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

Explanation

`theta = 2pi` revolution
`=2pi xx 2pi = 4pi ^(2)` red
`omega _(1) = 2pi f = 2pi xx (3)/(60) rad //s ( because f = ( 3)/(60) s ^(-1))`
`omega _(f) = 0`
Work energy theorem.
`W = (1)/(2) t (omega _(t) ^(2) - omega _(1) ^(2))`
`implies - r theta =(1)/(2)xx (1)/(2) mr ^(2) ( 0^(2) - omega _(1) ^(2))`
`implies - tau = ((1)/(2) xx (1)/(2) xx 2 ( 4 xx 10 ^(-2)) ^(2) ( -3 x (2pi)/(60 )) ^(2))/( 4pi ^(2))`
`implies tau =2 xx 10 ^(-6) Nm`