A thin circular hoop of mass 1.0 kg and radius 2.0 m is rotating about an axis through its center and perpendicular to its plane. It is slowing down at the rate of 7.0 rad/s². The net torque acting on it is: A) 7.0 N. m B) 14.0 N . m C) 28.0 N . m D) 44.0 N . m E) none of these

Question

Quizplus · Accepted Answer

The rotational equivalent of Newton's second law is τ = Iα, where τ is the net torque, I is the moment of inertia, and α is the angular acceleration. Rearranging this equation gives τ = I(α/t), where t is the time it takes for the angular velocity (ω) to change by α. In this case, ω is decreasing at a rate of 7.0 rad/s, so we have α = -7.0 rad/s^2.

The moment of inertia of a thin circular hoop is I = MR^2, where M is the mass and R is the radius. Substituting the given values, we get I = (1.0 kg)(2.0 m)^2 = 4.0 kg*m^2.

Finally, we can calculate the net torque by multiplying I and α: τ = (4.0 kg*m^2)(-7.0 rad/s^2) = -28.0 N*m = 28.0 N*m in the clockwise direction. Therefore, the correct answer is C.

A Thin Circular Hoop of Mass 1