In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

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

$A$ flywheel attains a speed of $540 \, r.p.m.$ in $6 \, s$. Its angular acceleration is:

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane. One of its ends is hinged on a vertical axis. $A$ horizontal force $F = Mg/2$ is applied at a distance of $5L/6$ from the hinged end. What will be the angular acceleration of the rod?

$A$ thin solid disk of $1 \ kg$ is rotating along its diameter axis at the speed of $1800 \ rpm$. By applying an external torque of $25 \pi \ Nm$ for $40 \ s$,the speed increases to $2100 \ rpm$. The diameter of the disk is . . . . . . $m$.

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be: