A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to an angular velocity of 120 rad/s in 2.5 s. b) Through what angle has it turned during that time? c) What is the work done by the torque?

a) τ = 0.672 N m, b) θ = 150 rad, c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

w = w₀ + α t

α = - w₀ / t

α = 120 / 2.5

α = 48 rad / s²

The moment of inertia of a cylinder is

I = ½ M R²

Let's calculate the torque

τ = I α

τ = ½ M R² α

τ = ½ 2.8 0.1² 48

τ = 0.672 N m

b) we look for the angle by kinematics

θ = w₀ t + ½ α t2

θ = ½ α t²

θ = ½ 48 2.5²

θ = 150 rad

c) work in angular movement

W = τ θ

W = 0.672 150

W = 100.8 J