Originally Posted by **monomoco** 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.
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 dont really know how to find the average force for part b. |

Angular momentum is conserved in this collision. After the putty sticks on to the disc, the moment of inertia changes and so does angular vel.

$\displaystyle I \omega = constant$

$\displaystyle I_1 \omega_1$ = $\displaystyle I_2 \omega_2$

We have to calculate $\displaystyle I_2$,

It equals $\displaystyle \frac{mr^2}{2} + 0.5 X 2.5^2$

$\displaystyle I_2$ = $\displaystyle 21.125 kg/m^2$

Substituting in the equation, $\displaystyle \omega_2$ is obtained as = $\displaystyle 8.52 rad/s$. Your answer is correct.

For the second part,

Use Impulse, $\displaystyle J = \Delta p = F \Delta t$