Let $F $ be the force acting on a particle having position vector $\vec r$ and $\vec T$ be the torque of this force about the origin. Then ..
$\vec r.\vec T = 0\,{\rm{ and\, }}\vec F.\vec T = 0$
$\vec r.\vec T = 0\,{\rm{ and\, }}\vec F.\vec T \ne 0$
$\vec r.\vec T \ne 0\,{\rm{ and \,}}\vec F.\vec T = 0$
$\vec r.\vec T \ne 0\,{\rm{ and \,}}\vec F.\vec T \ne 0$
A smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size, each of mass $m$ , which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity $\omega _0$ about its axis perpendicular to the rod and passing through its mid-point (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is
A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity $v\,\,m/s.$ If it is to climb the inclined surface then $v$ should be
A disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v,$ the height to which the disc will rise will be
If the density of material of a circular plate and a square plate as shown is same, the centre of mass of the composite system will be
Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed of $B$ is $2v$, the speed of centre of mass of the system is