The mass per unit length of a rod of length $l$ is given by : $\lambda = \frac{M_0x}{l}$ ,where $M_0$ is a constant and $x$ is the distance from one end of the rod. The position of centre of mass of the rod is
$\frac{4l}{3}$
$\frac{l}{3}$
$\frac{2l}{3}$
$\frac{5l}{3}$
A rod of length $L$ is held vertically on a smooth horizontal surface. The top end of the rod is given a gentle push. At a certain instant of time, when the rod makes an angle $\theta$ with horizontal the velocity of $COM$ of the rod is $v_0$ . The velocity of the end of the rod in contact with the surface at that instant is
For a rolling body, the velocity of $P_1$ and $P_2$ are ${\vec v_1}$ and ${\vec v_2}$ , respectively
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
If a solid sphere is rolling the ratio of its rotational energy to the total kinetic energy is given by