A uniform solid cylinder of mass 20 kg and radius 0.5 m is rotating about its central axis. The amount of work required to increase its angular speed from 120 rpm to 300 rpm is: (Take ^2 = 10 )

Step-by-step solution

Moment of inertia of the solid cylinder about its central axis is I = 1 2 mr^2 Substituting the given values: I = 1 2 20 (0.5)^2 = 2.5 kg m ^2 Initial angular speed, ₁ = 120 rpm = 120 2 60 = 4 rad/s Final angular speed, ₂ = 300 rpm = 300 2 60 = 10 rad/s According to the work-energy theorem, the work required is equal to the change in rotational kinetic energy: W = 1 2 I( ₂^2 - ₁^2) W = 1 2 2.5 ((10 )^2 - (4 )^2) W = 1.25 (100 ^2 - 16 ^2) W = 1.25 84 ^2 Given ^2 = 10 : W = 1.25 84 10 = 1050 J Answer: 1050 J