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Virat Munshi
Asked: 2022-11-11T00:29:52+05:30 In:

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 s
B. 11 s
C. 44 s
D. 12 s

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 s
B. 11 s
C. 44 s
D. 12 s
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  1. Correct Answer – a
    Here, `v_(0) = 420 rpm = 7 rps`
    `omega_(0)= 2 xx 22/7 xx 7 = 44 rads^(-1)`
    `omega=0, alpha=-2 rad s^(-2)`
    `therefore t=(omega-omega_(0))/(alpha) = -44/-2=22 s`

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Hira Charandeep Hegde
Asked: 2022-10-30T23:38:51+05:30 In:

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 s
B. 11 s
C. 44 s
D. 12 s

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 s
B. 11 s
C. 44 s
D. 12 s
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. 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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