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The angular position of a point over a r...

The angular position of a point over a rotating flywheel is changing according to the relation, `theta = (2t^3 - 3t^2 - 4t - 5)` radian. The angular acceleration of the flywheel at time, t = 1 s is

A

Zero

B

4 rad/`s^2`

C

6 rad/`s^2`

D

12 rad/`s^2`

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Knowledge Check

  • The instantaneous angular position of a point on a rotating wheel is given by the equation theta (t) = 2t^(3) - 6t^(2) . The torque on the wheel becomes zero at

    A
    t = 1 s
    B
    t = 0.5 s
    C
    t = 0.25 s
    D
    t = 2 s

  • The instantaneous angular position of a point on a rotating wheel is given by the equation theta(t) = 2t^(3) - 6 t^(2) The torque on the wheel becomes zero at

    A
    `t = 1 s`
    B
    `t = 0.5 s`
    C
    `t = 0.25 s`
    D
    `t = 2 s`

  • the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is

    A
    `10 rad/s^2`
    B
    `5rad/s^2`
    C
    Zero
    D
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