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A flywheel rotating at `420 rpm` slows down at a constant rate of `2 rad s^-2` The time required to stop the flywheel is.A. 22 sB. 11 sC. 44 sD. 12 s |
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Answer» Correct Answer - A (a) Here, `v_0 = 420 rpm = 7 rps` :. `omega_0 = 2 pi v_0 = 2 xx (22)/(7) xx 7 = 44 rad s^-1` `omega = 0, prop = -2 rad s^-2` `t = (omega - omega_0)/(prop) = (-44)/(-2) = 22 s`. |
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