In the given figure,$r_1$ and $r_2$ are $10 \ cm$ and $20 \ cm$ respectively. If the moment of inertia of the wheel is $1500 \ kg \cdot m^2$,find the angular acceleration.

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 $t = $ ...... $s$.

$A$ thin uniform rod of mass $M$ and length $L$ is pivoted at a height $\frac{L}{3}$ from its lower end as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is . . . . . . . ($g$ = gravitational acceleration)

$A$ thin uniform rod of mass $m$ and length $L$ is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is [consider negligible friction at the pivot] ($g=$ acceleration due to gravity).

$A$ flywheel gains a speed of $540\ r.p.m.$ in $6\ s$. Its angular acceleration will be

What is the value of the torque acting on an object moving in a circular path with a constant angular speed? Why?