A circular disc of radius R meter and mass M | vedclass

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(A) The angular position is given by $	heta(t) = 5t^2 - 8t$.Angular velocity $\omega$ is the derivative of angular position with respect to time: $\o

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) The angular position is given by $	heta(t) = 5t^2 - 8t$.Angular velocity $\omega$ is the derivative of angular position with respect to time: $\omega = \frac{d	heta}{dt} = 10t - 8$.Angular acceleration $\alpha$ is the derivative of angular velocity with respect to time: $\alpha = \frac{d\omega}{dt} = 10 	ext{ rad/s}^2$.The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is $I = \frac{1}{2}MR^2$.The torque applied is $	au = I\alpha = (\frac{1}{2}MR^2)(10) = 5MR^2$.Power $P$ delivered by the torque is $P = 	au \omega$.At $t = 2$ s,$\omega = 10(2) - 8 = 12 	ext{ rad/s}$.Therefore,$P = (5MR^2)(12) = 60 MR^2$ $W$.

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