A uniform rod $AB$ of length $l$ and mass $\mathrm{m}$ 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 $\frac{m{l}^2}{3}$, the initial angular acceleration of the rod will be

 

Consider a body, shown in figure, consisting of two identical balls, each of mass $\mathrm{M}$ connected by a light rigid rod. If an impulse $J=MV$ is imparted to the body at one of its ends, what would be its angular velocity ?

Let $\overrightarrow{F}$ be the force acting on a particle having position vector $\overrightarrow{r}$ and $\overrightarrow{\tau }$ be the torque of this force about the origin. Then

A thin uniform rod of length $l$ and mass $\mathrm{m}$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of

A small object of uniform density rolls up a curved surface with an initial velocity $\mathrm{v}$  It reaches up to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is

A uniform bar of length $6$ and mass $8\mathrm{m}$ lies on a smooth horizontal table. Two point masses $\mathrm{m}$ and $2 \mathrm{m}$ moving in the same horizontal with speed $2\mathrm{v}$ and $\mathrm{v}$ respectively, strike the bar as shown in the figure, and stick to the bar after the collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by $\omega , E$ and $v_{\mathit{c}}$ respectively, we have after the collision: