = x_o + \frac{v + v_ot}{2}\]
One more substitution for \(v = v_o+ at\) and we get:
\[x= x_o + v_ot + (½)at^2\]
We now have the four kinematic equations of motion.
\[v = v_o + at\]
\[x - x_o = (½)(v + v_o)t\]
\[x - x_o = v_ot + (½) at^2\]
\[v^2 = v_o^2 + 2a(x - x_o)\]
The first three equations relate displacement, velocity, and acceleration in terms of time, while the fourth equation does not contain the time.
Example 1
A train starts from rest and moves with constant acceleration. On first observation, the velocity is 20 m/s and 80 s later the velocity is 60 m/s. At 80 s, calculate the position, average velocity, and constant acceleration over the interval.
Calculate the acceleration:
\[a =\frac{v_f-v_o}{t} = \frac{60\text{ m/s} - 20\text{ m/s}}{80\text{ s}} = 0.50 \text{ m/s^2}\]
Calculate the distance traveled over this 80 s:
\[x = v_ot + (½)at^2 = (20\text{ m/s})(80\text{ s}) + (½)(0.50\text{ m/s^2})6400\text{ s^2}= 3200\text{ m}\]
The average velocity is:
\[v_{avg} = \frac{v_f+v_o}{2} = {20\text{ m/s}+60\text{ m/s}}{2} = 40\text{ m/s}\]
If the acceleration is constant, then the average velocity is the average of 20 m/s and 60 m/s, or 40 m/s, and at an average velocity of 40 m/s and 80 s, the distance traveled is 3200 m.
Example 2
Using the above problem, calculate the position of the train at 20 s.
\[x = v_ot + (½)at^2 = (20\text{ m/s})(20\text{ s}) + (½)(0.50\text{ m/s^2})(20\text{ s})^2 = 500 \text{ m}\]
Example 3
Using the above problem, find the time required for the train to reach 100m.
\[x-x_o = v_ot+(½)at^2 \longrightarrow 100\text{ m} = (20\text{ m/s})t + (½)(0.50\text{ m/s^2})t^2\]
\[t = \frac{-80{\pm} \sqrt{6400-4(1)(400)}}{2(1)} = \frac{-80{\pm}89}{2} = 4.5, -85\]
The negative answer is not correct because we do not want negative time, so t=4.5s
Example 4
Using the above problem, find the velocity of the train at 120 m.
\[v^2 = v_o^2 + 2ax = (20\text{ m/s})^2 + 2(0.50\text{ m/s^2})120m = 400\text{ m^2/s^2} + 120\text{ m^2/s^2} = 520\text{ m^2/s^2} = 23\text{ m/s}\]
Example 5
Two vehicles are at position x=0 and t=0. Vehicle 1 is moving at constant velocity of \(30 \text{ m/s}\). Vehicle 2, starting from rest, has acceleration of \(10 \text{ m/s^2}\). Where do they pass?
Translated into algebra, this is asking what is the value of x when they pass? This can be determined by writing equations for the position of each vehicle and equates
\[ x_1 = v_10 t = 30 \text{ m/s} \text{ t} \]
and
\[x_2=(½)a_2t^2=(5\text{ m/s^2})t^2\]
Setting x1=x2 gives the time when they pass as \(30t=5t^2\) or \(5t(t-6)=0\), so the vehicles pass at t=0 and t=6. Putting t=6 in either equation for x gives 180 m as the distance.
Example 6
In the above situation, when do the vehicles have the same velocity?
In algebra, this means to set the equations for velocity equal (v1=v2) and solve for the time. Remember that three of the four equations of motion are functions of time, so most questions are answered by first calculating the time for a certain condition to occur.
\[30\text{ m/s}=(10\text{ m/s^2})t = 3.0\text{ s}\]
Now that we know when, we can calculate where they have the same velocity. Use either equation for position and t=3.0s
\[x_1|{t=3.0s}=(30\text{ m/s})(3.0s)=90\text{ m}\]
Example 7
For the situation in problem 5, what are the position, velocity, and acceleration of each vehicle when vehicle 2 has traveled twice the distance of vehicle 1?
The time when this occurs is when x2=2x1 so:
\[5t^2=60t\]
Or
\[5t(t-12)=0\]
This gives the time of t=0 and t=12. The time t=0 is correct but t=12 is what we are looking for because we want to know something after some time has passed. At t=12,
\[x_1=(30\text{ m/s})(12\text{ s})=360\text{ m}\]
From the original statement of the problem,
\[v_1 0=30\text{ m/s}\]
and \(a_1=0\). Now solve for the remaining variables for the second vehicle by substitution.
\[x_2=(5\text{ m/s^2})(12\text{ s})=720\text{ m}\]
\[v_2=(10\text{ m/s^2})(12\text{ s})=120\text{ m/s}\]
From the original statement of the problem, \(a_2=10\text{ m/s}\).
Example 8
Two trains are traveling along a straight track, one behind the other. The first train is traveling at \(12\text{ m/s}\). The second train, approaching from the rear, is traveling at \(20\text{ m/s}\). When the second train is 200 m behind the first, the operator applies the brakes, producing a constant deceleration of \(0.20\text{ m/s^2}\). Will the trains collide, and if so, where and when?
First, diagram the situation. Write down the equations for each train and the information provided in the problem.
Train 1 | Train 2 |
\(x1=200m +(12\text{ m/s})t\) | \(x2=(20 \text{ m/s})t-(0.10 \text{ m/s^2})t^2\) |
\(v1=12 \text{ m/s}\) | \(v2=20 \text{ m/s}-(0.20 \text{ m/s^2})t\) |
\(a1=0\) | \(a2=-0.20 \text{ m/s^2}\) |
|
|
Take t=0 when the brakes are applied and the first train is 200 m ahead of the second. This makes the position of the first train 200 m at t=0.
The question as to whether the trains collide means, is there a real time solution to the equation resulting from setting x1=x2?
\[200m+(12 \text{ m/s})t=(20 \text{ m/s})t-(0.10 \text{ m/s^2})t^2\]
If there are no real solutions to the equation, then the trains do not collide. Drop the units, and write \(0.10t^2-8t+200=0\) which is solved by the quadratic formula. In a calculator, you can see there are no real solutions, so the trains never collide.
Example 9
Change the previous problem by giving the second train an initial velocity of 25 m/s. This will give a real time for the collision. Find the collision time.
Setting the expressions for x1 ands x2 equal and dropping the units produces
\[200+12t=25t-0.10t^2\]
Or
\[0.10t^2-13t+200=0\]
This equals \(17.8, 112\)
Train 1 | Train 2 |
\(x_1=200 \text{ m}+(12 \text{ m/s})t\) | \(x_2=(25 \text{ m/s})t-(0.10 \text{ m/s^2})t^2\) |
\(v_1=12 \text{ m/s}\) | \(v_2=25 \text{ m/s}-(0.20 \text{ m/s^2})t\) |
\(a_1=0\) | \(a_2=-0.20 \text{ m/s^2}\) |
The two times correspond to when x1=x2. The earliest time is the first coincidence and the end of the physical problem. The position at this time can be obtained from either expression for x.
\[x_1=200m+(12 \text{ m/s})17.8\text{ s}=414 \text{ m}\]
Verify this distance by using x_2. The velocity of the second train at collision is
\[v_2=25 \text{ m/s}-(0.20 \text{ m/s^2})17.8\text{ s}=21.4 \text{ m/s}\]
The relative velocity between the two trains is \(v=1.4 \text{ m/s}\). The two times are the result of the quadratic in t. The two solutions occur when the curves cross. The equation for \(x_1=200+12t\) is a straight line of slope 12 starting at 200. The equation for \(x_2=25t-0.10t^2\) is a parabola that opens down. While the mathematics produces two times, the reality of the problem dictates the earlier time as the one for the collision.
Falling Body Problems
The kinematic equations of motion for constant acceleration can be applied to a large collection of problems known as falling body problems. These problems are where the constant acceleration is the acceleration due to gravity on the surface of the Earth. I will list them here.
\[v_{avg} = \frac{v+v_{0}}{2}\]
\[v = v_{0} + at\]
\[x - x_{0} = (½)(v + v_{0})t\]
\[x - x_{0} = v_{0}t + (½)at^{2}\]
\[v^{2} = v_{o}^{2} + 2a(x - x_{o})\]
Example 1
Consider a ball dropped from the top of a 40 m tall building. Calculate everything possible.
You will not be able to calculate everything at first.
The trick is to calculate what you can and those answers will lead you to the rest of the calculations.
We know:
displacement=40 meters
\[x=0\text{ } v_{o}=0\text{ } a=9.8m/s^{2}\]
Since most of the kinematic equations contain the time, this is usually one of the first things to calculate.
Use the equation that has all the variables included in it.
\[x-x_{0}=v^{2}_{0}t+(½)at^{2}\]
\[40m = (½)(9.8m/s^{2})t{2}\]
\[= 2.9s\]
Knowing the time, we can calculate the velocity using the second equation.
\[v = v_{0} + at\]
\[v = 0 + 9.8m/s^{2}(2.9)\]
\[= 28m/s\]
Example 2
Now, using the original problem, let us add an initial velocity of 8.0m/s when the ball is thrown down instead of just dropped. Find the time for the ball to reach the ground and the velocity on impact.
First, we calculate the time for the ball to hit the ground as before.
\[x - x_{0} = v_{0}t + (½)at^{2}\]
\[40m = 8.0t + (½)(9.8t^{2}\]
\[40m = 8.0t + 4.9t^{2}\]
\[4.9t^{2} + 8.0t - 40m = 0\]
Use a quadratic formula.
\[t = \frac{-80\pm\sqrt{64-4(4.9)(-40)}}{2(4.9)}\]
\[t = 2.2s\]
The velocity at the ground level is:
\[v^{2} = (8.0m/s^{2})^{2} + 2(9.8m/s^{2})(40m) = 848m^{2}s^{2} = 29m/s\]
Example 3
Using the first problem, now throw the ball up from the top of the building with a velocity of 8.0 m/s. Find the time for the ball to reach the ground and the velocity on impact.
Take x as the displacement as positive down and the velocity as negative up. It is important to remember that the sign of the velocity is opposite that of the displacement. It does not matter whether the velocity is negative and the distance down is positive, only that they are opposite. Calculate the time of flight.
\[40m = (-8.0m/s)t + (4.9m/s^{2})t^{2}\]
\[4.9t^{2} - 8.0t - 40 = 0\]
\[t = \frac{8.0\pm\sqrt{64-4(4.9)(-40)}}{9.8} = 3.8s\]
The positive time is the correct choice.
The velocity when the ball strikes the ground is:
\[v^{2} = (-8.0m/s^{2}) + 2(9.8m/s^{2})40m = 848m^{2}/s^{2} = 29m/s\]
Notice that whether the v term is a positive number or negative number, the result is the same. If the ball is thrown up with a certain velocity or down with the same velocity, the velocity at impact is the same. This is to be expected from the symmetry of the equations. If the ball is thrown up with a certain velocity, then on the way down it passes the same level with that same velocity.
Example 4
For the situation in the above problem, calculate the maximum height above the top of the building and the time for the ball to reach maximum height.
The time for the ball to reach maximum height is from equation 2. Note that at maximum height, the velocity must be zero. Watch the signs closely. It does not matter how you choose the signs, but the acceleration has to be opposite the velocity.
\[v = v_{0} + at\]
\[0 = 8.0m/s - (9.8m/s^{2})\]
\[t = 0.82s\]
Because of the symmetry, it takes the ball the same amount of time to reach maximum height as it does for the ball to return to the original level.
Calculate the height above the top of the building from equation 5.
\[v^{2} = v_{o}^{2} + 2a(x - x_{o})\]
\[0^{2} = (8.0m/s^{2}) - 2(9.8m/s^{2})(x - x_{0}) = 3.3m\]
Example 5
A bottle of champagne is dropped by a balloonist. The balloon is rising at a constant velocity of 3.0 m/s. It takes 8.0 s for the bottle of champagne to reach the ground. Find the height of the balloon when the bottle was dropped, the height of the balloon when the bottle reached the ground, and the velocity with which the bottle strikes the ground.
There are several possibilities concerning the origin and direction of the coordinate system. Take the origin at the height of the balloon when the bottle is dropped and the position of the balloon at t=0. Take the displacement as positive down. The main reason for taking the displacement as positive down is that the acceleration is down and the initial velocity is up, making two positives and one negative. As time goes on, displacement, velocity, and acceleration will be positive.
First, write the equation for the height of the balloon starting from the time when the bottle is dropped.
\[x = (-3.0m/s)t + (4.9m/s^{2})t^{2} = (-3.0m/s)(8.0s) + (4.9m/s^{2})(64s^{2}) = 290m\]
The height of the balloon when the bottle was dropped plus the amount the balloon rose in the 8.0s it took the bottle to reach the ground:
\[290m + (3.0m/s)8.0s = 314m\]
The velocity on impact :
\[v = v_{0} + at = -3.0m/s + (9.8m/s^{2})(8.0s) = 75m/s\]
Example 6
A parachutist descending at a constant rate of 2.0 m/s drops a smoke canister at a height of 300m. Find the time for the smoke canister to reach the ground and its velocity when it strikes the ground. Then find the time for the parachutist to reach the ground, the position of the parachutist when the smoke canister strikes the ground, and an expression for the distance between the smoke canister and the parachutist.
The time for the smoke canister to reach the ground is from equation 4.
\[x = v_{0}t + (½)at^{2}\]
\[300m = (2.0m/s)t + (4.9m/s^{2})t^{2}\]
Without units, the equation is \(4.9t^{2} + 2.0t - 300 = 0\) with solutions.
\[t = \frac{-2\pm\sqrt{4-4(4.9)(-300)}}{2*4.9}\]
\[= \frac{-2\pm76.7}{9.8} = 7.6\]
The time for the canister to hit the ground is 7.6s.
The velocity when it strikes the ground is:
\[v^{2} = (2.0m/s)^{2} + 2(9.8m/s^{2})300m = 5884m^{2}/s^{2}= 77m/s\]
The time for the parachutist to reach the ground is from equation 3.
\[300m = (2.0m/s)t = 150s\]
When the canister strikes the ground, the parachutist has dropped:
\[(2.0m/s)7.6 = 15m\]
And is 285 m above the ground. The expression for the distance between the canister and the parachutist is:
\[x_{c} - x_{p} = (2.0m/s)t + (4.9m/s^{2})t^{2} - (2.0m/s)t = (4.9m/s^{2})t^{2}\]
Example 7
A coconut is dropped from a height of 60m. One second later, a second coconut is thrown down with an initial velocity. Both coconuts reach the ground at the same time. What was the initial velocity of the second coconut?
In problems where there is a time delay, it is usually best to calculate the position, velocity, and acceleration of the first particle at the time when the second particle starts to move. It is possible to do time delay problems with a time differential in one set of equations. The difficulty with this approach is that it is easy to get an algebraic sign wrong. If you say that the time for the second particle is the time for the first particle plus the difference between them, then it is essential that the algebraic sign of the difference be correct. It is much easier to do them, in this slower but inherently more accurate way. First, calculate the state of the first particle when the second one begins moving. Then write the two sets of equations describing the motion with this instant as t=0.
