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/s,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 angular momentum equation is L = Iω, where I is the moment of inertia and ω is the angular velocity. Differentiating this equation with respect to time, we get:

dL/dt = d(Iω)/dt

dL/dt = I(dω/dt) + ω(dI/dt)

We know that dL/dt = 25.0 kg m^2/s^2, I = 1/2MR^2, M = 10 kg, R = 0.3 m. Also, dI/dt = 0 since the mass and radius of the cylinder are constant.

Substituting these values, we get:

25.0 = (1/2)(10)(0.3)^2(dω/dt)

dω/dt = 55.6 rad/s^2, which is the angular acceleration.

Therefore, the answer is C.

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

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