Question

the moment of inertia of a ring is 0.40 kg metre square if it is rotating at the rate of 2100 revolutions per minute calculate the torque required to stop it in 2 seconds what will be the work done

Answer 1

Here's the answer....

Answer 2

We can calculate initial angular velocity (ω0) by,

     ω0 = revolutions per minute*2π/60

           = 2100*2π/60

     ω0 = 70 rad/s

After that we can calculate angular acceleration (α) by using a rotational motion equation.

      ω = ω0 + αt

Where, ω = final angular velocity = 0

           ω0=initial angular velocity = 70 rad/s

              α = angular acceleration

               t = time = 2 s

      ω = ω0 + αt

      0 = 70 + 2α

      α = -35 rad/s²

Torque = moment of inertia of the ring * angular acceleration

            = Iα

            = 0.4 * 35

            =14 Nm