Conservation of Angular Momentum

[veronicak5678]

Homework Statement

A 500 gram piece of putty falls straight down and sticks to a 4 kg rotating solid disk below. The uniform disk has a radius of 3 meters, and the putty hits the disk 2.5 meters from its center. You may neglect all forces other than that of the collision.
a- If the disk was initially rotating at 10.0 rad/s, find its final angular velocity.
b- The putty takes 15 ms to collide with the disk. Calculate the avergae force on the putty.

Homework Equations

I = mr^2 and .5 mr^2, L = I omega

The Attempt at a Solution

For part a,
I found moment of inertia of the disk (I_d) = 1/2 * 4kg* (3m) ^2 =
18 kg m^2
moment of inertia of putty = .5kg*(2.5m)^2 = 3.13 kgm^2
Since the putty falls straight down, i assumed the initial angular momentum of the putty was 0. So the initial angular momentum of the system is L_d = 18kgm^2*10 rad/s) = 180 kgm^2/s
Lf = Li, so (I_p + I_d) * angular velocity = 180, angular velocity = 8.52 m/s.

I'm really unsure about this, so I just wanted someone to doublecheck it. Also, I don't really know how to find the average force for part b.

 

[naresh]

Part (a) is right, except angular velocity isn't m/s.

Hint for part (b): Look up the impulse-momentum theorem

 

[veronicak5678]

I looked up the impulse-momentum theorem, and I see that force = change in velocity over time. How can I apply this if I don't know the initial velocity of the putty? Can I use angular velocity and say vf - vi = 8.52 rad/s- 0? So force on putty would just be 8.52/.015 seconds?

 

[naresh]

1. That is not an accurate statement of the theorem. Re-look at it.
https://www.physicsforums.com/library.php?do=view_item&itemid=53

2. No, you cannot simply use the angular velocity. You have to convert it into a linear velocity. How?

You do not know the vertical velocity, so you are right, you cannot say anything about the vertical force that acts on the putty.

 

[veronicak5678]

I converted the final angular to linear using omega = v/r
8.52 = v/ 2.5
v= 3.41

Since the putty is dropped straight down, can I say the initial velocity in the horizontal was 0 and the final is 3.41 m/s?

 

[naresh]

Answer to your question: Yes.

Whether 8.52 = 3.41 / 2.5: No.

 

[veronicak5678]

OK, so now I have vf = 8.52rad/s * 2.5 m = 21.3 m/s

p initial = 0
p final = mvfinal = 21.3 m/s(.5kg) = 10.7 kg m/s

pf - pi = 10.7 kg m/s = J

Force avg = J/ time = 107. kg m/s / .015 s = 713 N

Please let me know if I have done this correctly.