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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 isA. `1.5 s`B. `2 s`C. `0.5 s`D. `1 s` |
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Answer» 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` |
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