Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same $(M)$ and the mass of the threads are negligible,the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at $\frac{r}{2}$ distance from the end $A$ of the handle will be ....... $Mr^2$?

$A$ block of mass $m$ is attached to a pulley disc of equal mass $m$ and radius $r$ by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of $5\, m/s$. Its velocity when the string becomes taut will be

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_{0}$ in the $x-y$ plane as shown in the figure. For time $t < \frac{v_{0} \sin \theta}{g}$,the angular momentum of the particle is (where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$ axes respectively):

$A$ uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of $20^o$ to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above,is there a torque acting on it,and if so,in what direction?

$A$ stick of length $l$ and mass $M$ lies on a frictionless horizontal surface on which it is free to move in any way. $A$ ball of mass $m$ moving with speed $v$ collides elastically with the stick at one of its ends as shown in the figure. If after the collision the ball comes to rest,what should be the mass of the ball?

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: