$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown is figure. $A$ rod of identical mass is fixed along one of its diameter. The total kinetic energy of the system is :-
$\frac{7}{5}m{v^2}$
$\frac{2}{5}m{v^2}$
$\frac{5}{3}m{v^2}$
$\frac{5}{4}m{v^2}$
Two point masses of $0.3\ kg$ and $0.7\ kg$ are fixed at the ends of a rod of length $1.4\ m$ and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively as shown is the fig. The velocities of two discs at the points $A$ and $B$ of the inclined planes are related as
A $2 \,{kg}$ steel rod of length $0.6\, {m}$ is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \,{ms}^{-1}$. (Take $g = 10\, {ms}^{-2}$ )
Two discs of moment of inertia $I_1$ and $I_2$ and angular speeds ${\omega _1}\,{\rm{and }}{\omega _2}$ are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational $KE$ of system will be
If $L, M$ and $P$ are the angular momentum, mass and linear momentum of a particle respectively which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$