A flywheel of moment of inertia `3 xx 10^(2) kg m^(2)` is rotating with uniform angular speed of `4.6 rad s^(-1)`. If a torque of `6.9 xx 10^(2)Nm` retards the wheel, then the time in which the wheel comes to rest is
A. `1.5 s`
B. `2 s`
C. `0.5 s`
D. `1 s`
A. `1.5 s`
B. `2 s`
C. `0.5 s`
D. `1 s`
Correct Answer – B
Here, `I = 3 xx 10^(2) kg m^(2), omega_(1) = 4.6 rad s^(-1)`
`tau = 6.9 xx 10^(2) Nm, omega_(2) = 0, t = ?`
`alpha = (tau)/(I) = (-6.9 xx 10^(2))/(3 xx 10^(2)) = – 2.3 rad//s^(2)`
From `omega_(2) = omega_(1) + alpha t`
`0 = 4.6 – 2.3 t, t = 2 sec`