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 N∙m C)28 N∙m D)44 N∙m E)none of these

Question

Quizplus · Accepted Answer

The net torque acting on the hoop can be calculated using the formula:
τ = Iα
where τ is the net torque, I is the moment of inertia, and α is the angular acceleration.
The moment of inertia of a thin circular hoop is given by:
I = MR²
where M is the mass of the hoop and R is its radius.
Substituting the given values, we get:
I = (1.0 kg)(2.0 m)² = 4.0 kg∙m²
The angular acceleration can be calculated using the formula:
α = Δω/Δt
where Δω is the change in angular velocity and Δt is the time taken for this change.
Substituting the given values, we get:
α = (-7.0 rad/s)/(1 s) = -7.0 rad/s² (Note that the negative sign indicates that the hoop is slowing down.)
Now, we can calculate the net torque as:
τ = Iα = (4.0 kg∙m²)(-7.0 rad/s²) = -28 N∙m
(Note that the negative sign indicates that the torque is in the opposite direction to the initial rotation of the hoop.)
Therefore, the best choice is C) 28 N∙m.

A Thin Circular Hoop of Mass 1