Calculate the position and velocity of the first coconut at the end of 1s, the time when the second one starts.
\[x = (½)at^{2} = (½)(9.8m/s^{2})(1.0s)^{2} = 4.9m\]
\[v = at = 9.8m/s^{2}(1.0s) = 9.8m/s\]
Since position, velocity, and acceleration is positive down, orient the coordinate system for positive down with the origin at the top. At the instant the second coconut is thrown down, the first coconut has position 4.9m, velocity 9.8m/s and acceleration 9.8m/s^{2}.
Since both coconuts strike the ground at the same time, use the conditions of the first coconut to find the total time. First, calculate the velocity at impact of the first coconut.
\[v_{1}^{2} = v_{1o}^{2} + 2a(x_{1}-x_{1o}) = (9.8m/s)^{2} + 2(9.8m/s^{2})(60-4.9)m = 1176m^{2}/s^{2} = 34.3m/s\]
The time for the second coconut to reach the ground is the same as the time for the first coconut to go from 4.9 m to 60 m or the time for the first coconut to go from 9.8 m/s to 34.3 m/s.
This comes from \(v = v_{0} + at\) where the velocity of the first coconut when the second one is thrown down, and v is the velocity of the first coconut at the ground.
\[34.3m/s = 9.8 m/s _ (9.8m/s^{2})t = 2.5s\]
Now that we have the time for the second coconut to travel the 60 m, we can find its initial velocity from \(x-x_{0}=v_{2o}t + (½)at^{2}\) where t is the total time for the second coconut.
\[60m = v_{2o}(2.5s) + (4.9m/s^{2}(2.5s)^{2} =11.75m/s\]
Example 8
A boat is passing under a bridge. The deck of the boat is 15 m below the bridge. A small package is to be dropped from the bridge onto the deck of the boat when the boat is 25 m from just below the drop point. What boat speed is necessary to have the package land in the boat?
Calculate the time for the package to fall the 15.0 m using \(x-x_{0} = v_{o}t+(½)at^{2}\)
\[15m = (4.9m/s^{2})t^{2} = 1.7s\]
The boat must move forward at 25 m/ 1.7s = 14 m/s.
Example 9
You are observing steel balls falling at a constant velocity in a liquid filled tank. The window you are using is 1m high, and the bottom of the window is 12m from the bottom of the tank. You observe a ball falling past the window taking 3.0s to pass the window. Calculate the time required to reach the bottom of the tank after the ball has reached the bottom of the window.
The observed velocity is \(1.0m/3.0s so 12 m = (1.0m/3.0st = 36s\)
Example 10
In a situation similar to the last example, the tank is filled with a different liquid, causing the acceleration in the tank to be \(6.0 m/s^{2}\) and the time to traverse the window .40 s. Calculate the height of the liquid above the window, the time to reach the bottom of the tank, and the velocity of the ball when it reaches the bottom of the tank.
From the data about the window, calculate the velocity of the ball at the top of the window.
\[x-x_{o} = v_{t}t + (½)at^{2}\]
\[1.0m = v_{t}(.40s) + (3.0m/s^{2})(.40s)^{2} = 1.3m/s\]
The velocity at the bottom of the window is from:
\[v_{b}^{2} = v_{t}^{2} + 2a(x-x_{o}) = (1.3m/s)^{2} + 2(6.0m/s^{2})1.0m = 3.7m/s\]
The time to reach the bottom of the tank is from \(x-x_{o} = v_{b}t + (½)at^{2}\)
\[12m = (3.7m/s)t + (3.0m/s^{2})t^{2}\]
Eliminate the units
\[t = \frac{-3.7\pm \sqrt{(3.7)^{2} - 4(3.0)(-12)}}{2(3)} = \frac{-3.7\pm12.6}{6} = 1.5\]
This positive time is for the ball to travel from the bottom of the window to the bottom of the tank.
Assuming that the ball started at zero velocity, the distance from the top of the liquid to the top of the window will come from \(v_{t}^{2}=v_{o}^{2}+2ax_{t}\)
\[(1.3m/s)^{2} = 2(6.0m/s)x_{t} = .14m\]
The velocity with which the ball strikes the bottom of the tank is:
\[v = v_{b} + at = (3.7m/s) + (6.0m/s^{2})1.5s = 12.7m/s\]
Example 11
A ball is observed to pass a 1.4m tall window going up and later going down. The total observation time is .40s(.20s going up and .20s going down). How high does the ball rise above the window?
This is another example of a question that seems totally unrelated to the information in the problem. When you don’t know where to start and you do not see a clear path to the desired answer, simply start where you can, calculating what you can and hopefully learning enough to answer the specific question.
One of the first things we can calculate in this problem is the average velocity at the middle of the window.
\[v_{avg} = (1.4m/2.0s) = 7.0m/s\]
This average velocity is the velocity of the ball on the way up and on the way down at the middle of the window. Add this feature to the problem. With this information, we can use \(v^{2} = v_{o}^{2} + 2ax_{t}\) to find the distance the ball rises above the midpoint of the window.
\[(7.0m/s)^{2} = 2(9.8m/s^{2})x_{t} = 2.5m\]
So the ball rises 2.5-0.7=18 m above the top of the window.
View the ball as decelerating as it goes up past the window, ands find \(v_{b}\) at the bottom of the window from \(x-x_{o}=v_{o}t+(½)at^{2}\).
\[1.4m = v_{b}(.20s)-(4.9m/s^{2})(.20s)^{2} = 8.0m/s\]
Velocity and displacement are taken as positive, so acceleration is negative.
Now the distance to maximum height, where velocity is zero, is:
\[(8.0m/s)^{2} = 2(9.8m/s^{2})x_{m} = 3.2m\]
Again, the maximum height of the ball above the top of the window is 3.2 m - 1.4 m = 1.8m.
Projectile Motion In Physics
In order to understand projectile motion, you have to look at the motion in two directions with one direction oriented in the direction of constant acceleration and the other direction at a right angle to it as to form an x-y coordinate system. In most problems where an initial velocity and angle with the horizontal is given, the velocity is written in component form.
\[v_y = v_o sin \Theta\]
\[v_x = v_o cos \Theta\]
Note that motion is effectively separated into horizontal and vertical components. One is in the direction of constant acceleration and one is at a right angle to the acceleration.
The motion is a parabola in this coordinate system. The symmetry of parabolas is helpful in understanding the motion.
Usually, the velocity of the projectile is written in component form. The acceleration in the horizontal direction is zero and in the vertical direction is due to gravity. With the velocity components and the acceleration in the direction of an axis, we can write the six equations describing acceleration, velocity, and position in the x and y directions.
These equations are based on the kinematic equations of motion for constant acceleration:
\[a = \text{ constant}, v = v_o + at, \text{and} s = s_o + v_ot + (1/2)at^2\]
The six equations are:
\[a_x = 0\]
\[v_x = v_o \cos\Theta\]
\[x = v_o(\cos\Theta)t\]
\[a_y = -g\]
\[v_y = v_o \sin\theta - gt\]
\[y = v_o(\sin\Theta)t - (1/2)gt^2\]
The equations for x and y can be considered parametric equations in time. Parametric equations such as \(x = f(t)\) are equations that can be combined to produce \(y = f(x)\) or \(x = f(x)\). To find the position of the particle in x-y or y as a function of x, solve one equation for t and substitute into the other.
In this case, solve x =...for t because this is the simplest choice.
\[t = \frac{x}{v_o \cos\Theta}\]
Substitute into y=...to obtain:
\[y = (\tan\Theta)x - \frac{g}{2v^2_o \cos^2\Theta}x^2\]
This is of the form \(y=-ax^2+bx=x(-ax+b)\), which is a parabola that opens down and intercepts the y-axis (makes y=0) at x=0 and x=b/a.
The range, or value of x when y=0, can be determined from the factored form of this equation.
\[y=x{\tan\Theta-\frac{g}{2v^2_o \cos^2\Theta}x}\]
which tells us that y=0 at x=0 and :
\[x=\frac{2v^2_o}{g} \cos^2\Theta \tan\Theta= \frac{2v^2_o}{g}\]
The maximum range occurs for an angle of \(45\deg\), corresponding to \(\sin2\Theta=1\).
The main point is that the motion is a parabola, and the properties of parabolas can be used in solving problems in projectile motion. That the maximum range occurs at \(45\deg\) is not surprising and is generally not of interest in problems. The expression for maximum range is only of passing interest because the range is one of the easier things to calculate in a problem.
Do not consume precious memory space memorizing formulas for the range or time of flight. If you work problems by first writing down the six equations describing the motion, then the time of flight, range, and many other things are easily calculated.
In order to set up a problem in projectile motion, first orient one axis of a right angle coordinate system in the direction of constant acceleration. Remember to place the origin of the coordinate system and the positive direction for x and y for your convenience in solving the problems. Then write equations for acceleration, velocity, and position for each direction. Along with the symmetry for the motion, these equations can be used to find the characteristics of the projectile at any point in space or time.
Example 1
A soccer player kicks a ball at an initial velocity of 18 m/s at an angle of 36 degrees to the horizontal. Find the time of flight, range, max height, and velocity components at t=0, midrange, and impact.
The acceleration is down, so set up the coordinate system with x horizontal and y vertical, and place \(v_o\) on the graph. Write the velocity in component form, and calculate \(v_{xo}\) and \(v_{yo}\).
\[v_o=18 m/s\]
\[v_{yo}=v_o \sin 36 = 10.6 m/s \]
\[v_{xo}=v_o \cos 36 = 14.6 m/s \]
Now, write down the six equations governing the motion starting with the accelerations.
\[ a_x=0 \text{and }a_y=-9.8m/s^2 \]
\[v_x=14.6m/s \text{and } v_y=10.6m/s - (9.8m/s^2)t \]
\[ x=(14.6m/s)t^2 \text{and } y=(10.6m/s)t - (4.9m/s^2)t^2 \]
Now look at the motion, which is parabolic, keeping in mind the properties of parabolas.
The ball is on the ground at t=0 and t=t, the time of flight.
To find these times, set y=0
\[10.6t - 4.9t^2 = 0 \]
This gives \( t=0\) and \( t=(10.6/4.9)=2.16s\), the time of flight.
The range is the value of x at 2.16s.
\[x_{t=2.16} = (14.6m/s)2.16s = 31.5m \]
The max height and \(v_y=0\) occur at 1.08s. Therefore, the max height is y at t=1.08s.
\[y_{max} = y_{t=1.08} = (10.6m/s)1.08s - (4.9m/s^2)(1.08s)^2 = 5.73m \]
The velocity components are:
t=0: v(x)=14.6 m/s v(y)=10.6 m/s
t=1.08s v(x)=14.6 m/s v(y)=0
t=2.16s v(x)=14.6 m/s v(y)=-10.6 m/s
The procedure for doing the problem is to write the initial velocity in component form, then write the six equations for acceleration, velocity and position. Then perform the mathematical operations answering questions about the problem.
Example 2
for the situation in example 1, suppose that the field is covered with fog down to 3.0 m above the ground. What are the times for the ball's entry into and exit out of the fog?
The times for the ball's entry into and exit out of the fog, are found from the equation for y as a function of time with y set equal to 3.0 m.
\[3.0m = (10.6m/s)t - (4.9m/s^2)t^2 \]
This is a quadratic equation, and after removing the units, it reads \(4.9t^2 - 10.6t + 3.0 = 0\) and is solved with the quadratic formula.
\[\frac{10.6\pm \sqrt{10.6^2-4(4.9)3.0}}{2(4.9)} = \frac{10.6\pm7.3}{9.8} = 0.34,1.8 \]
Example 3
An airplane traveling at 100 m/s drops a bomb from a height of 1500m. Find the time of flight, distance traveled, and velocity components as the bomb strikes the ground.
Start by placing the origin of the coordinate system at the point where the bomb is released. Take the direction in which the bomb falls due to gravity as positive x and the horizontal position as y. This is different from the conventional orientation, but it is convenient in this problem because the six equations all come out positive.
Now write down the six equations governing motion.
\(a_x = 9.8m/s^2\) and \(a_y = 0 \)
\(v_x = (9.8 m/s^2)t\) and \(v_y = v_o = 100m/s\)
\(x = (4.9m/s^2)t^2\) and \(y = (100m/s)t\)
With these six equations we can answer all the questions in the problem. We need only translate the word questions into algebra questions.
A. The time for the bomb to reach the Earth means "Find the time when x=1500m".
\(1500m = (4.9m/s^2)t^2 = t = 17s\)
B. How far does the bomb travel horizontally means "Find the value of y when t=17s.
\(y_{t=17s} = (100m/s)17s = 1700m\)
C. Find the velocity components at impact means "Find v(x) and v(y) at t=17s.
\( v_{x,t=17} = (9.8m/s^2)17s = 167 m/s \)
\( v_{y,t=17} = (100m/s)\)
D. Where is the airplane when the bomb strikes the Earth means "What is y at 17s?
Remember that the plane and the bomb have the same velocity in the horizontal direction.
\(y_{plane} = (100m/s)17s = 1700m\)
Example 4
A baseball is hit at a 45 degree angle and a height of 0.90 m. The ball travels a total distance of 120m. What is the initial velocity of the ball?
What is the height of the ball above a 3.0 m fence from where the ball was hit?
This problem is unique in that it requests the initial velocity, a number usually given in the problem. Also note that the angle is 45 degrees, the angle for max range. The other interesting feature of this problem is the question concerning the height at some specific point down range.
Set up the problem in the conventional way assigning \(v_o\) to the initial velocity.
\(v_o \sin 45\deg\)
\(v_o \cos 45\deg\)
The next questions is how to handle the ball being hit at 0.90 m above the ground. Since \(v_o\) is specified at the point where the ball is hit, and this is where we start counting time, the origin for the coordinate system should be put here. We just need to keep in mind that this origin is 0.90 m above the ground level.
Now write the six equations.
\(a_x = 0\) and \(a_y = -9.8m/s^2\)
\(v_x = v_o \cos 45\deg\) and \(v_y = v_o \sin 45\deg - (9.8m/s^2)t\)
\(x = v_o (\cos 45\deg)t\) and \(y = v_o \sin 45\deg t - (4.9m/s^2)t^2\)
The first question reduces to finding \(v_o\) for the 120 m hit. Algebraically, this means that when \(y=-0.9m, x=120m\), so write the two equations with these conditions.
\(120m = v_o \cos 45\deg t\) and \(-0.9m = v_o(\sin 45\deg)t - (4.9 m/s^2)t^2\)
Note that \(v_o(\cos 45\deg)t = v_o(\sin 45\deg)t = 120m\)
\(0 = 0.90m + 120m - (4.9m/s^2)t^2\) and \(t=4.97s\)
This is the time of flight, so put this time into the equation for x.
\(120m = v_o(\cos 45\deg)(4.97s)\) to find \(v_o\) as \(v_o = \frac{120m}{\cos 45\deg*4.97s} = 34.1m/s\)
The next part of the problem asks for the height above the fence at 100m. Again, we need to find the time for x to be 100m down range and substitute this value of t into the y equation.
\(100m = (34.1m/s)(\cos 45\deg)t = 4.15s\)
This is the time for the ball to go 100m. The height of the ball at this time is:
\(y_{t=4.15s} = (34.1m/s)(\cos 45\deg)(4.15s) - (4.9m/s^2)(4.15s)^2 = 15.6m\)
Remember that this is 15.6m above the zero of the coordinate system, which is 0.90m above the ground, so the ball is 16.5m above the ground at this point and, for a 3m high fence, is 13.5 m above the fence.
Example 5
A physicist turned motorcycle stunt rider will jump a 20 m wide row of cars. The launch ramp is 30 degrees and 9.0 m high. The land ramp is also 30 degrees and is 6.0 m high. Find the minimum speed for the launch.
Diagram the problem
First, write down the six equations of motion.
\(a_x = 0\) and \(a_y = -9.8m/s^2\)
\(v_x = v_o \cos 30\deg\) and \(v_y = v_o \sin 30\deg - (9.8m/s^2)t\)
\(x = v_o(\cos 30\deg)t\) and \(y = v_o(\sin 30\deg)t - (4.9m/s^2)t^2\)
The minimum \(v_o\) is dictated by the condition that the rider be at x=20m and y=-3.0m.
So put these conditions into the equations for x and y.
\(20m = v_o(\cos 30\deg)t\) and \(-3.0m = v_o(\sin 30\deg)t - (4.9m/s^2)t^2\)
At this point we can solve for \(v_o\) by substituting \(t=20m/v_o \cos 30\deg\) from the first equation into the second equation.
\(-3.0m = (20m) \tan 30\deg - \frac{4.9*400m^3}{v^2_o(\cos^230\deg)s^2} = v_o=13.4m/s\)
This is the minimum velocity to make the jump.
Forces In Physics
Intro
In the development of mechanics, the first thing to learn is the interrelations
of position, velocity, and acceleration. These interrelations are described with
the four kinematic equations of motion.
In this section, we relate constant force to constant acceleration and then to
all of kinematics. Force can be defined as what makes masses accelerate.
Actually, it is the unbalanced force on a mass that makes it accelerate. So,
unbalanced forces make masses accelerate.
Force, acceleration, and mass are related by Newton's second law as F = ma.
Example 1
A 40 N force is applied to a 20 kg block resting on a horizontal frictionless
table. Find the acceleration.
Solution:
The acceleration is \(a=\frac{f}{m} = \frac{40N}{20kg} = 2.0m/s^2\)
Once the acceleration is known, the kinematic equations of motion can be applied
to find position, velocity, and acceleration as a function of time.
Example 2
Now place the 20 kg mass on a frictionless 50 degree inclined plane. What is the
acceleration of the mass?
Solution:
The force acting on the mass is due to gravity, so setup a vector diagram
starting with this 196 N force acting down. The mass is constrained to move down
the plane. The gravitational force is shown with components down the plane and
normal to the plane. Notice the geometry of the situation, where the 50 degree
angle of the plane is the same as the angle between the gravitational force and
the normal force.
So, the force straight down is \(F=20kg(9.8m/s^2)=196N\)
However, the mass is going down an incline plane at 50 degrees.
\(F_{p}=(196N)(sin50)=150N\)
The 150 N force acting down the plane causes the 20 kg mass to accelerate at:
\(a=\frac{150N}{20kg}=7.5m/s^2\)
Example 3
Place a 10 kg mass on a frictionless 35 degree incline plane, and attach a
second 20 kg mass via a cord to hang vertically. Calculate the acceleration of
the system.
Solution:
\(mg = 98N\)
\(98N*sin35 = 56N\)
\(98N*cos35 = 80N\)
Mass 1 is 10 kg at 35 degree down incline plane.
Mass 2 is 20 kg and hangs straight down. It has a downward force of 196N.
This problem introduces the concept of the tension in the connecting cord. The
most convenient way to visualize this tension is that if the cord were cut and a
force meter inserted, it would read a certain tension(force). This tension acts
on each mass. Notice the 35 degree angle between the normal and the direction of
mg.
To write equations relating the unbalanced force to the acceleration, we need to
make an assumption as to which way the masses move. Assume that the system moves
to the right or clockwise.
Start by writing equations relating the unbalanced force on each mass. On m1,
the unbalanced force is T-56N, and this force makes the 10 kg accelerate.
\(T-56N = 10kg*a\)
Likewise on mass2, the unbalanced force to make the mass accelerate is:
\(196N-T = 20kg*a\)
The assumption that that system accelerates clockwise dictates the directions of
the forces in these two equations. Look again at these equations, and notice
that if we had assumed the acceleration were counterclockwise, then the signs
would be different. In frictionless problems it is alright to have a negative
acceleration. The negative sign means that you guessed wrong in assuming the
acceleration direction. This is not true with friction problems. Add the two
equations, and take the acceleration as the same. The accelerations are the same
because the masses are linked together with the cord.
\(140N = 30kg * a | a = 4.7m/s^2\)
Now the tension in the cord is easily calculated from the equation for mass 1.
\(T = 56N + 10kg * 4.7m/s^2 = 103N\)
Example 4
The next complication in force problems with inclined planes is a double
inclined plane.
Mass 1 is 12 kg going down to the left of the apex at a 40 degree angle.
Mass 2 is 20 kg going down to the right of the apex at a 25 degree angle.
Solution: Create a vector diagram, it is helpful.
Assume the system moves clockwise.
We will start with mass 1 first.
The force straight down is 118 N, from (12 kg * 9.8 m/s^2).
Now, lets calculate the force down the incline plane to the left.
118 N * sin 40 = 76 N
Now, lets go to mass 2.
The force straight down is 196 N, from (20 kg * 9.8 m/s^2).
Now, lets calculate the force down the incline plane to the right.
196 N * sin 25 = 83 N
The equation for mass 1 is: T-76N = 12 kg * a
The equation for mass 2 is: 83N-T = 20 kg * a
Add the two equations
7.0N = 32 kg * a
a = 0.22 m/s^2
The tension in the cord is:
T = 76N + 12 kg * 0.22 m/s^2 = 79 N
Example 5
Another popular problem is the Atwood machine. Find the acceleration of the
system.
Solution: Assume that the system moves counterclockwise.
The equation for mass 1 is 98N-T = 10kg * a
The equation for mass 2 is T-147N = 15kg * a
Add the two equations
-49N = 25kg * a
a = -2.0 m/s^2
Force of mass 1 is 10kg * 9.8 m/s^2 = 98N
Force of mass 2 is 15kg * 9.8 m/s^2 = 147N
The system accelerates clockwise. The tension in the cord is from the equation
for mass 1.
T = 98N + 10kg * 2.0m/s^2 = 118N
Friction
A good first problem in friction is where the problem is first done without
friction and then with friction.
For the purposes of doing problems, there are two important properties of
frictional forces:
They oppose the motion
They are less than or equal to a constant (coefficient of friction), times the
normal force, the force at the frictioning surface.
Example 6
Two masses are arranged with one, of 50kg, on a frictionless table and the
other, of 30 kg, attached by a cord and hanging, over a frictionless pulley, off
the edge of a table. Find the acceleration of the system.
Solution: Assume that the system moves clockwise.
The equation for mass 1 is T=50kg * a
The equation for mass 2 is 294N-T = 30kg * a
Add the two equations:
294N = 80kg * a
a = 3.7 m/s^2
Example 7
For the situation in Example 6, add a coefficient of friction \mu=.20 between
the block and the table.
Solution:
As above, horizontal:
\mu N = mu*m*g = 0.20*50kg*9.8m/s^2 = 98N
Vertical:
294N
The equation for mass 1 is T-98N=50kg*a
The equation for mass 2 is 294N-T=30kg*a
Add the equations:
196N=80kg*a
a=2.4m/s^2
Example 8
For the situation in the previous problem, increase the coefficient of friction
to 0.80. What is the acceleration of the system?
Solution:
\(f=\mu N=\mu*m*g=0.80*50kg*9,8m/s^2=392N\)
\(294N\)
The equation for mass 1 is \(T-392N=50kg*a\)
The equation for mass 2 is \(294-T=30kg*a\)
Add the two equations
\(-98N=80kg*a\)
\(a=-1.2m/s^2\)
Clearly, the system does not accelerate counterclockwise. This is an
illustration of the second property of frictional forces: the frictional force
is less than or equal to \(\mu N\). In this case the frictional force just
balances the tension in the cord. The frictional force can be up to 392N. In
order for mass 2 not to move, the tension in the cord must be 294N. If mass 2 is
not moving, the forces must be in balance. Therefore, the forces on mass 1 must
be 294N due to the tension in the cord and 294N due to friction.
Another important point with frictional forces is the minimal force.
Example 9
A 50kg sled is pulled along a level surface at constant velocity by a constant
force of 200N at an angle of 30 degrees. What is the coefficient of friction
between the sled and the surface?
Solution: The first step in this problem is to find the components of the
applied force and the mg of the sled. These are shown in the vector diagram.
Notice the normal force, the actual force between the sled and the friction
surface, is mg minus the component of the applied force lifting up the sled.
Sometimes it is easy to miss this vertical component of the applied force. This
is a major mistake in this type of problem. To help visualize this vertical
component of force, try placing the origin for the vectors in the middle of the
sled rather than at one end, where a rope would be attached.
\(f = \mu*N\)
\(mg = 50kg*9.8m/s^2 = 490N\)
\(200N * sin 30 = 100N\)
\(200N * cos 30 = 173N\)
The normal force is mg minus the vertical component of the applied force, 100N,
or 390N total. Since the sled is moving at constant velocity, the forces must be
in equilibrium. The horizontal component of the applied force must equal the
frictional retarding force, the coefficient of friction times the normal force.
\(\mu*390N = 173N\)
\(\mu = 0.44\)
Example 10
A 65 degree inclined plane has a mass of 20kg on the plane where the coefficient
of friction is 0.40 and a mass of 50kg hanging free. Calculate the acceleration
of the system.
Solution: Setup the vector diagram assuming that the motion is counterclockwise.
In the diagram, the frictional force is calculated as a maximum of 33N.
The equation for mass2 is \(490N-T=50kg*a\)
The equation for mass1 is \(T-178N=50kg*a\)
Add the equations: \(279N=70kg*a\)
\(a=4.0m/s^2\)
\(mg = 50kg*9.8m/s^2 = 490N\)
Mass2 of 50kg is hanging straight down off a pulley.
Mass1 of 20kg is sliding down an incline plane. The plane is a 65 degree angle.
Coefficient of friction is 0.40
\(196N cos65=83N\)
\(196N sin65=178N\)
\(0.40*83N=33N\)
In doing friction problems like this, you have to be careful with the frictional
force. If the forces up and down the plane are nearly balanced, it is important
to remember that the frictional force is not 33N but that it can be up to 33N.
In a problem similar tot his, it could easily happen that the blocks would not
move.
Example 11
Consider a double inclined plane with angles, masses, and coefficients of
friction as shown. Calculate the acceleration of the system.
You have a sort of triangle with the apex at the top. To the left and right of
the apex are inclined planes. On the left inclined plane, you have mass1 of
6.0kg, coefficient of friction of 0.30, and going down at an angle of 30
degrees.
On the right inclined plane, you have mass2 of 9.0kg, coefficient of friction of
0.16, and sliding down at an angle of 45 degrees.
Solution: Assume that the system moves counterclockwise, and set up the vector
diagram.
The equation for mass1 is \((30-15-T)N=6.0kg*a\)
The equation for mass2 is \((T-62-10)N=9.0kg*a\)
Add the equations \(-57N=15kg*a\)
\(59N sin30=30N\)
\(59N cos30=51N\)
\(f=0.30*51N=15N\)
\(88N sin45=62N\)
\(88N cos45=62N\)
\(f=0.16*62N=10N\)
This is a negative acceleration. We cannot simply reverse the sign of "a" to
find the correct answer because if the system were moving the other way, the
frictional forces would have to be reversed. further analysis requires us to
consider another vector diagram assuming that the system moves in a clockwise
direction.
The new equations are \((T-30-15)N=6.0kg*a\)
\((62-T-10)N=9.0kg*a\)
Add the equations: \(7.0N=15kg*a\)
\(a=0.54m/s^2\)
The tension in the cord is from the first equation:
\(T=45N+6.0kg*0.54m/s^2)=48N\)
If the acceleration of the system were negative for both the clockwise and
counterclockwise calculations, then we would conclude that the system would not
move.
The final complication in these inclined plane problems is the introduction of
three masses and two connecting rods.
Example 12
Consider a flat surface with a coefficient of friction of 0.20 with a 2.0kg mass
and a 3kg mass connected together with a 6kg mass along a slant as shown. The
coefficient of friction on the slant is 0. The slant is going down at an angle
of 25 degrees. Calculate the acceleration of the system.
Solution:
In this problem, note that the tensions are different, leading to 3 unknowns,
T1, T2, and a. The solution will require three equations from three vector
diagrams. Assume that the acceleration of the system is clockwise.
The frictional retarding force on the 2kg block is:
\(f=\mu N=0.20*19.6N=3.92N\)
The frictional retarding force on the 3kg block is:
\(f=\mu N=0.20*29.4N=5.88N\)
\(58.8N sin 25 =24.8N\)
The equations for the masses are:
\(T2-3.92N=2.0kg*a\)
\(T1-T2-5.88N=3.0kg*a\)
\(24.8N-T1=6.0kg*a\)
The last two equations added together are:
\(18.9N-T2=9.0kg*a\)
Adding this to the first equation is:
\(15.0N=11kg*a\)
\(a=1.36m/s^2\)
The tensions come from the first and third equations for the masses:
\(T2=3.92N+2.0kg*1.36m/s^2=6.64N\)
\(T1=24.8N-6.0kg*1.36m/s^2=16.6N\)
Circular Motion
A particle moving at constant speed in a circle is moving in uniform circular
motion. There is an acceleration and a force associated with such a particle.
While the length of the velocity vector remains the same, the direction changes
continually. Acceleration points radially inward.
Example 13
A certain car driven in a circle can exert a maximum side force of 0.95g. What
is the maximum speed for this car driven in a circle of 160m radius.
Solution:
A side force of 0.95g means a side force or acceleration directed at right
angles to the direction of travel of 0.95*9.8=9.3m/s^2. Using \(a_{rad}=v^2/r\):
\(v=\sqrt{a_{rad} r} = \sqrt{9.3m/s^2*160m} = 38.6m/s\)
At a speed greater than 38.6 m/s, the car will slide out of the circle.
The acceleration directed toward the center of the circle for a mass in uniform
circular motion is called centripetal acceleration. The force associated with
moving a mass in a circle is \(ma_{rad}\) and is called centripetal force.
Example 14
What is the centripetal force produced by the tires acting on the pavement for
the car of the previous problem if the car is 1200kg of mass.
Solution:
\(F=ma_{rad}=1200kg*9.3m/s^2=11200N\)
Example 15
A 0.60kg rubber stopper is whirled in a horizontal circle of 0.80m radius at a
rate of 3.0 revolutions per second. What is the tension in the string?
Solution:
Three revolutions per second means three \(2\pi r\) circumferences per second,
so: \(v=\frac{3.2 \pi *0.80}{1.0s}=15m/s\)
The tension in the string, which is providing the centripetal force, is:
\(F=m\frac{v^2}{r} = 0.60kg*\frac{(15m/s)^2}{0.80m} = 170N\)
Example 16
A 1 oz gold coin is placed on a turntable at 33 1/3 rpm. What is the coefficient
of friction between the coin and the turntable if the maximum radius, before the
coin slips, is 0.14m?
