Question

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?

Answer 1

Answer:

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