In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus

T = 6.43 x 10⁻⁵ N.m

Explanation:

First, we will calculate the deceleration of the apparatus by using the third equation of motion:

where,

α = angular decelration = ?

θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad

ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s

ωf = final angular speed = 0 rad/s

Therefore,

negative sign shows deceleration

Now, for torque:

T = Iα

where,

T = Torque = ?

I = moment of inertia = 0.076 kg.m²

Therefore,

T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)

T = 6.43 x 10⁻⁵ N.m

your answer is c! hope this : )