Solution:
The frictional force between the coin and turntable provides the center directed
force to keep the coin on the turntable. This center directed force must equal
\(mv^2/r\).
\(f=\mu N = \mu mg\)
This is a force balance problem. In equation form: frictional force=centripetal
force, or: \(\mu mg=mv^2/r\)
The velocity is 2\pi r\) times the number of \(2\pi r's\) per minute, 100/3 or:
\( v=2\pi*0.14m*(100/3)*(1/min)*(min/60s)=0.49m/s\)
Therefore: \( \mu = \frac{v^2}{rg} = \frac{(0.49m/s)^2}{0.14m*9.8m/s^2}=0.18\)
Example 17
A conical pendulum is a mass on the end of a cord, where the mass moves at
constant speed in a circle with the cord tracing out a cone. A conical pendulum
of length 1.2m moves in a circle of radius 0.20m. What is the period of the
pendulum?
Solution:
Note that the mass is not given. Start with a vector diagram of the forces on
the mass.
The mg must equal the vertical component of the tension in the string \(T cos
\theta = mg\). The horizontal component is due to the centripetal force, so \(T
sin \theta = mv^2/r\). Dividing the second equation by the first eliminates T and
m. The angle is from \(sin \theta=r/L\), making \(tan \theta = 0.17\) and:
\(v=\sqrt{rg tan \theta} = \sqrt{.20m*9.8*.17}=.57m/s\)
The period is from \(2 \pi r/T=v\)
\(T=\frac{2\pi r}{v} = \frac{2\pi*0.20m}{0.57m/s}=2.2s\)
Example 18
What is the speed, at which no side force due to friction is required, for a car
traveling at a radius of 240m on a banked 20 degree road?
Solution:
Start with mg acting down. The normal force(between the car and the road) is
normal to the surface, with the vertical component equal to mg and the
horizontal component equal to \(mv^2/r\).
We have: \(N cos 20 = mg\) and \(N sin 20 = mv^2/r\)
Then: \(\frac{N sin 20}{N cos 20} = tan 20 = \frac{v^2}{rg} \)
So: \(v = \sqrt{rg tan 20} = \sqrt{240m*9.8*tan 20} = 29m/s \)
Notice that in this problem the mass of the car does not enter into the
calculation.
Go back over problems 13-18 and notice how each problem could be written asking
for a different variable. For example, in problem 13, give the side force and
the velocity, and ask for the radius. In problem 16, give the speed and the
coefficient of friction, and ask for the radius. Changing the problems like this
is an excellent way to generate practice problems.
Example 19
A 2kg ball is whirled in a vertical circle of radius 1m at a constant velocity
of 6.28m/s. Calculate the tension in the cord at the top of the motion, the
tension in the cord at the bottom of the motion, and the minimum velocity to
keep the mass from falling out at the top of the circle.
Solution:
We are considering this problem from the reference frame of the ball, not from
the point of view of the person whirling the ball. From the person's point of
view, the tension is pulling radially outward. From the ball's reference frame,
the tension is radially inward. If the ball is to move in a circle, this must be
the case. There must be a net center directed force.
Operationally, a good method for solving centripetal force problems is to find
the net center directed force and set this equal to \(mv^2/r\).
\(T+mg=\frac{mv^2}{r}\)
\(T=\frac{mv^2}{r}-mg\)
\(T=\frac{2.0kg*6,28m/s^2}{1.0m} - 2.0kg(9.8m/s^2)=59.4N\)
\(T-mg=\frac{mv^2}{r}\)
\(T=\frac{mv^2}{r}+mg\)
\(T=\frac{2.0kg*6.28m/s^2}{1.0m} + 2.9kg*9.8m/s^2 = 98.6N\)
For part c, the minimum velocity required can be found by considering the
condition where mv^2 is less than or equal mg. In this case, at the top of the
motion, the mass will fall out of the loop. So let's set the two forces equal
and solve for the minimum velocity needed to keep the mass in the loop.
\(\frac{mv^2_{min}}{r} = mg \rightarrow v_{min} = \sqrt{rg} =
\sqrt{1.0m*9.8}=3.13m/s\)
Example 20
A person on a dirt bike travels over a semicircular shaped bridge with a radius
of curvature of 45m. Calculate the max speed of the bike if its road wheels are
to stay in contact with the bridge.
Solution:
Again, let us find the net center directed force and set this equal to
\(mv^2/r\). So: \(mg-R=\frac{mv^2}{r}\) and: \(R=mg-\frac{mv^2}{r}\).
As v increases, R must decrease because mg is constant. In the limiting case,
when the wheels are just about to leave the ground, R=0, so
\(mg=\frac{mv^2}{r}\).
The mass m cancels out and is not required. So max speed v is given by
\(v^2=rg\).
Plugging in the numbers: \(v=\sqrt{rg} = \sqrt{441} = 21m/s\)
At this point, you might be thinking that from the point of view of the person
on the dirt bike, the \(mv^2/r\) force is radially outward. The rider is
momentarily weightless as the wheels leave the ground. This is correct- from the
reference frame of the person inside the circular motion. Astronauts in the
space station orbiting Earth are weightless. To them, there is no net force, and
the effect of the \(mv^2/r\) force is radially outward to counterbalance
gravity.
If you go around a corner in a car, you seem to feel a force outward from the
circle, not inward. But from the reference frame of a stationary observer, there
must be a net center directed force on both the astronaut(gravity) and the
person toward the center of the circle; otherwise, the person could not be
moving in a circle. Is one force correct and the other incorrect? Is either of
these an illusion? No, they are all real forces.
Take a thoughtful trip in an airplane flying in a horizontal circle with its
wings vertical. The force acting on you is toward the center of the circle, but
you exert an equal and opposite force on the seat that is radially out of the
circle. In some airplanes, if the banking is done too hard, this radially
outward force can be so great as to the blood in the pilot to remain in the
lower part of the body, resulting in a loss of oxygen to the brain and what is
called a blackout. So the blackout is the result of a radially outward force.
Is one reference frame correct and the other incorrect? When you study
Einstein's theory or relativity in physics, you will come across the idea that
there is no one absolute reference frame. You may think that a reference frame
at rest relative to the ground is correct, but the Earth spins on its axis and
rotates around the Sun. The Sun moves around the galactic center. In relativity,
we always have to specify the reference frame we are considering. And the
observed effects of many things change depending on the chosen reference frame.
So, when doing circular motion problems, always think about what reference
frame you are considering. For the method in the preceding problem, our
reference frame is outside the motion, and in this perspective, we take the net
center directed force and set it equal to \(mv^2/r\). This is the default
reference frame that is commonly used to solve circular motion problems.
Apparent Weight
Apparent Weight
You usually encounter apparent weight when working force problems. One of the
first points to get straight is the difference between mass and weight. Mass is
the property that makes different objects accelerate differently when equal
forces are applied. operationally, mass is the m in the equation \(F=ma\). Mass
is measured in kilograms. Weight in equation form is \(W=mg\) and is the force
that something exerts. Weight is measured in newtons or pounds.
The first step in keeping all this straight is to remember that weight is force.
A scale that measures weight in pounds or newtons is just a force meter. Tension
in a rope is a force.
A person of 80 kg mass standing on the surface of the Earth is subject to the
force of gravity that acts between the 80 kg person and the mass of the earth.
This force is expressed as an acceleration due to gravity. On the earth, this
acceleration is \(9.8 m/s^2\). The force on the person, also called weight, is
\[F=80kg*9.8 m/s^2 = 784N\]
A force meter placed between the person and the earth would read 784 newtons.
To understand apparent weight, think of this analogy. Imagine a force meter
between you and the floor of a elevator. As you step on the elevator, the force
meter reads the same as when you are standing on the ground. As the elevator
accelerates upward, the force meter registers higher, reading maximum at maximum
acceleration.
As the acceleration decreases and the elevator assumes a constant
velocity(upward), the force meter reads the same as when the elevator was not
moving. This constant velocity condition is equivalent to forces in equilibrium,
so the only thing contributing to the reading of your force meter is the mg due
to the earth. As the elevator slows, by decelerating, the force meter reads
less than the force for constant velocity. As the elevator comes to a rest at a
higher level, the force meter again reads the same as when you were at ground
level.
Let's go back over this elevator ride and put in some numbers. At zero or
constant velocity, the force meter reads 784 N. If the elevator were accelerating
upward at \(2.0 m/s^2\), then the force meter would read:
\[F=784N+2.0 m/s^2(80kg)=944N\]
If the elevator were accelerating downward at \(3.0 m/s^2\), then the force
meter would read:
\[F=784N-3.0m/s^2(80)kg=544N\]
If the elevator were in free fall, then the force meter would read zero. The
acceleration acting on the elevator is the same as on the person.
Example 1
A 12 kg flower pot is hanging by a cord from the roof of an elevator. What is
the tension in the cord when the elevator is stationary and when it is
accelerating upward at \(3.0 m/s^2\)?
Solution:
When the elevator is stationary, the tension in the cord is:
\[T=12kg(9.8m/s^2)=118N\]
When the elevator is accelerating upward, the tension in the cord is increased
by an amount equal to ma:
\[T=118N+12kg(3.0m/s^2)=154N\]
Example 2
A rope fastened at the top is hanging over a cliff. What is the tension in the
rope with a 70 kg mountain climber sliding down the rope at a constant
acceleration of \(6.0 m/s^2\)?
Solution:
The tension in the rope is less than if the climber were at zero velocity or
sliding at constant velocity. In this instance, the \(6.0m/s^2\) must be
subtracted from the \(9.8 m/s^2\).
\[T=(9.8-6.0) m/s^2(79 kg)=266N\]
When in doubt whether to add or subtract the acceleration from g, look to the
extreme situation. In this problem, the climber sliding down the rope at
constant velocity would produce \(9.8m/s^2(70kg)=686N\) of tension in the rope.
If the climber were in free fall, or accelerating \(9.8m/s^2\), he or she would
produce no tension in the rope. So any acceleration down the rope should be
subtracted from the \(9.8m/s^2\).
Example 3
What is the tension in a rope with the 70 kg mountain climber accelerating up
the rope at a constant acceleration of \(1.0 m/s^2\)?
Solution:
In this case, the tension in the rope is increased by ma:
\[T=(9.8+1.0)m/s^2(70 kg)=756 N\]
Example 4
A 100 kg astronaut produces a force(weight) on the surface of the earth of 980
N. What force(weight) would the astronaut produce on the surface of the moon,
where g is about 1/6 the g on earth?
Solution:
The force or weight would be:
\[F=100 kg(9.8 m/s^2)1/6 = 163 N\]
Example 5
A 16 kg monkey wishes to raise a 20 kg mass by climbing(accelerating) up a rope
that passes over a pulley attached to the mass. How much must the monkey
accelerate up the rope to raise the mass?
Solution:
The mass produces a force or tension in the rope of:
\[T=20kg(9.8m/s^2)=196N\]
The mass of the monkey hanging on the rope produces a force of:
\[T=16kg(9.8m/s^2)=157N\]
To just balance the mass, the monkey must accelerate up the rope to produce a
force of:
\[(196-157)N=39N\]
So, \(39N=(16kg)a \rightarrow a=2.4m/s^2\)
At this acceleration, the tension in the rope is 196 N. To raise the mass, the
monkey must accelerate at a rate greater than \(2.4 m/s^2\). Any acceleration
greater than 2.4 will increase the tension in the rope by an amount equal to the
additional acceleration times the mass of the monkey.
Calculating Work
Work is defined as the force, in the direction of a displacement, times that
displacement.
According to the definition, work is \(W=F_{o}x_{o}\). This is seen as a shaded
rectangular area. In calculus terminology, the work is the area under the curve.
The area under this curve is the area of the rectangle. The unit of work is the
joule (J=N*m).
Example 1
Calculate the work performed by a constant force of 60 N acting in the direction
of a displacement of 3.0 m.
Solution:
The force is acting in the direction of the displacement, so the work is the
product of force times displacement.
\[W=F*x=60N*3.0m=180J\]
The force is in the direction of the displacement, and the work is simply the
product of force and displacement. The simplest way to obtain the component of
the force in the direction of the displacement is to form the dot product of
force and displacement. Using s as a general symbol for displacement, work is
defined as the dot product of the force vector times the displacement vector.
\[W = F * s\]
Work is a scalar, the dot product of force and displacement.
Example 2
Calculate the work performed by a constant force of 40 N pushing a block a
distance of 3.0 m along a horizontal surface at an angle of 20 degrees.
Solution:
The work performed is calculated by:
\[W=F*s=40N \cos 20 * 3.0m=113J\]
Example 3
Calculate the work done by the force of gravity acting on a 70 kg student
sliding down a 30 degree incline slide with a slant distance of 10 m.
Solution:
The first thing to calculate is the force due to gravity mg, which is 70 kg*9.8
m/s^2 = 686 N. This is the force of the student who is sliding down a 30 degree
incline that is 10 m long.
Calculate the force in the direction of the incline:
F=686 N cos 60 = 343 N
The work performed by the gravitational force is:
W = F * s = 343 N * 10 m = 3430 J
Now consider variable forces and look first at a force that is proportional to
x, the displacement, or F=kx, where k is a constant. This form of force is
encountered in springs where one unit of force compresses (or elongates) the
spring a certain distance and the next unit of force doubles the compression,
and so on.
The area under any curve of force versus distance can be approximated by taking
a sum of forces times increments of distances over the curve. In this instance,
take each increment as \(\Delta x\) and the associated forces as F1, F2, F3, up
to N forces and increments. In mathematical language, this sum of the forces
would be written as \(\sum F \Delta x\) where F represents the N discrete forces
measured at intervals of \(\Delta x\) from 0 to \(x_{0}\). This sum is a
collection of rectangles of height F1, F2, and so on until Fn and with width
\(\Delta x\).
Better approximations are obtained by increasing the number of intervals
(decreasing the width of each one). Geometrically, this limit of the sum
represents the area under the curve between x=0 and x=xo and physically
represents the work performed by this linear force over x. For a linear spring,
the work performed is the shaded triangle which is (1/2 base * width) or
\(kx^2/2\).
Example 4
Calculate the work performed in compressing a spring with a force constant of
200 N/m the first 2.0 cm.
Solution:
The work performed is calculated by:
\[W=\frac{kx^2}{2}=\frac{200N}{m}*\frac{(30*10^{-2}m)^2}{2}=9.0*10^{-2}J\]
Example 5
Calculate the work performed in compressing the spring the next 3.0 cm
Solution:
The general expression for the work performed in is \(kx^2/2\). The work to
compress the spring from 3.0 cm to 6.0 cm is the work to compress it 6.0 cm less
the work necessary to compress is 3.0 cm.
\[W=\frac{200N}{m}*\frac{1}{2}[(6.0*10^{-2}m)^2-(3.0*10^{-2}m)^2]=0.27J\]
This process of taking an infinite number of sums of infinitesimally small
increments defines an integral.
For the linear or Hooke's Law spring where F=kx, the integral becomes \(kx^2/2\)
which corresponds to the area under the curve of kx in the interval from 0 to
x0.
The area is the area of the triangle and is equal to \(kx^2/2\). We will not
develop the general technique for obtaining integrals of power law functions. We
will present a justification for the general formula for power law integrals.
