Question

Two solid uniform cylinders are connected with a belt wrapped around both cylinders, as shown in the figure below. The mass and radius of the smallest cylinder is mA and rA, respectively, and the mass and radius of the largest cylinder is mB and rB, respectively. When the cylinders are rotating, what is the ratio of the kinetic energy of the largest cylinder to the kinetic energy of the smallest cylinder?

Answer 1

The kinetic energy of the smallest cylinder is KEA = (1/2)IAwA2, where IA is the rotational inertia about the center, and wA is the angular velocity. Now, IA = (1/2)mArA2. Similarly, for the largest cylinder, the kinetic energy is KEB = (1/2)IBwB2. Now, IB = (1/2)mBrB2. Since the two cylinders are connected with a belt, their linear velocities are equal, which means that wArA = wBrB. The desired kinetic energy ratio is KEB/KEA. After substituting the above relations, and simplifying, the kinetic energy ratio becomes KEB/KEA = mB/mA.

Answer 2

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