a solid cylinder of mass 2kg and radius 4cm rotating about axis at the rate of 3rpm. the torque required to stop after 2pi revolutions is
Answer (1)
3rd Feb, 2020
Hello,
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
So, I = 1/2 × 2 × (0.04)² = 16 × 10^-4 Kgm²
= 1.6 × 10^-3 kgm²
Angular frequency, = 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
So, 1/2 I = torque × 4π²
So, 1/2 × 1.6 × 10^-3 × 1/100 = torque × 4
Hence, Torque = 2 × 10^-6 Nm
Best Wishes.
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