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

Answer

B

Explanation

`omega_(0) = 3 rpm= 3 xx (2pi)/(60)` rad/sec `= (pi)/(10)`
`omega^(2) = omega_(0)^(2) + 2 alpha theta`
`0^(2) = ((pi)/(10))^(2) + 2 (alpha) (2pi xx 2pi)`
`alpha= - (1)/(800) "rad/sec"^(2)`
`I= (mR^(2))/(2) = ((2) ((4)/(100))^(2))/(2)= (16)/(10^(4))`
`tau = I alpha= ((16)/(10^(4))) xx (-(1)/(800)) = -2 xx 10^(-6 ) Nm`