$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 A_0$ in $4$ seconds. The magnitude of this torque is ...........

$A$ rigid massless rod of length $3l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from the initial horizontal position,its instantaneous angular acceleration will be

$A$ rectangular solid rod of length $0.3\, m$ is held horizontally,with one of its sides on the edge of a platform of height $5\, m$. When released,it slips off the table in a very short time $\Delta t = 0.01\, s$,remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

If a constant torque acts on an object,then the object:

The moment of inertia of a body about a given axis is $1.2 \; kg \cdot m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1500 \; J$,an angular acceleration of $25 \; rad/s^2$ must be applied about that axis for a duration of: (in $; s$)

Write Newton's second law for a system of particles performing rotational motion.