The angle turned by a body undergoing circular motion depends on time as $\theta = \theta_0 + \theta_1 t + \theta_2 t^2$. Then the angular acceleration of the body is

$A$ 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?

The angular speed of the wheel of a vehicle is increased from $360 \; rpm$ to $1200 \; rpm$ in $14 \; s$. Its angular acceleration is,

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 ......... $rad/s$.

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

$A$ particle is rotating in a circular path and at any instant its motion can be described as $\theta = \frac{5t^4}{40} - \frac{t^3}{3}$. The angular acceleration of the particle after $10 \text{ s}$ is . . . . . . $\text{rad/s}^2$.