CoCalc Shared FilesNotes / Day 8: Angular Momentum / Angular Momentum.ipynb

# Angular Momentum

## Puzzle of the Day

The spool and the block have equal masses and the strings are pulled with equal forces along a frictionless surface. The string is attached to the middle of the block while it is wrapped tightly around the spool (so that it doesn't slip). Which one will win the race?

## Angular Momentum

• angular momentum ($\vec{L}$ or $\vec{\ell}$) is kind of like how hard it is to stop something from spinning
• for a point particle, $\vec{\ell}=\vec{r}\times\vec{p}$, where $\vec{r}$ points from the rotation axis to the particle
• for an extended body, $\vec{L}=I\vec{\omega}$, where $I$ is moment of inertia and $\vec{\omega}$ is angular velocity
• torque $\vec{\Gamma}=\vec{r}\times\vec{F}$ changes angular momentum $\vec{\Gamma}_\text{net}=\frac{d\vec{L}}{dt}$
• if $\vec{\Gamma}_\text{net}=0$, then the angular momentum of the system is conserved
• rotating objects can't be treated as point particles

## Practice

For each of the situations below, indicate for what choice of system the angular momentum will be conserved (if any). Then indicate whether linear momentum will be conserved.

1. A disk is dropped on top of another spinning disk as shown below.
2. A block is dropped on top of a spinning disk.
3. A sticky ball of putty is launched at a hanging pendulum.
4. A baseball spins as it flies through the air.
5. A spinning figure skater brings her arms closer to her body.
6. A bottle rocket is attached to a disk and fires its thrusters.

### Solution

1. The two disks interaction with each other with both normal forces and friction forces. By choosing both disks as the system, those forces are considered internal forces and therefore don't count in calculating $\vec{\Gamma}_\text{net}$. The gravitational force from the earth is external. However, it does not produce a torque because it acts on the center of mass (it doesn't affect the way the disks spin). Linear momentum, by the way, is not conserved (assuming that the bottom disk is sitting on some sort of table, which I didn't include in the above analysis).

2. The answer to this depends on how we drop it. Choosing both the block and the disk as the system does allow us to not count the internal forces between them. However, the gravitational force on the block would cause the left edge of the disk to rotate downward and the right side to rotate upward, which would change the angular momentum of the system (specifically, it would add rotation in a new direction while leaving the old rotation unaffected). If we drop the block on the center of the disk, then that wouldn't happen and angular momentum would be conserved. And just like the previous example, linear momentum is not conserved.

3. Angular momentum is conserved for the combination of the putty and the pendulum. The fulcrum here is the pivot point of the pendulum, and while that does exert a force on the pendulum (it supports it vertically and also keeps it from flying to the right), it doesn't exert a torque on the pendulum (i.e. it doesn't cause a rotation). The gravitational force is also an external force, but that also doesn't produce a torque at the moment of collision for the same reason that you can't open a door by pulling in the direction of the door. It should be noted, however, that after the collision angular momentum will not be conserved because then gravity will start slowing down the rotation. For that reason, we could say that this system is momentarily isolated from external torques. Linear momentum, by the way, is not conserved at all here. This is due to the fact that the pendulum is hinged at the top, which will prevent any net rightward motion of the system. It may seem strange to you that angular momentum is conserved even though the putty wasn't necessarily moving in a circle before the collision. The truth is that even objects moving in a straight line can have angular momentum. Imagine, for example, that there was a tether between the putty and the fulcrum; as the putty gets closer to the pendulum, the tether gets shorter and shorter. In that sense, the putty is "rotating" around the fulcrum, just not with a constant radius.

4. Since gravity is the only force acting on the baseball, and since gravity acts on the center of mass, there is no external torque acting on the baseball. Therefore the angular momentum of the ball alone is conserved. Linear momentum, however, is not conserved (though it would be if we included earth as part of the system as well).

5. The figure skater interacts with the ice and with earth, neither of which exerts a torque (unless you consider the small amount of friction between the skates and the ice). Therefore the angular momentum of the figure skater alone will be conserved. When she brings her arms in, by the way, her angular speed will increase because her moment of inertia decreases. Linear momentum is conserved in this case.

6. This situation is similar to the putty hitting the pendulum, but in reverse. The angular momentum of the disk by itself it not conserved since the external torque created by the rocket causes the angular momentum of the disk to increase. If the rocket (and the expelled bits of fuel) are treated as part of the system, though, then the only external interactions are with the spindle (i.e. the rotation axis) and the earth. Neither of these creates a torque on the disk. It could be argued that the gravitational force on the rocket does create a torque (just think about what would happen if the rocket thrusters were not on). In practice, though, that torque is likely very small in comparison to the thrust of the rocket. The linear momentum of the system is not conserved because the forces on the diagram do not cancel out.

## Practice

The spool and the block have equal masses and the strings are pulled with equal forces along a frictionless surface. The string is attached to the middle of the block while it is wrapped tightly around the spool (so that it doesn't slip). Which one will win the race?

## Solution

Intuitively it may seem like the block should win because some of the pulling force of the spool goes into rotating the spool instead of moving its center of mass. To answer the question definitely, let's look at force diagrams for each (as viewed from above).

The spool will spin because the force from the string applies a torque about the center of mass of the spool. The block will not spin. Linear motion however is only determined by the net force, which is the same for the spool and the block. Given that the net force acting on each is the same, and given that they have the same mass, the center of mass of each will move in exactly the same way. They will tie.

If you guessed that the block would win because some of the force would go into rotating the spool, you are partially correct. It's not the force that gets divided between the linear motion and the rotational motion (after all, how do you divide an interaction?) but rather the energy. Where does the extra energy come from? As the string is pulled, the spool will unravel. So even though the block and the spool end up traveling the same distance, the string pulling the spool has to be pulled farther (which is where the extra energy comes from).

## Practice

A ball of putty is thrown horizontally at a hanging rod that hangs from its end as a pendulum. The putty sticks to the end of rod and the two rotate together. Determine an expression for the angular speed immediately after the collision. (The moment of inertia of a rod about its end point is $I=\frac{1}{3}mL^2$, where $L$ is its length.)

## Solution

### Assumptions

• the putty is a point particle with mass $m_p$, and it travels horizontally with velocity $\vec{v}$ before the collision
• the pendulum rod has length $L$ and mass $m_r$
• the rod and the putty are momentarily isolated during the collision.
• after the collision, the putty and rod rotate together with angular speed $\omega$

### Analysis

The putty+pendulum system is isolated from external torques, so the angular momentum of the system stays the same.

$\vec{r}\times\vec{p}=I\vec{\omega}$

Since we are treating the putty as a point particle, we use $\vec{\ell}=\vec{r}\times\vec{p}$ to describe the angular momentum before the collision. From the right-hand rule for cross products, we also find that the angular momentum points out of the page (the +z direction). The right-hand rule for rotation says that if you curl your fingers in the diretion of rotation, your thumb points in the direction of the angular velocity. In this case, that is the +z direction.

Recall that $\left | \vec{a}\times\vec{b}\right |=ab\sin\phi$ for a cross product, where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$. This form of the cross product is convenient to use in a lot of situations (like this one).

$rm_pv\sin\theta=\left (\frac{1}{3}m_rL^2+m_pL^2\right )\omega$

Notice that $r\sin\theta=L$ no matter the value of $\theta$. Here I have also used the fact that the moment of inertia of a point particle is $I=mr^2$, where $r$ is the distance of the point particle from the rotation point.

$m_pvL=\left (\frac{1}{3}m_rL^2+m_pL^2\right )\omega$

$\omega=\frac{m_pvL}{\frac{1}{3}m_rL^2+m_pL^2}=\frac{m_pv}{\frac{1}{3}m_rL+m_pL}$

### Check

The SI units of the final answer are $\frac{[kg][m/s]}{[kg][m]}=\frac{1}{[s]}$, which is what we expect for an angular speed.

Let's check the limits where $m_r\rightarrow \infty$ and $m_p\rightarrow \infty$. In the former case, we would expect that $\omega=0$ because the rod wouldn't move. In the latter case the rod shouldn't slow down the putty at all.

$\lim_\limits{m_r\to\infty}\frac{m_pv}{\frac{1}{3}m_rL+m_pL}=\frac{m_pv}{\infty}=0$

$\lim_\limits{m_p\to\infty}\frac{m_pv}{\frac{1}{3}m_rL+m_pL}=\frac{\cancel{m_p}v}{\cancel{m_p}L}=\frac{v}{L}$

The latter result is a little harder to interpret, but it is what the angular speed of the putty was just before the collision (tangential speed is $v_t=\omega r$).

### Interpretation

Unsurprisingly, the angular speed increases with $v$ and decreases with $L$ and $m_r$.

## Practice

According to Kepler's Second Law, an orbiting object will sweep out equal areas in equal times as shown below. Prove this. (Hint, show that it is true for an infinitesimal displacement along the orbit.)

## Solution

As the hint suggests, let's look at infinitesimal displacements (and therefore infinitesimal time intervals) along the arc. If the displacement is infinitesimal, then the shaded areas will basically be isosceles triangles where the height of the triangle is $r$ and the base of the triangle is $vdt$. We are looking to prove that

$\frac{1}{2}r_1(v_1dt)\overset{?}{=}\frac{1}{2}r_2(v_2dt)$,

where subscripts indicate different triangles. After canceling out terms, we get

$r_1v_1=r_2v_2$.

