$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitude of its centripetal acceleration and tangential acceleration be equal?

$A$ flywheel at rest is to reach an angular velocity of $24 \ rad \ s^{-1}$ in $8 \ s$ with constant angular acceleration. The total angle turned through during this interval is (in $rad$)

Angular displacement $(\theta)$ of a flywheel varies with time as $\theta = at + bt^2 + ct^3$. The angular acceleration is given by:

Two bodies $A$ and $B$ rotate about an axis,such that the angle $\theta_A$ (in radians) covered by the first body is proportional to the square of time,and $\theta_B$ (in radians) covered by the second body varies linearly with time. At $t = 0$,$\theta_A = \theta_B = 0$. If $A$ completes its first revolution in $\sqrt{\pi} \ s$ and $B$ needs $4\pi \ s$ to complete half a revolution,then the ratio of their angular velocities $\omega_A : \omega_B$ at $t = 5 \ s$ is:

$A$ ball is spun with angular acceleration $\alpha = 6t^2 - 2t$,where $t$ is in seconds and $\alpha$ is in $rad/s^2$. At $t = 0$,the ball has an angular velocity of $10 \ rad/s$ and an angular position of $4 \ rad$. The most appropriate expression for the angular position of the ball is:

$A$ flywheel rotates with a uniform angular acceleration. Its angular velocity increases from $20\pi \, rad/s$ to $40\pi \, rad/s$ in $10 \, s$. How many rotations did it make in this period?