A rod has a uniform mass density of 0.05 kg/cm it has a moment of inertia equal to 0.1ML^2. The rod sweeps out an area equal to 0.25* pi meters squared in 4 seconds. If you apply a net torque of -1/16 Nm to the rotating rod, how many revolutions will the rod make before coming to rest.
As i don't have a scanner on me i'll do my best to describe the rod. It's rotating about it's center in a counterclockwise direction. Therefore the -1/16 Nm should slow it down.
Torque = I*Angular Acceleration
I=0.1 *M* L^2
[tex]\omega[/tex] = V/r
The Attempt at a Solution
Okay, i know that if the rod sweeps out an area equal to 0.25 * [tex]\pi[/tex] in 4 seconds, that each rod is sweeping an area of half of that. Therefore, the linear velocity of the system right before the application of torque is (1/32) * [tex]\pi[/tex] m/s.
I know that i have to solve for the radius of the rod now if order to convert linear velocity into angular velocity, and to solve for the moment of inertia so that i can find the angular acceleration. However, my teacher has never given us a problem with uniform mass density before, so i don't understand how to use it in order to solve for this radius.
If anyone can help that would be great :)