Basics of Physics
Measurements and dealing with their uncertainties are some first skills you learn in a lab. In this article, I will go over what you need to know when dealing with data.
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.