Constant F \(F_{0}X_{0}\)
Linear kx \(kx^2/2\)
Quadratic kx \(kx^3/3\)
Power kx^n \(kx^{n+1}/n+1\)
The first two entries in the table are based on the area of a rectangle and a
triangle. The third entry, for the quadratic, follows the pattern of the
previous two and can be verified by numerically integrating the quadratic by
taking small increments and calculating the associated force and multiplying to
find the area of each little approximating rectangle. This process can also be
done for higher power curves to come to the conclusion shown in the last entry
in the table, the one for the general power law curve.
We now come to another definition of work as the integral of the dot product of
force and differential distance:
\[W=\int F*ds\]
Most of the situations encountered in an elementary course are for constant or
linear forces. The linear force is encountered in springs. Linear springs are
characterized by the constant k. The k in the equation F=kx has the units of
N/m. The work performed in compressing a kx spring is \(kx^2/2\).
Example 6
What is the work performed in compressing a quadratic spring \(F=kx^2\) with
k=2000 N/m^2 from 0.30 m to 0.40 m?
Solution:
The force is in the direction of the displacement.
\[W = \int F*ds \int_{0.3}^{0.4} kx^2 dx = kx^3/3 = 2000N/m^2
1/3(0.4)^3-(0.3m)^3 = 25N*m = 25J\]
Notice the development of the definition of work. The first definition was the
product of the force in the direction of displacement times displacement. In
order to handle forces not in the direction of the displacement, the dot product
was introduced. Finally, for non-constant forces, the integral was introduced,
giving the last definition.
Work and Energy
Work-energy analysis is a very powerful tool for solving problems. Force
analysis works well in problems where there is a constant or where there is a
balance of forces.
But when the force acts over a distance, especially when that force is not
constant, then work-energy analysis is the correct analytic tool. To handle the
transformation of force over a distance, we first need to define the concepts of
kinetic and potential energy.
Kinetic Energy
A constant force applied to a mass over a distance produces an acceleration
according to F=ma. This acceleration over a distance changes the velocity of the
mass in accord with the kinetic equation \(v^2=v^2_{o}+2a(x-x_{o})\). If an in
this equation is replaced with F/m, then \(v^2=v^2_{o}+2(F/m)(x-x_{o})\). If the
constant force over a distance is identified as the work, then:
\[F(x-x_{o})=W=mv^2/2-mv^2_{o}/2\]
This simple derivation suggests a more formal look at the work performed on a
mass by a force.
The work performed on a mass by a force is defined as the integral of the force
times the distance over a specified distance. Mathematically, this is \(\int
F*ds\). If the force is replaced by ma and the distance taken in one dimension,
then the work integral becomes:
\[W=\int_{x1}^{x2} madx\]
The acceleration can be written as a chain derivative:
\[a=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\]
Then:
\[adx=\frac{dy}{dx}\frac{dx}{dt}dx=\frac{dx}{dt}dv=vdv\]
Which makes the integral:
\[W=\int_{v1}^{v2} mvdv=\frac{mv_{2}^{2}}{2}-\frac{mv_{1}^{2}}{2}\]
When a force is applied to a mass over a distance and that mass accelerates,
then there is a difference in velocity of the mass between before and after the
application of the force. The work performed is manifest in this velocity, and
the \(mv^2/2\) form is called the kinetic energy. Work is transformed into
kinetic energy.
Gravitational Potential Energy
Gravitational potential energy comes from the acceleration due to gravity and
the accompanying force on a mass on the surface of the Earth. If a mass is
raised from a height\( y_{1}\), to a height \(y_{2}\), directly opposite the
acceleration due to gravity, then work performed appears as gravitational
potential energy.
\[W=\int F*ds=\int_{y1}^{y2} mgdy=mgy_{2}-mgy_{1}\]
This mgy term is the gravitational potential. In mechanical systems, the total
energy of the system is the sum of the kinetic plus potential energies. In a
mechanical system on the surface of the Earth, work performed on the system can
appear as kinetic or potential energy.
work performed = change in kinetic energy + change in potential energy
A convenient way of understanding this is to say that work goes into kinetic or
potential energy.
Look first at a block sliding down a frictionless inclined plane. The velocity
at the bottom of the plane is related to the height h from which the block fell.
The kinetic energy at the bottom of the plane is \(mv^2/2\), and this must equal
mgh, the potential energy at the top of the plane. The gravitational potential
energy goes into kinetic energy.
\[\frac{mv^2}{2}=mgh\]
If mass =3.0 kg
Inclined plane =4.0 m
Angle at bottom of inclined plane and ground is 58 degrees
Height =3.4 m
We get:
\[v=\sqrt{2(9.8m/s^2)3.4m}=8.2m/s\]
If the block were to continue to slide horizontally along a frictionless
surface, it would maintain this 8.2 m/s velocity.
Frictional Forces
Frictional forces are proportional to the normal force(the force between two
surfaces) and a constant characteristic of the interface(the surfaces involved).
They also act to oppose the motion. Frictional forces acting over a distance
result in energy lost due to friction.
A coefficient of friction between the block and a flat surface at the bottom of
the plane would result in energy lost to friction and the block eventually
coming to rest. In this problem, the potential energy at the top of the incline
plane goes into kinetic energy at the bottom of the plane, and this energy goes
into work to overcome friction.
Give the flat surface a coefficient of 0.20, and the frictional retarding force
is:
\[f=\mu mg=0.20(3kg)9.8m/s^2=5.9N\]
To find the distance the block slides, we can go back to the kinetic energy at
the bottom of the plane or the potential energy at the top of the plane. Since
there are no lossed, thes enumbers must be the same. Take the potential energy
at the top of the plane because it is easier to calculate and is original data
in the problem. This energy must equal the work due to friction fl or:
\[mgh=fL=\mu mgL\]
\[3.0kg(9.8m/s^2)3.4m=100J=0.20(3kg)(9.8m/s^2)L\]
\[L=17m\]
Example 1
Place a 3 kg block at the top of a 3.4 m high frictionless incline. At the
bottom of the incline, the block encounters a spring with a constant of 400 N/m
on a horizontal surface. No energy is lost to friction. how far is the spring
compressed?
Solution:
The energy at the top of the plane, which is the same as the energy at the
bottom of the plane, goes into compressing the spring. The energy at the top of
the plane is:
\[100J=\frac{kx^2}{2}\]
\[x=\frac{200J*m}{400N}^{1/2}=0.71m\]
Example 2
Now complicate the first problem by adding a coefficient of friction of 0.20 for
the horizontal surface. How far does the block slide while compressing the
spring?
Solution:
Now the potential energy goes into compressing the spring and overcoming
friction. This statement is very helpful in writing the equation. In words, the
potential energy mgh equals the energy to compress the spring \(kx^2/2\) plus
the work to overcome friction \(\mu mgx\) or:
\[mgh=\frac{kx^2}{2}+\mu mgx\]
\[100J=(200N/m)x^2+0.20(3kg)(9.8m/s^2)x\]
\[200x^2+5.9x-100=0\]
\[x=\frac{-5.9 \pm \sqrt{(5.9)^2-4(200)(-100)}}{2(200)}=\frac{-5.9 \pm
283}{400}=0.70m\]
The negative root will not work for this problem, it does not make sense. The
block compresses the spring 0.70 m while sliding on this frictional surface.
It is instructive to do a problem first with force analysis and then with
work-energy analysis. Consider the case of a block sliding down an inclined
plane with a coefficient of friction. As will become evident, it is possible to
do this problem with force analysis techniques. Work-energy analysis is
conceptually and computationally easier.
Example 3
Consider the inclined plane with friction. Calculate the velocity of the block
at the bottom of the plane using force analysis and then work-energy analysis.
Solution:
The unbalanced force of 20 N acts on the 5 kg block, causing it to accelerate at 4 m/s^2 down the plane. The slant height of the plane is 6m*sin35 degrees = 10.5 m.
This acceleration over the 10.5 m results in a velocity of:
\[v^2=2a(x-x_{o})=2(4m/s^2)10.5m=9.2m/s\]
Solution2:
Now do the same using work-energy analysis. When the block is at the top of the
plane, the energy is:
\[mgh=5kg(9.8m/s^2)(6m=294J\]
At the bottom of the plane, all this potential energy has gone into kinetic
energy(velocity of the block) except for the amount used to do work against the
frictional force. The work(against the frictional force) is:
\[\mu NL=0.20(40N)10.5=84J\]
The energy left after the block has slid down the plane is the 294 J at the top
of the plane minus the 84 J lost to friction, and this energy(210J) is manifest
in the velocity of the block(kinetic energy). This energy is equal to the
\(mv^2/2\), the kinetic energy of the block at the bottom of the plane.
\[\frac{(5kg)v^2}{2}=210J=9.2m/s\]
Example 4
A 3 kg block sliding at 12 m/s along a frictionless surface encounters a spring
with a force constant of 500 N/m resting on a surface with a coefficient of
friction of 0.25. The friction surface is only under the spring. What is the
maximum compression of the spring?
Solution:
When the block encounters the spring and the friction surface, all the energy is
kinetic, \(mv^2/2\). This energy goes into compressing the spring and sliding
along the friction surface. The equation looks like this:
\[\frac{mv^2}{2}+\frac{kx^2}{2}+\mu mgx\]
be careful in writing this statement. Many people write the kinetic energy
correctly, the energy to compress the spring correctly, and then write the force
to slide along the friction surface. Be aware of this mistake, and check the
units to be sure that the units on one side of the equation are the same as the
units on the other side of the equation.
You may have noticed that in all the problems the first equation is written with
units just as a reminder that checking units will reduce the number of errors.
It is very frustrating to make a simple mistake that could have been avoided had
units been checked in the problem before proceeding with the numerical
calculations.
\[\frac{3kg(12m/s)^2}{2}=\frac{500N}{m}\frac{x^2}{2}+0.25(3kg)(9.8m/s^2)x\]
The units are correct, so without units the equation is:
\[250x^2+7.35x-216=0\]
\[x=\frac{-7.35 \pm \sqrt{7.35^2-4(250)(-216)}}{2*250}=\frac{-7.4 \pm
464.8}{500}=0.91m\]
This is the amount the spring is compressed as the block comes to a stop.
Example 5
For the situation described previously, find the velocity with which the block
leaves the spring.
Solution:
Start this problem with the spring compresses the 0.91 m. At this point the
potential energy stored in the spring is used to push the block 0.91 m back
across the friction surface. The energy remaining as the block leaves the spring
is the original kinetic energy less the energy expended in two 0.91 trips across
this friction surface. In word equation form, this is:
\[KE_{initial}-2(\mu mgx's)=KE_{final}\]
The total work performed in the block moving across the surface is:
\[2[0.25(3kg)(9.8m/s^2)0.91m]=13.4J\]
The velocity of the block as it flies off the spring is:
\[\frac{3kg(144m/s^2)}{2}-13.4J=\frac{(3.0kg)v^2_{f}}{2}=v_{f}=11.6m/s\]
Example 6
Go back to an inclined plane problem but with the addition of a spring with a
force constant of 300 N/m placed at the bottom of the incline. Calculate the
compression of the spring when the block slides down the plane.
Solution:
An approximate answer can be obtained by setting mgh, where h is the vertical
distance from the block to the spring, equal to \(kx^2/2\), then compresison of
the spring then becomes:
\[2.5kg(9.8m/s^2)4m=\frac{300N}{m}\frac{x^2}{2} \rightarrow x=0.81m\]
The compression of the spring gives the block a little more potential energy, so
a more correct statement would take into account the additional height h.
\[mgh+(mgx)\sin40=\frac{kx^2}{2}\]
\[h'=x(\sin40)\]
\[2.5kg(9.8m/s^2)4m+2.5kg(9.8m/s^2)x(\sin40)=\frac{300N}{m}\frac{x^2}{2}\]
Eliminating the units and cleaning up the numbers yields a quadratic.
\[150x^2-15.7x-98=0\]
\[x=\frac{15.7 \pm \sqrt{15.7^2-4(150)(-98)}}{2(150)}=\frac{15.7 \pm 243}{300}
=0.86m\]
This value is larger than what was calculated earlier due to the additional
height.
The procedure of finding an approximate solution and then refining it to the
exact solution is very helpful. First, you get an approximate solution that is
close to the exact one, and second, you educate yourself as to how to proceed to
the exact solution.
Example 7
Place a mass on a track made up of a flat section L with coefficient of friction
\(\mu\) and two frictionless semicircular surfaces of radius R. Let the mass
start from the top of one of the semicircles and calculate where it comes to
rest.
Solution:
The initial potential energy is mgR. When the mass encounters the friction
surface, this potential energy is dissipated in doing work to overcome friction.
Assuming that the energy lost due to friction in one traverse is less than the
initial potential energy, the mass will rise to a height(on the opposite
semicircle) R' dictated by the energy statement:
\[mgR'=mgR-\mu mgL\]
After another traverse of the flat portion of the track, the height will be
dictated by:
\[mgR''=mgR-2 \mu gL\]
and so on until all the original potential energy is dissipated.
Example 8
For the track in the previous example, take \(\mu =0.10\), R=1, and L=2m, and
find where the mass stops.
Solution:
The only way the mass loses energy is by sliding along the flat friction part of
the track. The initial potential energy goes into frictional work.
\[(mg)1.0m=0.10(mgx) \rightarrow x=10m\]
The mass crosses the friction area five times and ends up at the edge of the
friction surface opposite from where it started.
Example 9
Consider a 10g bullet passing through a 3kg block resting on a table. The
velocity of the bullet is 400 m/s on entering the block and 250 m/s on exit.
Calculate the energy lost by the bullet in passing through the block.
Solution:
The energy is all kinetic, so calculating before and after gives;
\[KE_{i}=(1/2)(10*10^{-3}kg)(400m/s^2)=800J\]
\[KE_{f}=(1/2)(10*10^{-3}kg)(250m/s^2)=312J\]
The kinetic energy lost = 488J
Example 10
If 30 percent of the energy lost in the previous example is available to move
the block along a surface with a coefficient of friction of 0.80, how far will
it move?
Solution:
Equate the energy available to the work performed.
\[0.30\Delta KE=\mu mgL\]
\[0.30(488J)=0.80(3kg)(9.8m/s^2)L \rightarrow L=6.2m\]
Example 11
Consider an elevator weighing 4000 N held 5 m above a spring with a force
constant of 8000 N/m. The elevator falls onto the spring while subject to a
frictional retarding force(brake) of 1000 N. Describe the motion of the
elevator.
Solution:
The potential energy of the elevator with respect to the top of the spring is:
\[mgh=4000N(5m)=20000J\]
When the elevator falls, this energy, less the energy lost due to the frictional
brake, is available to compress the spring.
\[20000J-1000N(5m)=15000J=energy-to-compress-spring\]
The friction brake stays on while the elevator is in contact with the spring, so
this energy goes into \(kx^2/2\). In equation form, this is:
\[15000J=k(x^2/2)+fx\]
This expression is only approximately correct because if the spring is
compressed a distance x, then this mgx provides an additional amount of
potential energy. The correct equation that determines the compression of the
spring is:
\[mg(h+x)=kx^2/2+f(h+x)\]
Putting numbers into the equation:
\[(4000N/m)x^2+(1000N-4000N)x+(1000N-4000N)5m=0\]
The units are correct so \(4000x^2-3000x-15000=0\)
\[4x^2-3x-15=0\]
\[x=\frac{3 \pm \sqrt{9-4(4)(-15)}}{2(4)}=2.3m\]
The energy stored in the spring at this compressed point is:
\[kx^2/2=8000N/m\frac{2.3m^2}{2}=22040J\]
This amount of energy will raise the elevator and do work against friction
according to the equation:
\[22040J=mgx'+fx'\]
\[22040J=(5000N)x'\]
\[x'=4.4m\]
This places the elevator 2.1 m above the spring.
This analysis can be repeated until the elevator comes to a stop.
Center of Mass
A discussion of the application of the law of conservation of momentum starts
with a consideration of the center of mass of a collection of particles. For
discrete mass points, the center of mass is defined as:
\[r_{cm} = \frac{1}{m} \sum_{i} m_{i} r_{i}\]
Example 1
Calculate the center of mass for a distribution of mass points.
20kg at (3,3), 12kg at (-1,-1), and 18kg at (-2,-2)
Solution:
The \(r_{cm}\) form indicates that the calculation is to be done in vector
notation, so the units are often left out of calculations involving unit vectors
and added at the end.
\[r_{cm} = \frac{1}{50}[20(3i+3j) + 18(-2i-2j) + 12(i-j)]\]
\[r_{cm} = \frac{1}{50}[36i+12j] = \frac{36}{50}i + \frac{12}{50}j = 0.72i +
0.24j\]
These mass points act as if all their mass (50kg) were at the point (0.72,
0.24). If these three masses were placed on a plate of negligible mass, the
balance point would be at (0.72, 0.24).
The law of conservation of momentum can be viewed as a consequence of the
statement: the total mass of a collection of particles times the acceleration of
the center of mass equals the applied or external force, or the sum of the
forces on all the individual component masses. This is a vector equation:
\[Ma_{cm} = F_{ext}\]
Example 2
For the same collection of masses at the same points, add forces to each mass,
and find the resulting acceleration of the center of mass.
20 kg at 45 degrees with a force of 30 N at (3,3)
18 kg at 180 degrees with a force of 24 N at (-2,-2)
12 kg at 270 degrees with a force of 40 N at (-1,-1)
Solution:
Remember, the acceleration of the center of mass is the total mass times the
vector sum of all these forces.
\[(50kg)a_{cm} = [30(\cos 45)i + 30(\sin 45)j - 24i - 40j]N = (-3i - 19j)N\]
\[a_{cm} = (\frac{-3}{50}i - \frac{19}{50}j) m/s^2\]
If the preceding statement \(F_{ext}=ma_{cm}\) is viewed as
\(F_{ext}=\frac{d}{dt}(mv_{cm}\), then if \(F_{ext}=0\), then \(mv_{cm}\) must
be a constant. The derivative of a constant is zero, or viewed graphically, if
the curve of \(mv\) versus time is a constant, then the slope is zero.
Stated another way, for a system with no external forces, the sum of the
momentum vectors \(m1v1+m2v2+...\), which add to \(mv_{cm}\) must all add to
zero.
Example 3
A 5.0 g pellet is compressed against a spring in a gun of mass 300 g. The spring
is released and the gun allowed to recoil with no friction as the pellet leaves
the gun. If the speed of the recoiling gun is 8.0 m/s, what is the speed of the
pellet?
Solution:
The problem is solved by application of the law of conservation of momentum.
This law can be applied because there is no external force. Since there is no
external force, all the mv's must add to zero.
\[m_{g}v_{g} = m_{p}v_{p}\]
or
\[m_{g}v_{g} - m_{p}v_{p} = 0\]
The conservation of momentum statement on the left is based on the simple
observation "bullet goes one way, gun goes the other" while the formal statement
that the mv's add to zero is on the right. With a well labeled diagram of the
situation, the statement on the left is probably easier to visualize. The
momenta are equal and opposite. Putting in the numbers we have :
\[300g(8.0m/s) = 5.0g*v_{p} \text{ so } v_{p} = 480m/s\]
As a check, note that the momentum of the gun \(P_{g} = m_{g}v_{g} = 2.4kg*m/s\)
and the momentum of the pellet \(P_{p} = m_{p}v_{p} = 2.4kg*m/s\) are
numerically equal because they are in the opposite direction, add to zero.
The energy of each is \(mv^2/2\) or \(p^2/2m\). so for the gun,
\[KE_{gun} = \frac{(2.4kg*m/s)^2}{(2*0.30kg)} = 9.6j\]
Performing the same calculation for the pellet:
\[KE_{pellet} = \frac{(2.4kg*m/s)^2}{(2*0.01kg)} = 288l\]
The total energy stored in the spring is the sum of these energies.
Example 4
Make the pellet gun of previous example fully automatic and capable of firing 10
pellets per second. Calculate the force these pellets make on a target where the
pellets do not bounce.
Solution:
This problem is solved by calculating the average momentum transferred to the
target per unit of time. The momentum of each pellet is \(2.4kg*m/s\). The force
on the target is calculated from the simple expression:
\[F = \frac{\Delta p}{\Delta t} = \frac{10(2.4kg*m/s)}{1.0s} = 24N\]
The total momentum transferred each second is 10 individual momenta of each
pellet.
Example 5
A 75 kg hockey player traveling at 12 m/s collides with a 90 kg player
traveling, at right angles to the first, at 15 m/s. The players stick together.
Find their resulting velocity and direction. Assume the ice surface to be
frictionless.
Solution:
This problem can be analyzed by conservation of momentum. Calculate the momenta,
and draw a vector diagram.
\[p_{1} = 75kg(12m/s) = 900kg*m/s\]
and
\[p_{2} = 90kg(15m/s) = 1350kg*m/s\]
The angle of the two hockey players is from:
\[\tan \theta = 1350/900 = 1.5 \text{ or } \theta = 56 \]
and the resulting momentum is:
\[p = \sqrt{1350^2 + 900^2} kg*m/s = 1620kg*m/s\]
The players move off with velocity:
\[v = \frac{P}{m1+m2} = 9.83m/s\]
at an angle of 56 degrees to the original direction of the 75 kg player.
Example 6
James Bond is skiing along being pursued by Goldfinger, also on skis. Assume no
friction. mr. bond, at 100 kg, fires backward a 40 g bullet at 800 m/s.
Goldfinger, at 120 kg, fires forward at Bond with a similar weapon. What is the
relative velocity change after the exchange of six shots each. No bullets hit
bond or Goldfinger.
Solution:
The problem is analyzed with conservation of momentum. The \(m_{b}v_{b}\) of the
bullet fired by Bond increases his momentum by \(m_{B} \Delta v_{B}\). Remember
that each bullet Bond fires increases his velocity. Set \(m_{b}v_{b} = m_{B}
\Delta v_{B}\) and solve for \(\Delta v_{B}\).
\[40*10^{-3}kg(800m/s) = (100kg) \Delta v_{B}\]
\[\Delta v_{B} = 0.32 m/s\]
The \(40*10^{-3}kg\) bullet is small compared to the 100 kg of Bond, and it
would not affect the calculation. The \(\Delta v_{B}\) notation is used to
indicate that each bullet fired by Bond causes a change in his velocity.
Goldfinger has his momentum decreased. In his case, \(m_{b}v_{b} = m_{G} \Delta
v_{G}\). Putting in the numbers:
\[32kg*m/s = (120kg) \Delta v_{G}\]
\[\Delta v_{G} = 0.26m/s\]
Bond goes faster and Goldfinger goes slower, with the total change in velocity
0.58 m/s for each pair of shots fired. For six shots, this amounts to a
difference of 3.48 m/s. If Bond and Goldfinger have been traveling at the same
speed, then after this exchange Bond would have a relative speed advantage of
3.48 m/s.
Example 7
A 3000 kg closed boxcar traveling at 3.0 m/s is overtaken by a 1000 kg open
boxcar traveling at 5.0 m/s. The cars couple together. Find the resulting speed
of the combination.
Solution:
The momentum before coupling is the same as the momentum after coupling.
\[3000kg(3.0m/s) + 1000kg(5.0m/s) = (4000kg)v\]
\[v = 3.5 m/s\]
Example 8
Continuing with the previous example, rain falls into the open boxcar so that
the mass increases at 1.0 kg/s. What is the velocity of the boxcars at 500 s?
Solution:
The total momentum of the boxcars is 4000kg(3.5m/s)=14000kg*m/s. Assume that
there is no horizontal component of the rain to change the momentum in the
direction of motion of the boxcars. The mass increases by (1.0kg/s)3.5m/s=500kg.
The momentum is a constant, so the new velocity is:
\[14000kg*m/s = (4500kg)v_{R}\]
\[v_{R} = 3.11 m/s\]
Example 9
For the situation described in the previous problem, what is the rate of change
in velocity for the boxcars?
Solution:
This is a very interesting calculus problem that involves taking the total
derivative. Since there are no external forces, the total change in mv must
equal zero.
\[d(mv) = mdv + vdm = 0\]
or
\[mdv = -vdm\]
Now write m as a function of time.
\[m = m_{o} + rt = 4000kg + (1.0kg/s)t\]
The derivative of m is dm = rdt.
Using the two previous equation and rearranging:
\[\frac{dv}{v} = -\frac{dm}{m} = -\frac{r}{m_{o}+rt}dt\]
Introduce a change of variables:
\[u = m_{o} + rt\]
with du = rdt
\[\frac{dv}{v} = -\frac{du}{u}\]
integrating, ln v = -ln u + ln K because it is a convenient form for the
constant. Now rearrange:
\[ln v + ln u = K\]
and:
\[ln uv = ln K\]
\[uv = K\]
Change the variable back to t, so that \((m_{o} + rt)v = k\)
Evaluate the constant from the condition that at t=0,
\[m_{o} = 14000kg*m/s\]
So:
\[K = 14000kg*m/s\]
The relation between v and t is:
\[v = \frac{K}{m_{o} + rt} = \frac{14000kg*m/s}{(4000kg + 1.0kg/s*t)}\]
The velocity at t=500s is:
\[v_{t=500} = \frac{14000kg*m/s}{4500kg} = 3.11 m/s\]
conservation of momentum and a little calculus produce the v versus t relation.
Example 10
A 3.0 kg cat is in a 24 kg boat. The cat is 10 m from the shore. The cat walks
3.0 m toward the shore. How far is the cat from the shore? Assume no friction
between boat and water.
Solution:
There are no external forces, so the center of mass of the cat-boat system is
constant. Knowing that the center of mass doesn't move is all that is necessary
to do this problem.
Write the center of mass of the cat-boat system before the cat walks. M is the
mass of the boat and m is the cat. Then write the center of mass of the cat-boat
system after the cat walks.
\[x_{cm} = \frac{Mx_{b}+mx_{c}}{M+m}\]
Because there are no external forces, the centers of mass are the same, so:
\[Mx_{b} + mx_{c} = Mx'_{b} + mx'_{c}\]
Watch the algebra, and solve for \(x'_{c}\).
\[x'_{c} = 10m-3.0m+(x'_{b}-x_{b})\]
Now substitute from \(M(x'_{b}-x_{b}) = m(x_{c}-x'_{c}\)
\(x'_{c} = 7+(1/8)(10-x'_{c})\) and \(8x'_{c} = 56+(10-x'_{c})\)
\[x'_{c} = 7.33m\]
Collisions and Impulse
Momentum provides a new analysis technique. With force analysis and work-energy
analysis, application of the appropriate analysis technique was reasonably
straightforward. Knowing when to use work-energy analysis and when to use
momentum analysis adds a whole new dimension to problem solving.
There are certain types of collisions where conservation of momentum and
conservation of energy can be applied and certain problems where where only
conservation of momentum can be applied. This is nowhere better illustrated than
in the ballistic pendulum problem where conservation of momentum is applied in
one part of the problem and conservation of energy in another part of the
problem. Learning where to apply these two conservation laws is difficult. In
the upcoming problems, be aware of which law is being applied and why.
Before doing specific problems, take a general look at collision in one
dimension. A and B will designate two particles, and 1 and 2 will designate
before and after the collision.
There are two types of collisions. In elastic collisions, both momentum and
energy are conserved. Examples of elastic collisions are billiard balls or any
collision where the participants bounce. In inelastic collisions, only momentum
is conserved. Example of inelastic collisions are railroad cars coupling or a
steel ball thrown into a piece of clay or any collision where the participants
stick together and do not bounce.
It is this sticking or non-sticking that determine if energy is conserved or not.
When we say that energy is not conserved in a collision where the participants
stick together, we mean only tat mechanical energy \(1/2mv^2\) is not conserved.
Inelastic Collisions
Consider first inelastic collisions, where the particles stick together. Using
the notation:
\[v_{A2}=v_{B2}=v_{2}\]
This equation states that after the collision, the velocity of the A particle is
the same as the velocity of the b particle. Applying the law of conservation of
momentum and this condition:
\[m_{a}v_{A1}+m_{B}v_{B1}=(m_{A}+m_{B})v_{2}\]
The left side of the equation is the momentum before collision, and the right
side is the momentum of the two masses stuck together at the same velocity after
the collision.
Take the special case where the B particle is initially stationary. While
mechanical energy is not conserved in the collision, the kinetic energies can be
written as:
\[K_{1}=(1/2)m_{A}v_{A1}^2 \text{ and } K_{2}=(1/2)(m_{A}+m_{B})v_{2}^2\]
\(K_{2}\) can be rewritten using the conservation of momentum statement.
\[K_{2}=1/2(m_{A}+m_{B})\frac{m_{A}v_{A1}}{m_{A}+m_{B}}^2=1/2
\frac{m_{A}^2}{m_{A}+m_{B}} v_{A1}^2\]
Comparing these two equations gives a relationship for the energy before and
after collision.
\[K_{2}=\frac{m_{A}}{m_{A}+m_{B}} K_{1}\]
Example 1
A ballistic pendulum, a device for measuring the speed of a bullet, consists of
block of wood suspended by cords. When the bullet is fired into the block, the
block is free to rise. How high does a 5.0 kg block rise when a 12 g bullet
traveling at 350 m/s is fired into it?
Solution:
This is a most interesting and instructive problem. The collision between the
bullet and the block is clearly inelastic(the bullet comes to rest in the
block). Part of the kinetic energy of the bullet goes into friction as the
bullet burrows its way into the block. Therefore, mechanical energy is not
conserved.
Because the collision is inelastic, apply conservation of momentum to the
collision. Before the collision, all the momentum is in the mv of the bullet.
After the collision, the momentum is in the (m+M)V of the block and bullet. We
assume that the bullet comes to rest(transfers all its momentum) before there
is appreciable motion of the bullet-block combination.
\[mv=(m+M)V\]
After the collision, the rise of the block is determined by energy analysis. The
kinetic energy of the block goes into potential energy.
\[(1/2)(m+M)V^2=(m+M)gh\]
Substituting for V from mv =(m+M)V
\[(1/2)(\frac{m}{m+M})^2 v^2=gh\]
\[v=\frac{m+M}{m} \sqrt{2gh}\]
This gives the relation between the velocity of the bullet and the height the
block and bullet rise.
\[h=\frac{(350m/s)^2}{2*9.8m/s^2}*(\frac{0.012}{5.012})^2=3.6cm\]
In this problem, the 0.012 can be neglected in comparison with 5.0. This is not
always the case, so we write m+M as 5.012 as a reminder to include both m+M in
the calculation.
