# Homework Help: Angular momentum and velocty

1. Apr 3, 2009

### guitarman

1. The problem statement, all variables and given/known data
A rotating uniform-density disk of radius 0.7 m is mounted in the vertical plane. The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 3.4 kg. A lump of clay with mass 0.5 kg falls and sticks to the outer edge of the wheel at the location < -0.455, 0.532, 0 > m. Just before the impact the clay has a speed 6 m/s, and the disk is rotating clockwise with angular speed 0.82 radians/s.

(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C? (As usual, x is to the right, y is up, and z is out of the screen, toward you.)
C,i = < , , > kg · m2/s
(b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the center C?
C,f = < , , > kg · m2/s
(c) Just after the impact, what is the angular velocity of the wheel?
f = < , , > radians/s

2. Relevant equations

for parts a and b, I know I should be using Rcos(theta)mv
for c I should use ((Rcos(theta)mv)/((M+m)R^2)

3. The attempt at a solution

Firstly, I found that the hypotenuse was equal to 0.7,
then I did cos^-1(-0.455/0.7) and obtained that the angle is 130.542 degrees.
So for part a I plugged in 0.7 m*cos(130.542)*0.5 kg*7 m/s and obtained -1.37 kg *m^2/s, which is apparently wrong. Can someone please let me know what I am doing wrong? Thanks in advance!

2. Apr 3, 2009

### Staff: Mentor

That would give you the angular momentum of the clay, but not of the system. What about the rotating disk?

3. Apr 3, 2009

### guitarman

So would I want to do Rcos(theta)*(m+M)*(v+V) so as to take into account both masses and velocities? And if so, do I convert both speeds to m/s or radians/s?

4. Apr 3, 2009

### LowlyPion

You have 2 expressions for angular momentum don't you?

You have the disk:

L = I * ω where ω is clockwise or <-z>

Then you have the angular momentum about C of the clay ball:

L = r X P

where P is the momentum vector of the clay ball <-y> and the projection of the r which is directed <-x>.
That means that this cross product is <-x> X <-y> = <z> which is counter-clockwise.

By taking the projection of the momentum to the ⊥ of the -x axis, you don't even need to use the angle. Just use the |r| as being |x|.
Of course you need to use r as .7 for figuring the I later on.

When they are combined you need to consider M and m as part of figuring the combined I.

Last edited: Apr 3, 2009