# Spinning spokes and clay

1. Apr 26, 2015

### Westin

1. The problem statement, all variables and given/known data

A device consists of eight balls each of mass 0.78kg attached to the ends of low-mass spokes of length 2.2m, so that the radius of rotation of the balls is 1.1m. The device is mounted in the vertical plane. The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with mass 0.4kg falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is at 45 to the horizontal. Just before the impact the clay has a speed 7m s, and the wheel is rotating counterclockwise with angular speed 0.4rad s.

1)Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device?

Select all that are True.
a)The angular momentum of the device and clay system just after the collision is equal to the angular momentum of the device and clay system just before the collision.
b)The angular momentum of the falling clay is zero because the clay is moving in a straight line.
c)The angular momentum of the device is the same before and after the collision
d)Just before the collision the angular momentum of the wheel is zero.
e) The angular momentum of the device is the sum of the angular momenta of all eight balls.

2) Just before the impact, what is the angular momentum of the combined system of device plus clay about the center? (As usual, x is to the right, y is up, and z is out of the page.)

Li = , ,

3) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center?

Lf = , ,

4) Just after impact, what is the angular velocity of this device?

f = , ,

2. Relevant equations

L = mvr
L = I w
w = L/I

3. The attempt at a solution

For #1, I believe the answer is a) and e)

For #2, I tried Initial angular momentum of the system is the sum of the disk and clay.
L Disk = IW = .5MR^2*W (you know this quantity will be subtracted in the end since it points in the -z direction via RightHandRule)
L Clay = r x p = R(mv)
Li = Lc - Ld

For #3 Since there is no net torque acting on the system, the angular momentum after will be the same as before.

For #4 Final angular velocity: W = L/I. You need to update your I though, since the system is now Disk+clay
I final = .5MR^2 + mR^2

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2. Apr 26, 2015

### BvU

Hello Westin,

Is there a question you want answered ?

If not, I have some comments:

You have no relevant equation for I . You mention $L_{disc} = I\vec\omega$ and then treat the balls as a disc. No good!

$\vec L_{clay} = \vec r \times \vec p$ is good, Rmv is not correct

Something is going wrong with the signs of your $L$.

$\vec L_i = \vec L_{balls} + \vec L_{clay}$, not $-$

And a small detail for 1e: this is correct, provided you take the same distance between axis of rotation and center of ball.
Reason I bring it up is that usually (e.g. in tables) the moment of inertia around an axis through the center is given.​

3. Apr 26, 2015

### Westin

Is the first part right with a) and e) being true? and also what do i do for #2 instead?

4. Apr 26, 2015

### haruspex

True, but $\vec L_{balls}$ is negative, so the following line is OK (except for the error in Idevice that you already noted).
Not sure what you're saying there. Are you pointing out that unless the balls are small compared with the spoke length the parallel axis theorem should be used?

5. Apr 26, 2015

### haruspex

First, get the I correct for the device. It is not a uniform disk.

6. Apr 26, 2015

### Westin

is it a rod? My homework is due at 8pm..

7. Apr 26, 2015

### haruspex

You answered correctly that e) is true. What is the moment of inertia for one spoke+ball? Note this: