If the equation for the angular displacement of a particle moving on a circular path is given by $\theta = 2t^3 + 0.5$,where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle after $2 \ s$ from its start is ......... $rad/s$.

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

$A$ particle starting from rest moves along the circumference of a circle of radius $r$ with constant angular acceleration $\alpha$. The magnitude of the average velocity,in the time it completes a small angular displacement $\theta$,is:

$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

The angular speed of a wheel increases from $600 \, rpm$ to $1200 \, rpm$ in $10 \, s$. What is the angular acceleration?

$A$ 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: