Question

a solid cylinder of mass 2kg and radius 4 cm rotating about its axis at the rate of 3 rpm .the torque required to stop after 2π revolutions is

Answer 1

mass of solid cylinder , m = 2kg

radius of cylinder , r = 4cm = 0.04m

a solid cylinder is rotating about its axis,

so, moment of inertia , I = 1/2 mr²

= 1/2 × 2 × (0.04)² = 16 × 10^-4 Kgm²

= 1.6 × 10^-3 kgm²

angular frequency, $\omega$ = 3rpm = 3 × 2π/60 rad/s = π/10 rad/s

angular displacement = 2π revolution

= 2π × 2π = 4π² [ as 1 revolution = 2π ]

from conservation of energy theorem,

change in rotational kinetic energy = torque × angular displacement

or, 1/2 I$\omega^2$ = torque × 4π²

or, 1/2 × 1.6 × 10^-3 × 1/100 = torque × 4

torque = 2 × 10^-6 Nm

hence, torque = 2 × 10^-6 Nm