Center of Mass in Physics
These are my notes on the center of mass in physics.
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 \(8x'_{c} = 56+(10-x'_{c})\)
\[x'_{c} = 7.33m\]