Where do we proceed from here? The system consisting of the orbiting object and the sun is isolated, so the angular momentum stays the same. If we assume that the orbiting object is much smaller than the sun, then the sun will not move. Therefore

$\vec{\ell}_1=\vec{\ell}_2$

$\vec{r}_1\times\vec{p}_1=\vec{r}_2\times\vec{p}_2$

$\vec{r}_1\times \cancel{m}\vec{v}_1=\vec{r}_2\times \cancel{m}\vec{v}_2$

$\vec{r}_1\times \vec{v}_1=\vec{r}_2\times \vec{v_2}$.

For a triangle with an infinitesimally small base, $\vec{r}\perp\vec{v}$, so if we take the magnitude of both sides we get

$\left |\vec{r}_1\times \vec{v}_1\right | = \left | \vec{r}_2\times\vec{v}_2\right |$

$r_1v_1\sin 90^\circ=r_2v_2\sin 90^\circ$

$r_1v_1=r_2v_2$,

which is equivalent to the statement that the areas are the same. Even though we proved this for infinitesimal displacements, its still true for larger displacements because the larger displacements are made from many infinitesimal displacements.

## Practice

A cylinder rolls down a ramp without slipping. Determine a mathematical model for the motion (both linear and rotational). (Hint: $I_\text{cylinder}=\frac{1}{2}mR^2$)

## Solution

### Assumptions

• the cylinder has mass $m$ and radius $R$
• the ramp makes an angle $\theta$ with the horizontal
• static friction is the only friction acting on the cylinder

### Diagrams

The friction is static rather than kinetic because the cylinder is not slipping.

### Analysis

According to the force diagram,

$F_{ec}^g\sin\theta-F_{rc}^s=m\ddot{x}$

and

$F_{rc}^n-F_{ec}^g\cos\theta=m\cancelto{0}{\ddot{y}}$.

These simplify to

$mg\sin\theta-F_{rc}^s=m\ddot{x}$

and

$F_{rc}^n=F_{ec}^g\cos\theta=mg\cos\theta$.

In many cases we would use the normal force to determine the friction, but we cannot do that here. Recall that $F^s\le \mu_sF^n$ for static friction, and we have no way of knowing in what part of that range the cylinder is. In order to determine $F_{rc}^s$, we can try to look at the rotational version of Newton's Second Law, $\vec{\Gamma}=\frac{d\vec{L}}{dt}$. Of the three forces acting on the cylinder, only the friction force exerts a torque.

$\vec{R}\times \vec{F}_{rc}^s=\frac{d\vec{L}}{dt}$

Take the magnitude of both sides.

$RF_{rc}^s\sin 90^\circ=\frac{d}{dt}\left (I\omega \right )$

$RF_{rc}^s=I\frac{d\omega}{dt}$

$F_{rc}^s=\frac{I}{R}\frac{d\omega}{dt}$

Substitute this into Newton's Second Law in the x-direction.

$mg\sin\theta-\frac{I}{R}\frac{d\omega}{dt}=m\ddot{x}$

How does this help? Recall from introductory mechanics that $\frac{d\omega}{dt}=\alpha$ and $\alpha=\frac{a_{cm}}{R}$ (the latter only applies to an object that rolls without slipping).

$mg\sin\theta-\frac{I\ddot{x}}{R^2}=m\ddot{x}$

$mg\sin\theta=m\ddot{x}+\frac{I\ddot{x}}{R^2}=\ddot{x}\left (m+\frac{I}{R^2}\right )$

$\ddot{x}=\frac{mg\sin\theta}{m+\frac{I}{R^2}}$

For a cylinder, $I=\frac{1}{2}mR^2$.

$\ddot{x}=\frac{\cancel{m}g\sin\theta}{\cancel{m}+\frac{1}{2}\frac{\cancel{m}\cancel{R^2}}{\cancel{R^2}}}=\frac{2}{3}g\sin\theta$

We integrate to find $v(t)$ and $x(t)$.

$\ddot{x}=\frac{dv}{dt}=\frac{2}{3}g\sin\theta \Rightarrow \int_\limits{v_0}^{v(t)}dv=\int_\limits{0}^t\frac{2}{3}g\sin\theta dt'$

$v(t)-v_0=\frac{2}{3}g\sin\theta t$

$v(t)=\frac{2}{3}g\sin\theta t+v_0$

$v(t)=\frac{dx}{dt}=\frac{2}{3}g\sin\theta t+v_0\Rightarrow \int_\limits{x_0}^{x(t)}dx=\int_\limits{0}^t(\frac{2}{3}g\sin\theta t'+v_0)dt'$

$x(t)-x_0=\frac{1}{3}g\sin\theta t^2+v_0t$

$x(t)=\frac{1}{3}g\sin\theta t^2+v_0t+x_0$

We can use these to find the corresponding rotational motion.

