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

$A$ wheel rotates at $720 \ rpm$ about an axis passing through its center. It is brought to rest by applying a constant torque for $8 \ s$. What is the value of the torque in $Nm$? (Given: $I = \frac{24}{\pi} \ kg \cdot m^2$)

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ torque of $50\, Nm$ acting on a wheel at rest rotates it through $200\, rad$ in $5\, s$. Calculate the angular acceleration produced. ........... $rad\, s^{-2}$.

$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

If there is a change of angular momentum from $1 \, J \cdot s$ to $4 \, J \cdot s$ in $4 \, s$, then the torque is