Solving for Net Torque on Uniform Mass Density Rod

[taloz]

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 :)

 

[anathema]

Hi there! I would approach this problem by first understanding the concept of moment of inertia and how it relates to the distribution of mass in an object. In this case, the rod has a uniform mass density, which means that the mass is evenly distributed along the length of the rod.

To solve for the radius, you can use the formula for moment of inertia, I = mr^2, where m is the mass and r is the radius. Since the rod has a uniform mass density, you can rewrite the formula as I = (mass density * length) * r^2.

Now, we can plug in the given values to solve for r. We know that the mass density is 0.05 kg/cm, and the length of the rod is not given, so we can use a variable, L, to represent it. Therefore, our formula becomes I = 0.05 * L * r^2.

Next, we can plug in the given value for the moment of inertia, 0.1ML^2, and solve for r. This gives us 0.1ML^2 = 0.05 * L * r^2. Simplifying, we get r = sqrt(0.2L).

Now that we have the radius, we can use the angular velocity formula, \omega = V/r, to solve for the angular velocity, which is equal to the linear velocity divided by the radius. Plugging in the given value for linear velocity, (1/32) * \pi m/s, and the calculated value for r, sqrt(0.2L), we get \omega = (1/32) * \pi / sqrt(0.2L).

Finally, we can use the torque formula, Torque = I * Angular Acceleration, to solve for the angular acceleration. Plugging in the given value for torque, -1/16 Nm, and the calculated value for moment of inertia, 0.1 * M * L^2, we get -1/16 = (0.1 * M * L^2) * Angular Acceleration.

Now, we can solve for the angular acceleration, which will give us the rate at which the rod is slowing down. Once we have this, we can use the formula for angular velocity, \omega = \omega0 + \alpha * t, to find the final angular velocity at the point when the rod comes to