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

Options

A. `2 xx 10^(6) Nm`
B. `2 xx 10^(6)` N m
C. `2 xx 10^(-3)` N m
D. `12 xx 10^(-4)` N m

Answer

B

Explanation

Given : Mass M = 2 kg , Radius R = 4 cm
Initial angular speed `omega_(0) = 3 rpm = 3 xx (2pi)/(60)` rad/s = `(pi)/(10)` rad/s
We know that , `omega_(2) = omega_(0)^(2) + 2 alpha theta`
`implies 0 = ((pi)/(10))^(2) + 2 xx alpha xx 2pi xx 2pi implies alpha = (-1)/(800) "rad/s"^(2)`
Moment of inertia of a solid cylinder ,
`I = (MR^(2))/(2) = (2 xx ((4)/(100))^(2))/(2) = (16)/(10^(4))`
Torque `tau = I alpha = ((16)/(10^4)) xx (- (1)/(800)) = -2 xx 10^(-6) N m`