# Conservation of angular momentum pre lab help

1. Apr 9, 2012

### Triathlete

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

4) In order to determine the moment of inertia I of a rotating platform, a string is wrapped
around a spool of radius r = 2.0 cm beneath the platform. The string is then fed over a
pulley with a hanging mass attached to its end. The hanging mass is then released from rest, and its linear acceleration is measured.

A.) If the hanging mass is M = 100 g, and its linear acceleration is a = 2.5 m/s2, what is the moment of inertia I of the rotating platform?

B.) Using the same rotating platform as in problem 1, a ball of mass m = 50 g is launched
into the catcher on top of the platform. After the ball is caught by the catcher, the angular velocity of the system is ω = 2.2 rad/s. If the catcher is R = 20 cm away from the axis of rotation of the platform, what is the linear velocity v of the ball before it is caught?

2. Relevant equations

FT = M(g-a)

α = a/r

I = (rFT)/α

vo = ($\omega$r2M(g-a))/amR

3. The attempt at a solution

For part a, I used the first two equations to solve for the tension force and angular acceleration, then plugged the values into the third equation to solve for inertia. The answer I got was 1.17x10-4 kgm2 (If you could verify this, that would be great!

For part b I am not sure where to begin, because there are too many unknowns. I can't figure out a way to combine any of the equations to solve for any of the unknowns either.

2. Apr 9, 2012

### Staff: Mentor

Your result for part (a) looks good.

For part (b), think of the problem in terms of an inelastic collision taking place, and in this case you're dealing with angular motion so conservation of angular momentum applies.

3. Apr 10, 2012

### Triathlete

I figured as much, but i'm confused on how exactly to apply it since I don't have a value for inertia or final or initial momentum. Since Linitial = Lfinal = Iω

4. Apr 10, 2012

### Staff: Mentor

You've got the moment of inertia of the platform from (a) and you've got the final angular velocity of the combined platform and ball. The ball adds its moment of inertia to that of the platform when it's "caught" (small ball sitting 20cm from the axis of rotation...). So what's the total angular momentum? Where did the angular momentum come from before the collision?

5. Apr 10, 2012

### Triathlete

Oh okay. So I used I=mR to get the inertia of the ball, then added it to part a to get the inertia of the system. To get the momentum, I just multiplied angular velocity by inertia, then plugged it into v = L/mR to get initial velocity of the ball. I got 0.466 m/s, but I think I'm doing something wrong.

6. Apr 10, 2012

### Staff: Mentor

Moment of inertia of a point mass m at a distance r from the axis of rotation is $I = m r^2$. Note that the distance is squared

7. Apr 10, 2012

### Triathlete

Oops I forgot to add the squared on here but I did that in my calculations.

8. Apr 10, 2012

### Staff: Mentor

So the question is then, why do you think you're doing something wrong? It seems that your combined moment of inertia is fine, and thus the final angular momentum (Iω) should be, too. You've also got the right idea about the initial angular momentum and the ball's speed.

Of course, the problem doesn't state what the trajectory of the ball is... we've just assumed it's a straight line that intersects the platform tangentially.

9. Apr 10, 2012

### Triathlete

I guess I'm just not very sure of myself. Thanks so much for your help!