Example 2
A 6.0 g bullet is fired horizontally into a 2.8 kg block resting on a horizontal
surface with a coefficient of friction of 0.30. The bullet comes to rest in the
block, and the block slides 0.65 m before coming to a stop. What is the velocity
of the bullet?
Solution:
Assume that the bullet comes to rest in the block before the block moves
appreciably and that all the momentum in the bullet is transferred to the
bullet-block combination.
\[mv=(m+M)V\]
Once the bullet-block combination is moving at V, the kinetic energy
\((1/2)(m+M)V^2\) goes into work to overcome friction, \(\mu (m+M)gx\).
\[(1/2)(m+M)V^2-\mu (m+M)gx\]
Substituting:
\[1/2(\frac{mv}{m+M})^2=\mu
gx=\frac{2.806}{0.006}*\sqrt{2(0.30)(9.8m/s^2)0.65m}=914m/s\]
Elastic Collisions
Now look at elastic collisions, where the particles bounce. In elastic
collisions no energy is lost to permanent deformation of the particles. Write a
conservation of momentum statement for the collision.
\[m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}\]
Now write a conservation of energy statement.
\[1/2m_{A}v^2_{A1}+1/2m_{B}v^2_{B1}=1/2m_{A}v^2_{A2}+1/2m_{B}v^@_{B2}\]
These two statements can be rewritten as:
\(m_{A}(v_{A1}-v_{A2})=m_{B}(v_{B2}-v_{B1})\) and:
\(m_{A}(v^2_{A1}-v^2_{A2})=m_{B}(v^2_{B2}-v^2_{B1})\)
Now divide the rewritten conservation of energy statement by the conservation of
momentum statement to find:
\[v_{A1}+v_{A2}=v_{B2}+v_{B1}\]
The term \(v_{A1}-v_{B1}\) is the speed of the approach, the speed of A relative
to B measured by an observer on B. The term \(v_{A2}-v_{B2}\) is the speed of
departure. So, in an elastic collision, the speed of approach is equal to the
speed of departure.
Now find expressions for the velocities after the collision in terms of the
masses and the velocities before the collision. From the velocity statement:
\[v_{B2}=v_{A1}+v_{A2}-v_{B1}\]
From the conservation of momentum statement, substitute the equations for
\(v_{B2}\):
\[m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}(v_{A1}+v_{A2}-v_{B1})\]
With a little algebra:
\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}+\frac{2m_{B}}{m_{A}+m_{B}}*v_{B1}\]
With these two statements, \(v_{A2}\) and then \(v_{B2}\) can be predicted from
the initial masses and velocities.
Look at some special cases: If \(m_{A}=m_{B}\) the masses are equal and then:
\(v_{A2}=v_{B1} \text{ and } v_{B2}=v_{A1}\)
The particles exchange velocities. This is what happens in billiards.
If \(v_{B1}=0\) the struck mass is at rest and then:
\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}\]
and:
\[v_{B2}=v_{A1}+\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}=\frac{2m}{m_{A}+m_{B}}*v_{A1}\]
For some realistic cases where the masses are equal or one particle is at rest,
the resulting expressions for the velocities are easily calculated.
Example 3
A 1.5 kg block traveling at 6.0 m/s strikes a 2.5 kg block at rest. After an
elastic collision, what are the velocities of the blocks?
Solution:
This is the special case where the struck block is at rest, so:
\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}=\frac{1.5-2.5}{4.0}*6.0m/s=-1.5m/s\]
\[v_{B2}=\frac{2m_{A}}{m_{A}+m_{B}}*v_{A1}=\frac{2*1.5}{4.0}*6.0m/s=4.5m/s\]
The striking block rebounds at -1.5 m/s and the stuck block moves off at 4.5
m/s.
Example 4
A 1.0 kg steel ball is attached to a lightweight 1.0 m long rod pivoted at the
other end. The ball is released at the horizontal and strikes a 3.0 kg steel
block resting on a surface with a coefficient of friction of 0.25. How far does
the block travel?
Solution:
This problem is similar to the ballistic pendulum problem in that conservation
of energy and conservation of momentum and then work energy have to be applied
correctly. From the description of the collision, assume that it is elastic.
First, calculate the velocity of the ball as it hits the block. Potential energy
goes into kinetic energy.
\[mgh=(1/2)mv^2\]
\[v=\sqrt{2gh}=\sqrt{2(9.8m/s^2)1.0m}=4.4m/s\]
The collision is elastic and \(v_{B1}=0\), so:
\[v_{B2}=\frac{2m_{A}}{m_{A}+m_{B}}*v_{A1}=\frac{2m_{A}}{m_{A}+m_{B}}*\sqrt{2gh}=1/2\sqrt{2gh}=2.2m/s\]
This gives the initial velocity of the struck block. The kinetic energy goes
into work against friction.
\[1/2m_{B}v^2_{B2}=\mu m_{B}gL\]
\[L=\frac{v^2_{B2}}{2\mu g}=\frac{1}{2\mu
g}*\frac{2gh}{4}=\frac{h}{4\mu}=\frac{1.0m}{4*0.25}=1.0m\]
Go back over this problem and see where energy analysis is applied, conservation
of momentum is applied, and work-energy analysis is applied. Knowing what laws
to apply where in the problems is the hard part of collision problems.
Impulse
Impulse is the name given to a force that acts for a very short period of time.
A struck baseball or golf ball are examples of impulses or impulse forces. In
most impulses, it is impossible to graph forces versus time, though we can often
estimate how the force varies with time.
Force is defined in terms of change in momentum as
\[F=\frac{\Delta (mv)}{\Delta t}\]
If the force is time dependent, F(t),then:
\[\Delta (mv)=F(t)\Delta t\]
With calculus notation this statement would be written as:
\[\int dp=\int F(t) dt\]
The left side is the change in momentum, and the right side is the area under
the F(t) versus t curve. This integral is called the impulse or impulse
integral. If the curve of F(t) versus t were a parabola, (reasonable for an
impulse), then the right side of the equation would be the area under the curve.
Example 5
A 1.2 kg croquet ball moving at 2.0 m/s is struck from behind by an impulse
force, what is the final velocity of the ball?
Solution:
The initial momentum is \(p_{i}=1.2 kg(2.0)m/s=2.4 kg*m/s\)
The impulse integral, the difference in momentum, can be calculated by
inspection. The area under the curve(1/2 base times height) is:
\[\Delta (mv)=\int F(t) dt=(1/2)(0.20s)100N=10kg*m/s\]
Th final momentum then is \[
p_{i}=12.4kg*m/s\]
Final velocity is:
\[v_{f}=\frac{12.4kg*m/s}{1.2kg}=10.3m/s\]
Rotational Motion
Rotational motion is the relationship between angular velocity and angular
acceleration. The radian measure is the arc length over the radius.
Another relationship in rotational motion is that between linear motion and
angular motion. These relate the motion of a particle along a circular path to
the angular motion. The position along the circular path is called the linear
position.
There is one more relationship in angular motion. If a point is rotating on a
circle with constant angular velocity, the linear velocity is numerically
constant. Velocity is a vector, and while the number associated with the linear
velocity is not changing, the direction is changing.
The vectors v and v' are separated by an angle. The acceleration is the
difference between the vectors. The vectors v and v' are at right angles to the
radius, so the angle between v and v' is the same as the angle between the r's.
For small angles, \(\Theta s\) approaches a straight line, so comparing the
similar triangles, write: \(\frac{\theta v}{v} = \frac{\theta s}{r}\) and from
this relation write: \(\frac{\theta v}{\theta t} = \frac{v \theta s}{r \theta
t}\).
In the limit, \(\frac{\theta s}{\theta t}\) is v and \(\frac{\theta v}{\theta t}\)
is a, so the acceleration is \(a = \frac{v^{2}}{r}\).
The acceleration vector points inward along r. In the limit, everything points
toward the center of the circle.
Angular velocity is usually written with the units (rad/s) and angular
acceleration as (rad/s^2), which is why radian is called a phantom unit. Radians
and revolutions are usually introduced as units for convenience in understanding
a problem and not as units to be carried and manipulated.
Example 1
An automobile is traveling at 60 km/h. Calculate the angular velocity of the
0.35 m radius wheels.
Solution:
The linear velocity of the contact point on the wheel is:
\[\frac{60km}{h}*\frac{1h}{3600s}*\frac{10^{3}m}{km}=\frac{100}{6}m/s\]
The linear velocity of the contact point on the wheel is the same as the
velocity of the axle. the angular velocity is:
\[w=\frac{v}{r}=\frac{100m}{6s}*\frac{1}{0.35m}=47.6-rad/s\]
Example 2
If the automobile in the previous example accelerates uniformly from 60km/h to
80km/h in 3.0s, what is the angular acceleration?
Solution:
This acceleration is the acceleration of the point where the wheel meets the
road. First, calculate the change in velocity:
\[\theta v = (80-60)km/h =
20\frac{km}{h}*\frac{1h}{60m}*\frac{1min}{60s}*\frac{10^{3}m}{km} =
\frac{200}{36}m/s\]
This is the change in velocity of the point on the wheel. The linear
acceleration is:
\[a = \frac{\theta v}{\theta t} = \frac{200m}{36s}*\frac{1}{3.0s} = 1.85 m/s^2\]
The angular acceleration is:
\[a = \frac{a}{r} = \frac{1.85m}{s^{2}}*\frac{1}{0.35m} = 5.29\frac{1}{s^{2}} =
5.29 rad/s^2\]
Example 3
A disk rotating at 30 rad/s slows to 20 rad/s while turning through 60
revolutions. How long does this take?
Solution:
Since the acceleration is constant, we can use:
\[t = \frac{2\theta}{w_{o} + w} = \frac{2*2\pi*60rad}{30 rad/s + 20 rad/s} =
\frac{240\pi}{50} s = 15.1 s\]
Example 4
For the acceleration of the previous problem, how many revolutions does it take
for the disk to stop?
Solution:
\[a = \frac{\theta w}{\theta t} = \frac{-10rad/s}{15.1s} = -0.662 rad/s^{2}\]
Now use \(w^2 = w_{o}^{2} + 2a\theta\) with \(w = 0\) and \(w = 30 rad/s\) to
find \(\theta\).
\[w_{o}^{2} = -2a\theta \rightarrow \theta = -\frac{w_{o}^{2}}{2a} =
\frac{30rad/s^{2}}{2*0.662rad/s^{2}} = 680 rad \rightarrow
680rad*\frac{1rev}{2\pi rad} = 108rev\]
Example 5
Consider a space station in the form of a donut with a rectangular cross section
connected by spokes to a central axis. The floor is the inside of the outer
wall. How fast would a 300 m radius station have to rotate to duplicate the
acceleration due to gravity on the surface of the Earth?
Solution:
The required acceleration is the \(v^2/r\) acceleration associated with
rotational motion. Since v is related to w by \(v = rw\), then the required
acceleration \(a = v^2/r = rw^2\). Set this acceleration equal to 9.8m/s^2 and
solve for w.
\[w=\sqrt{a}{r}=\sqrt{9.8m/s^2}{300m}=0.181rad/s.\]
The frequency in revolutions per minute is:
\[f=\frac{w}{2\pi}=\frac{1rev}{2\pi
rad}*\frac{0.181rad}{s}=\frac{0.0288rev}{s}*\frac{60s}{min}=1.72rev/min\]
Example 6
If someone of weight 800 N living in the space station described in previous
problem were to move one-half the radius in toward the axis of rotation, what
would be his weight at this radius?
Solution:
The person's weight, what a force meter placed between the person and the floor
would read, in this environment would be:
\[W=\frac{mv^{2}}{r}=mrw^{2}=\frac{800N}{9.8m/s^{2}}*150m*0.18^{2}=400N\]
Example 7
Consider a mass of 2.0 kg being whirled in a sling in a horizontal circle. The
period of rotation is 1.0s and the radius is 1.0m. What is the tension in the
cord?
Solution:
The force that makes the mass travel in a circle must be equal to
\(\frac{mv^{2}}{r}\). The velocity of the mass is the distance traveled divided
by the time to travel one circumference.
\[v=\frac{2\pi r}{T}=\frac{2\pi*1.0m}{1.0s}=2.0\pi m/s\]
The force in the cord is:
\[F=\frac{mv^{2}}{r}=\frac{2.0kg}{1.0m}*(2.0 \pi m/s)^{2}=79.0N\]
Example 8
Take the same mass, radius, and period as in previous problem but with the
rotation in a vertical circle. What is the force between the mass and the sling?
Solution:
In this case, gravity has to be taken into account. The mass exerts a force on
the sling radially out and equal to \(\frac{mv^{2}}{r}\). Gravity exerts a force
mg down. The gravitational force is \(mg=(2.0kg)*9.8m/s^{2}=19.6N\). The force
between the mass and the sling is \(79.0+19.6=98.6N\) at the bottom of the circle
and \(79.0-19.6=59.4N\) at the top of the circle.
Example 9
For the situation in the previous problem, calculate the minimum velocity and
period of rotation to keep the mass from falling out at the top of the circle.
Solution:
Zero tension in the cord occurs at the top when \(\frac{mv^{2}}{r}=mg\).
Notice that the mass drops out of the equation. Using the 1.0 m radius,
\[v_{min}=\sqrt{1.0m(9.8m/s^2)}=3.13m/s\]
The period is calculated by replacing the velocity by \(\frac{2\pi}{T}\).
\[T=2\pi\sqrt{(r)(g)}=2\pi*\sqrt{(1.0m)(9.8m/s^{2})}=2.0s\]
Example 10
A glider pilot wishing to fly in a vertical loop dives to attain a speed
sufficient to keep the glider from falling out of the top of the loop. What is
the minimum entry speed for a 200 m radius loop?
Solution:
This problem is solved with energy analysis. At the bottom of the loop, the
glider must have sufficient kinetic energy to climb the diameter of the loop and
have enough left to satisfy the \(\frac{mv^{2}}{r}=mg\) condition.
The energy statement is in word form: kinetic energy at entry equals the
potential energy to reach the top of the loop plus kinetic energy at the top of
the loop.
\[\frac{1}{2}mv^{2}=2rmg+\frac{1}{2}mv^{2}\]
The minimum velocity condition is:
\[\frac{mv^{2}}{r}=mg\]
Substituting for v, the minimum energy statement is:
\[\frac{1}{2}mv^{2}=2rmg+\frac{1}{2}mrg=\frac{5}{2}mrg\]
yielding a minimum entry speed of:
\[v=\sqrt{5rg}\]
For a 200 m radius loop, the minimum v for the glider is:
\[v=\sqrt{(5*(200m))*(9.8m/s^{2})}=99m/s=356km/h\]
For a 100 m radius loop, \(v=70m/s=252km/h\)
Example 11
A grinding wheel of 30 cm diameter is rotating with angular velocity of 3.0
rad/s and slowing under the constant acceleration of -3.o rad/s. Calculate
everything possible about the motion at several different times.
Solution:
Circular motion problems even more than linear motion problems often present a
challenge as to how to proceed from the data to the specific question. When
confused about the specified route to follow in a problem, the best thing to do
is calculate something simple and let this first calculation lead you to others
and ultimately the answer. After doing the problem once the hard way, you will
learn more direct routes through the problem. Given the data in this problem,
start calculating some simple things.
What is the angular velocity at 0.40 s?
\[w=w_{o}+at=3.0rad/s-3.0rad/s^{2}(0.40s)=1.8rad/s\]
What angle has been turned through in this 0.40 s?
\[\theta=w_{o}t+\frac{1}{2}at^{2}=3.0rad/s*0.40s-\frac{1}{2}3.0rad/s^{2}*0.16s^{2}=0.96rad\]
What is the velocity of a point on the rim at 0.20s?
First, find the angular velocity.
\[w=w_{o}+at=3.0rad/s-3.0rad/s^{2}*0.20s=2.4rad/s\]
Then find the linear velocity from \(v=rw=15*10^{-2}m*2.4 rad/s=0.36m/s\)
With this acceleration, how long does it take for the wheel to come to rest?
\[w=w_{o}+at \longrightarrow 0=3.0*(1/s)-3.0(1/s^{2}) \longrightarrow t=1.0s\]
How much angle has turned through in this much time?
\[w^{2}=w_{o}^{2}+2a\theta \rightarrow 0=9.0rad/s^{2}-2*3.0rad/s^{2}*\theta
\rightarrow \theta=1.5rad\]
How far does a spot on the rim travel in this time?
\[s=r\theta=15*10^{-2}m*1.5rad=0.225m\]
Rotational Dynamics
First, let us review rotation.
The average angular speed is:
\(w = \frac{\theta_{2} - \theta_{1}}{t_{2} - t_{1}} =
\frac{\Delta\theta}{\Delta t}\)
The instantaneous angular speed is:
\(w = lim_{\Delta t \rightarrow 0}\frac{\Delta\theta}{\delta t} =
\frac{d\theta}{dt}\)
The average angular acceleration is:
\(a = \frac{w_{2} - w_{1}}{t_{2} - t_{1}} = \frac{\Delta w}{\Delta t}\)
Instantaneous acceleration is:
\(a = lim_{\Delta t \rightarrow t}\frac{\Delta w}{\Delta t} = \frac{dw}{dt}\)
The relationships between the point and the angle all start with \(s =
r\theta\), then with successive derivatives comes:
\(\frac{ds}{dt} = r\frac{d\theta}{dt}\) and \(v = rw\)
\(\frac{dv}{dt} = r\frac{dw}{dt}\) or \(a = ra\)
Remember that the radial acceleration is \(a_{rad} = v^{2}/r = rw^{2}\)
A force has to be associated with the angular acceleration. Force does not lend
itself well to motion of a particle constrained to move in a circle because most
forces are not tangent to a circle. Torque works much better. Torque is defined
in a rather unusual manner.
Torque is the vector product of r, the vector from the axis of rotation to the
rotating point, and F, the force applied at that point. \(t = r * F\)
You may want to review the definitions of the cross product. Most problems in
torque can be done with the geometric interpretation of the cross product.
Torque, \(r * F\), is the product of r and the component of F perpendicular to
r.
The torque vector is perpendicular to the plane of r and F and follows the
right-hand rule pf rotating r into F.
Example 1
Calculate the torque on a 2.5 kg mass constrained to rotate in a 1.6 m radius
circle with a 250 N force applied at a constant angle of 35 degrees between r
and F.
Solution:
The definition of r * F gives the direction of the torque. Application of the
crossing of r into F is similar to generating a right-handed coordinate system
by rotating x into y with the right hand, with the thumb giving the direction of
z. The magnitude of the torque is:
\(T = r * F sin\theta = 1.5m(250N)sin35 = 215N*m\)
Torque is r cross force like angular momentum is r cross momentum or \(L = r *
p\)
Angular momentum is a vector in a direction perpendicular to the plane of r and
p and magnitude equal to the product of r and the component of p perpendicular
to r.
Example 2
For the situation in the first problem, add that the linear velocity is 2.8m/s
and find the angular momentum.
Solution:
The linear momentum is \(mv = 2.5kg(2.8m/s) = 7.0kg*m/s\). The linear velocity
is the tangential velocity, which is always at right angles to the radius. the
magnitude of the cross product is simplified because the sin of 90 = 1.
\(L = rp = 1.6m(7.0kg * m/s) = 11.2 kg * m^{2}/s^{2}\)
With these definitions for torque and momentum, several relationships can be
derived.
\(t = r * F = r * \frac{dp}{dt}\) and \( \frac{dL}{dt} = r * \frac{dp}{dt} +
\frac{dr}{dt} * p\)
so: \(\frac{dr}{dt} = v\) and \(p = mv\)
and: \(\frac{dr}{dt} * p = v * mv\) and \(v * v = 0\)
Finally: \(\frac{dL}{dt} = r * \frac{dp}{dt}\) and \(t = \frac{dL}{dt}\)
Torque is the time derivative of angular momentum as force is the time
derivative of linear momentum.
The total angular momentum of a system is the sum of the angular momenta of the
individual pieces. Internal torques are equal and opposite, so only external
torques change the angular momentum of the system.
\(\frac{dL}{dt} = t_{ext}\) is parallel to \(\frac{dP}{dt} = F_{ext}\)
If the derivative of the total angular momentum is zero, the condition for no
external torques, then the angular momentum is a constant.
Now look to the kinetic energy associated with rotating particles. Each
particles of a rotating mass has a linear speed \(v = rw\) and the kinetic
energy for each piece is a sum.
\(KE = 1/2(m_{1}v^{2}_{1} + m_{2}v^{2}_{2} +...) = 1/2(m_{1}r^{2}_{1} +
m_{2}r^{2}_{2} +...)w^{2}\)
This sum of the \(mr^{2}\)'s for the collection of pieces is called the
rotational inertia I, so the \(KE\) can be written as:
\(KE = \frac{1}{2}Iw^{2}\)
The rotational inertia I is easy to calculate for individual particles and has
been calculated for many shapes.
Example 3
Calculate the rotational inertia for one 2.0 kg mass at the end of a 3.0 m rod
of negligible mass and then the rotational inertia of a dumbbell consisting of
two 2.0 kg masses on the ends of a 6.0 m long rod of negligible mass pivoted
about the center of the rod.
Solution:
For the one mass, the sum of the \(mr^{2}\)'s is just \(2.0kg(9.0m^{2}) = 18 *
kg * m ^{2}\).
For the dumbbell, there are two masses, and the rotational inertia would be \(36
kg * m^{2}\). Notice that if the dumbbell were rotated about one mass, then the
rotational inertia would be \(2.0kg*36m^{2} = 72kg * m^{2}\). This is, of
course, neglecting the extent of the mass close around the rotating axis.
The total momentum of a rotating mass is mvr. If the mass is rotating, then v is
perpendicular to r, leading to a simple statement of angular momentum.
Angular momentum is: \(L = mvr = mr^{2}w = Iw\)
By definition, \(t = \frac{dL}{dt} = I\frac{dw}{dt} = IA\)
The power is: \(\frac{d(KE)}{dt} = \frac{d}{dt}(\frac{1}{2}Iw^{2}) =
Iw\frac{dw}{dt} = Iwa = tw\)
Angular momentum is conserved and is a powerful tool in solving certain problems
in rotational motion. First, calculate a torque using the vector form for
position and force and the determinant for the cross product.
\(r = xi + yj + zk\) and \(F = F_{x}i + F_{y}j + F_{z}k\)
\(t = r * F = (F_{z}y - F_{y}z)i + (F_{x}z - F_{z}x)j + (F_{y}x - F_{x}y)k\)
Most problems do not require a calculation like this. For a position and force
vector in the x-y plane, the torque reduces to a single z component.
Example 4
Calculate the angular momentum of a 3.0 kg mass at 3.0 m in the x direction and
-2.0 m in the y direction and with velocity components of 20 m/s in the x
direction and -30 m/s in the y direction.
\(\theta = -56.3 - (-33.7) = -22.6degrees\)
\(tan^{-1}(3/2) = 56.3\)
Solution:
Write r and p in vector form:
\(r = (3.0i - 2.0j)m\) and \(p = (60i - 90j)kg * m/s\)
The angular momentum is: \(L = r * p = (-270 + 120)k = -150k kg * m^{2}/2\)
The angular momentum vector is \(150 kg * m^{2}\)
Example 5
A hoop of mass 1.0 kg and radius 0.25 m is rotating in a horizontal plane with
an angular momentum of 4.0 kg*m^{2}/s. A lump of clay of mass 0.20 kg is placed
on the hoop. What happens to the angular velocity of the hoop?
Solution:
The moment of inertia for a hoop is \(I = mr^{2}\)
\(I = 1.0kg(0.25m)^{2} = 0.0625 kg * m^{2}\)
The angular momentum L is Iw, so the initial angular velocity can be calculated
from L=Iw.
\(4.0 kg*m^{2}/s = (0.0625kg*m^{2})w_{i}\) or \(w_{i} = 64rad/s\)
Since the lump of clay is placed on the hoop, placing the clay only adds mass to
the hoop. This additional mass adds another moment of inertia \(mr^{2}\) so with
the angular momentum the same, no external torque has acted to change it.
\(4.0kg*m^{2}/2 = [1.0kg(0.25m)^{2} + 0.2kg(0.25m)^{2}]w_{f}\)
\(w_{f} = 53.3rad/s\)
Example 6
An amusement park game consists of a paddle wheel arrangement where you shoot
at the paddles with a pellet gun, thereby turning the wheel. The paddle wheel is
set in motion with an initial angular momentum of 200 kg*m^{2}/s and angular
frequency of 4.0 rad/s. You shoot 8 40 g pellets at a speed of 200 m/s at the
paddles. The pellets hit the paddles at 0.80 m radius, stick, and impart all
their momentum to the wheel. Find the new angular momentum and the new angular
frequency.
Solution:
First, calculate the initial moment of inertia of the paddle wheel from l = Iw.
\(200kg*m^{2}/s = I(4.orad/s)\)
\(I = 50kg*m^{2}\)
Now calculate the w after the wheel has absorbed the momentum in the eight
pellets. The angular momentum of each pellet is the linear momentum times the
0.80 m radius.
\(L_{each-pellet} = mvr = 0.040kg(200m/s)0.80m = 6.4kg*m^{2}/s\)
For eight pellets, the absorbed angular momentum is:
\(L_{all-pellets} = 51.2kg*m^{2}/s\)
The new angular momentum is the original plus \(251.2kg*m^{2}/s\)
The \(I\) also has increased because of the additional mass at the 0.80 m
radius. This additional \(I\) is:
\(I_{pellets} = 8*0.040kg(0.80m)^{2} = 0.205kg*m^{2}\)
The equation for calculating the new angular velocity \(L = IW\) is:
\(251.2kg*m^{2}/s = (50.2kg*m^{2})w_{f}\)
\(w_{f} = 5.00rad/s\)
Example 7
Now consider a 10 kg solid cylindrical drum with radius 1.0 m rotating about its
cylindrical axis under the influence of a force produced by a 30 kg mass
attached to a cord wound around the drum. Calculate the torque, moment of
inertia, and angular acceleration.
Solution:
The moment of inertia for a solid drum rotating about its axis is \(mr^{2}/2\)
\(I = mr^{2}/2 = 10kg(1.0m)^{2}/s = 5.0kg*m^{2}\)
The torque on the drum is the tension in the cable times the radius of the drum.
This tension is much like the tension encountered in the problems of multiple
blocks sliding on tables. The system accelerates clockwise, so the unbalanced
force on M makes it accelerate according to:
\(Mg - F = Ma\) or \(F = \frac{ma}{2}\)
\(\frac{ma}{2} = Mg - Ma\)
\(a = \frac{g}{1+m/2M}\)
\(a = \frac{9.8m/s^{2}}{1+10/60} = 8.4m/s^{2}\)
angular acceleration is: \(a = \frac{a}{r} = \frac{8.4m/s^{2}}{1.0m} =
8.4rad/s^{2}\)
Torque is: \(t = Ia = (5.0kg*m^{2})8.4rad/s^{2} = 42 N*m\)
Example 8
When the 30 kg mass in the previous problem has fallen through 4.0 m, what are
the linear velocity, angular velocity of the drum, and the time for this to
occur?
Solution:
This part of the problem is approached from an energy point of view. The 30 kg
mass falls through a distance of 4.0 m. The work performed by gravity goes into
translational kinetic energy of the 30 kg mass and rotational kinetic energy of
the cylinder. The main difficulty in problems like this is remembering that some
of the work performed results in rotational kinetic energy and some in
translational kinetic energy.
\(Mgh = 1/2Mv^{2} + 1/2Iw^{2} = 1/2Mv^{2} + 1/2
\frac{mr^{2}}{2}\frac{v^{2}}{r^{2}}\)
\(Mgh = v^{2}\frac{M}{2} + \frac{M}{4}\)
\(30kg*\frac{9.8m}{s^{2}}4.0m = v^{2}*(\frac{30kg}{2} + \frac{10kg}{4})\)
\(v^{2} = 67m^{2}/s^{2}\)
\(v = 8.2m/s\)
From v=rw, the angular velocity is:
\(w = v/r = \frac{7.7m/s}{1.0m} = 7.7rad/s\)
The time for this system to reach this velocity is, from w=w + at:
\(t = \frac{w}{a} = \frac{7.7rad/s}{8.4rad/s^{2}} = 0.92s\)
Example 9
A solid sphere is constrained to rotate about a vertical axis passing through
the center of the sphere. A cord is wrapped around what would be the equator,
passing over a pulley of negligible mass, and is attached to a mass that is
allowed to fall under the influence of gravity. Write a conservation of energy
statement for the system.
Solution:
If the mass falls a distance h, then the energy gained by the system is Mgh.
This energy goes into translational kinetic energy of the mass and rotational
kinetic energy of the sphere. A conservation of energy statement is:
\(Mgh = 1/2Iw^{2} + 1/2 Mv^{2}\)
The velocity of a point on the string and the velocity of the mass are the same
and are related to the angular velocity through v = rw, so the energy statement
can be written in terms of v. Additionally, the formula for \(I = 2mr^{2}/5\)
can be substituted, so:
\(Mgh = 1/2I\frac{v^{2}}{r^{2}} + 1/2Mv^{2} = (\frac{M}{5} + \frac{M}{2})v^{2}\)
Example 10
For the previous problem, take the mass of the sphere as m = 2.0 kg, r = 0.30 m,
the hanging mass as M = 0.80 kg, and the height the mass falls through as h =
1.5 m. Find the velocity of the hanging mass and the angular momentum of the
sphere.
Solution:
The velocity of the mass, the rope, and a point on the equator of the sphere are
from the energy statement:
\(0.80kg*\frac{9.8m}{s^{2}}*1.5m = (\frac{2.0kg}{5} + \frac{0.80kg}{2})v^{2}\)
\(v = 3.8m/s\)
The angular velocity of the sphere is:
\(w = \frac{v}{r} = \frac{3.8m/s}{0.30m} = 13 rad/s\)
The angular momentum of the sphere is:
\(L = Iw(\frac{2*2.0kg*0.30m^{2}}{5})*(\frac{13rad}{s} = 0.94kg*m^{2}/s\)