A fly wheel of moment of inertia `3 xx 10^(2)` kg mạis rotating with uniform angular speed of 4.6 rad `s^(-1)`. If a torque of `6.9 xx 10^2` N m retards he wheel, then the time in which the wheel comes to rest is

Options

A. 1.5 s
B. 2 s
C. 0.5 s
D. 1 s

Answer

B

Explanation

Here , moment of inertia , `I = 3 xx 10^(2) kg m^(2)`
Torque `tau = 6.9 xx 10^(2)` N m
Initial angular speed `omega_(0) = 4.6 rad s^(-1)`
Final angular speed , `omega = 0 rad s^(-1)`
As `omega = omega_(0) + alpha t`
`therefore alpha = (omega - omega_(0))/(t) = (0 - 4.6)/(t) = - (4.6)/(t) rad s^(-2)`
Negative sign is for deceleration .
Torque , `tau = I alpha`
`6.9 xx 10^(2) = 3 xx 10^(2) xx (4.6)/(t) or t = (3 xx 10^(2) xx 4.6)/(6.9 xx 10^(2)) = 2 s`