# Torque/Rotational Motion-How many revolutions will an object make before stopping?

## Homework Statement

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.

## Homework Equations

Torque = I*Angular Acceleration

I=0.1 *M* L^2

$$\omega$$ = V/r

## The Attempt at a Solution

Okay, i know that if the rod sweeps out an area equal to 0.25 * $$\pi$$ 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) * $$\pi$$ 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 :)

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