Question

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is :
(1) 2 × 10⁻⁶ N m
(2) 2 × 10⁻³ N m
(3) 12 × 10⁻⁴ N m
(4) 2 × 10⁶ N m

Answer 1

Answer:

1. 2×10-^6Nm

Explanation:

I hope you're answers this

Answer 2

Answer:

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, \omegaω = 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^2ω

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