Physics Help Forum Conservation of Angular Momentum

 Nov 24th 2008, 05:52 PM #1 Junior Member   Join Date: Sep 2008 Posts: 7 Conservation of Angular Momentum 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.
Mar 15th 2009, 10:45 PM   #2
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Join Date: Feb 2009
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 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, $\displaystyle J = \Delta p = F \Delta t$

 Tags angular, conservation, momentum

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