$A$ ring of radius $R$ rolls without sliding with a constant velocity $u$. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be

$A$ rod $AB$ is free to rotate in a vertical plane about a horizontal axis through $A$ as shown in the figure. It is slightly disturbed from rest in its position of unstable equilibrium and when it is next vertical,the end $B$ collides with a fixed peg and rebounds. If the rod comes to instantaneous rest when $AB$ is horizontal (as shown in the figure),then:

Four point masses are fastened to the corners of a frame of negligible mass lying in the $xy$ plane as shown in the figure. Let $\omega$ be the angular speed of rotation. Then:

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

The torque acting on a body about a point is equal to $\overrightarrow{A} \times \overrightarrow{L}$,where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum about that point. This implies that:

Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $O$. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown,two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y-z$ plane and is same for both pebbles with respect to the disc. Assume that $(i)$ they land back on the disc before the disc completes $\frac{1}{8}$ rotation,$(ii)$ their range is less than half disc radius,and $(iii)$ $\omega$ remains constant throughout. Then