A 10 kg solid cylinder (I = 1/2 MR²) with a radius of 30.0 cm is rotating about a vertical axis through its center. If the angular momentum is increasing at the rate of 25.0 kg m²/s², then what is the angular acceleration? A) 75.3 rad/s² B) 65.9 rad/s² C) 55.6 rad/s² D) 40.5 rad/s² E) 35.2 rad/s²

Question

Quizplus · Accepted Answer

The formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular velocity. Taking the time derivative of both sides, we get dL/dt = I(dω/dt) + ω(dI/dt).

Since the cylinder is a solid object rotating around a central axis, we can use the formula for the moment of inertia of a solid cylinder to find I: I = (1/2)MR^2.

We are given that M = 10 kg, R = 0.3 m, and dL/dt = 25 kg m^2/s. Taking the derivative of I with respect to time, we get dI/dt = 0 since M and R are constant. Therefore, we can solve for the angular acceleration (dω/dt) as follows:

dL/dt = I(dω/dt)
25 kg m^2/s = (1/2)(10 kg)(0.3 m)^2(dω/dt)
dω/dt = 55.6 rad/s^2

Thus, the answer is choice (C), with an angular acceleration of 55.6 rad/s^2.

A 10 Kg Solid Cylinder (I = 1/2 MR2) with a Radius

A 10 Kg Solid Cylinder (I = 1/2 MR²) with a Radius