# Homework Help: Angular Momentum of clay ball

1. Dec 17, 2006

### lizzyb

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

A wad of sticky clay (mass m, velocity v), is fired at a solid cylinder (mass M, radius R). The cylinder is initially at rest and rotates on an axis through the center of mass. The line of motion of the projectile is perpendicular to the axis of rotation and at a distance d < R from the center.

Find the angular speed of the cylinder just as the clay hits the surface.

2. Relevant equations

Conservation of Angular Momentum

3. The attempt at a solution

I did:

$$L_i = d m v_i$$
$$L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2$$
$$L_i = L_f$$
$$d m v_i = d m r \omega + \frac{1}{2} M r^2 \omega$$
$$\omega = \frac{d m v_i}{(d m r + \frac{M r^2}{2})}$$

Which isn't correct; it may have something to do with the momentum of the wad of sticky clay; the reason I used L = m v d is because d would be the moment arm but perhaps I'm confusing this with torque? What value of r should I use when the force isn't perpindicular?

2. Dec 17, 2006

### OlderDan

You lost an ω in your second equation, so I added it in the quote. Why do you think it is wrong? Could it be that changing r to R is causing a problem?

Last edited: Dec 17, 2006
3. Dec 17, 2006

### lizzyb

The solution in the back of the book is:
$$\omega = \frac{2 m v_i d}{(M + 2m) R^2}$$
so using what I did:
$$\omega = \frac{d m v_i}{(dmr + \frac{M r^2}{2})} = \frac{2 m v_i d}{ 2 d m r + M r^2 }$$
which is close but not it.

I don't think changing R to r makes a difference (let r = R).

But there is a relationship between R and d, viz. $$\cos \theta = \frac{d}{r}$$ but while that will take care of the r^2 it adds the cos.

Last edited: Dec 17, 2006
4. Dec 17, 2006

### OlderDan

OOPs.. I missed it. Sorry. They are correct. In your second equation on the right hand side, the distance between the line of the velocity and the axis for the clay is R, not d.

I better go get some coffee.:yuck:

Last edited: Dec 17, 2006
5. Dec 17, 2006

### lizzyb

How do you figure?

$$L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2 \omega$$

The clay is coming in on the top part of the cylinder yet below the cylinder's radius, at a height d, which would be the distance between the velocity vector and the cylinder's axis.

oh wait, you mean because this is after the clay has stuck to the cylinder that the initial velocity vector has changed? In which case the distance between the line of velocity is r and not d as before?

$$d m v_i = m r^2 \omega + \frac{1}{2} M r^2 \omega$$
$$\omega r^2(m + \frac{1}{2} M) = \omega \frac{r^2}{2}(2m + M) = d m v_i$$
$$\omega = \frac{2 d m v_i}{(2 m + M) r^2}$$

Last edited: Dec 17, 2006
6. Dec 17, 2006

### OlderDan

That's it. The clay cannot penetrate the cylinder, so it has to abrubtly change direction at impact to move tangent to the cylinder.

7. Dec 17, 2006

### lizzyb

ok, that makes sense! thanks (again)!