$\alpha=\frac{\ddot{x}}{R}=\frac{\frac{2}{3}g\sin\theta}{R}$

$\alpha=\frac{d\omega}{dt}=\frac{\frac{2}{3}g\sin\theta}{R}\Rightarrow \int_\limits{\omega_0}^{\omega (t)}d\omega =\int_\limits{0}^t\frac{\frac{2}{3}g\sin\theta}{R}dt'$

$\omega (t)-\omega_0=\frac{\frac{2}{3}g\sin\theta}{R}t$

$\omega (t)=\frac{\frac{2}{3}g\sin\theta}{R}t+\omega_0$

$\omega (t)=\frac{d\phi}{dt}=\frac{\frac{2}{3}g\sin\theta}{R}t+\omega_0\Rightarrow \int_\limits{\phi_0}^{\phi (t)}d\phi = \int_\limits{0}^t\left (\frac{\frac{2}{3}g\sin\theta}{R}t'+\omega_0\right )dt'$

$\phi (t)-\phi_0=\frac{g\sin\theta}{3R} t^2+\omega_0t$

$\phi (t)=\frac{g\sin\theta}{3R} t^2+\omega_0t+\phi_0$

### Check

The SI units of $x(t)$, $v(t)$, and $a(t)$ work out to be $[m]$, $[m/s]$, and $[m/s^2]$ (respectively) as we expect. The SI units of $\alpha (t)$ are $\frac{[m/s^2]}{[m]}=\frac{1}{[s^2]}$, which is what we expect for an angular acceleration. The SI units of $\omega (t)$ work out to be $\frac{[m/s^2]}{[m]}[s]=\frac{1}{[s]}$, which is what we expect for an angular speed. the SI units of $\phi (t)$ work out to be $\frac{[m/s^2]}{[m]}[s]^2=\text{ no units}$, which is what we expect for an angle in radians.

In the limit that $\theta\rightarrow 0$, we would expect that there is no acceleration at all.

$\lim_\limits{\theta\to 0} x(t)=0+v_0t+x_0=v_0t+x_0$

$\lim_\limits{\theta\to 0}\phi (t)=0+\omega_0t+\phi_0=\omega_0t+\phi_0$

Both of these results are what we expected. The limit $\theta\rightarrow 90^\circ$ is a little trickier. You might think that the result should be free fall, but that isn't quite the case because of the friction. It's actually similar to a yo-yo, which falls differently that a freely falling object.

We could also go back and see what happens when we set $I=0$, which would correspond to a non-rotating point particle. If we do that, we get $a=g\sin\theta$, which is the same result you get for a point particle sliding down a ramp without friction.

### Interpretation

The linear acceleration of a rolling object is smaller than that of an object that is sliding without friction, which makes sense.

Realistically our result would likely break down at a high enough angle because the cylinder would begin to slide instead of roll.

In [1]:
from __future__ import division, print_function
from vpython import *

#initial conditions
L=5 #length of ramp
theta=10*pi/180 #angle of ramp in radians
h=L*sin(theta) #height of ramp
m=1 #mass of cylinder
I=0.5*m*R**2 #moment of inertia of cylinder
x=0 #tilted x-coordinate of cylinder
y=0 #tilted y-coordinate of cylinder
v=0 #initial velocity of cylinder
t=0
phi=0 #initial angle of cylinder (rotation)
omega=v/R #initial angular speed of cylinder
g=9.8 #gravitational field strength
dt=0.01 #time step

#set the scene
scene=canvas(center=vec(0.5*L*cos(theta),0.5*h,0))
#draw the ramp
tri=[[0,0],[0,L*sin(theta)],[L*cos(theta),0],[0,0]]
tripath=[vec(0,0,1),vec(0,0,-1)]
ramp=extrusion(path=tripath, shape=tri)
#draw the cylinder

while x<=L: #continue until cylinder reaches the end of the ramp
rate(1/dt) #make sure looping rate does not exceed real time
a=g*sin(theta)/(m+I/R**2) #calculate acceleration
v=v+a*dt #update speed
x=x+v*dt #update tilted x-coordinate
#perform the coordinate transformation to turn tilted coordinate system into vpython coordinate system
cyl.pos.x=x*cos(theta)
cyl.pos.y=h+R-x*sin(theta)
#angular motion
alpha=a/R #angular acceleration
omega=omega+alpha*dt #update angular speed
cyl.rotate(angle=-omega*dt) #rotate the cylinder (I put in the negative because the cylinder rotated the wrong way otherwise)
t=t+dt #increment the time


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In [ ]: