The total kinetic energy of a body of mass $10 \ kg$ and radius $0.5 \ m$ moving with a velocity of $2 \ m/s$ without slipping is $32.8 \ J$. The radius of gyration of the body is .......... $m$.

$A$ cubical block of side $a = 30\,cm$ is moving with velocity $v = 2\,m/s$ on a smooth horizontal surface. The surface has a small bump at a point $O$ as shown in the figure. The angular velocity (in $rad/s$) of the block immediately after it hits the bump is:

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$(1)$ If $|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}$,then the angle between $\vec{A}$ and $\vec{B}$ is ............ .
$(2)$ The angle between the angular momentum and linear momentum of a particle in rotational motion is ............ .
$(3)$ $A$ force $F\hat{k}$ acts on a particle having position vector $(2\hat{i} + \hat{j})$. The torque acting on the particle is ............ .

$A$ 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:

Consider regular polygons with number of sides $n=3, 4, 5, \ldots$ as shown in the figure. The center of mass of all the polygons is at height $h$ from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is $\Delta$. Then $\Delta$ depends on $n$ and $h$ as

$A$ solid sphere is in purely rotational motion about its diameter. The ratio of its angular momentum $(L)$ and kinetic energy $(K)$ is $\frac{\pi}{22}$. Find the angular velocity $(\omega)$ of the sphere. (Take $\pi = \frac{22}{7}$) (in $rad/s$)