A slender uniform rod of mass M and length $$l$$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. 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

The moment of inertia of a uniform cylinder of length $$l$$ and radius R about its perpendicular bisector is $$I$$. What is the ratio $${l \over R}$$ such that the moment of inertia is minimum?

A body of mass m = 10^–2 kg is moving in a medium and experiences a frictional force F = –kv². Its initial speed is v₀ = 10 ms^–1. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be:

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be: