question -rotational inertia

Question

I did:

$v = 0.067 m$ ; $m = 0520$ kg ; I = $\frac{1}{2}mr^2$

Then, $I = \frac{1}{2}0.520kg *0.0620m^2$

Then the answer is: $9.99 * 10^{-4} kg · m^2$

Is that right?

Answer 1

The moment of inertia of a hollow cylindrical shell of radius $0.0620$ of mass $0.520,$ whose mass is all on the curved surface of the cylinder, is $9.99 \times 10^{-4}.$

But you were not told that the object rolling down the incline is a cylinder whose mass is all on its curved surface. The mass might all be there, or it might mostly be in the center of the object, or it might be distributed in any other manner throughout the object.

The outer radius of the object and its mass are simply not enough information to tell you its rotational inertia.

Notice that you did not use the velocity of the object at all, nor any information about the incline.

There are a couple of ways to approach this problem:

• By considering the force accelerating the object down the incline, the rate of acceleration of the object (both the linear acceleration and the rotational acceleration), and the forces required to produce both of those accelerations (which must add up to the force exerted). If you set up the relevant equation correctly, the rotational inertia will be the only unknown; solve for it.
• By considering conservation of energy. The object starts from zero linear and rotational velocity, and after descending some vertical distance, has traded some potential energy for kinetic energy. If you set up the energy equation correctly, the rotational inertia will be the only unknown; solve for it.

In either case you will need to use the slope of the incline and the graph of the speed in order to find the necessary data.