$A$ bullet of mass $10 \; g$ and speed $500 \; m/s$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0 \; m$ wide and weighs $12 \; kg$. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

Two discs of moment of inertia $I_1$ and $I_2$ and angular speeds $\omega_1$ 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 the system will be:

$A$ horizontal disc of moment of inertia $4.25 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning counter-clockwise at $15 \,rps$ about its axis,as viewed from above. $A$ second disc of moment of inertia $1.80 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning clockwise at $25 \,rps$ as viewed from above about the same axis and is dropped on top of the first disc. The two discs stick together and rotate as one about their axis of symmetry. The new angular velocity of the system as viewed from above is close to

If the earth were to suddenly contract to $\frac{1}{n}^{th}$ of its present radius without any change in its mass,the duration of the new day will be nearly:

The angular velocity of a body changes from $\omega_1$ to $\omega_2$ without the application of any external torque,but this change occurs due to a change in its moment of inertia. The ratio of the radii of gyration in the two cases will be:

$A$ particle of mass $1 \ kg$ is subjected to a force which depends on the position as $\vec{F} = -k(x \hat{i} + y \hat{j}) \ N$ with $k = 1 \ kg \ s^{-2}$. At time $t = 0$,the particle's position is $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) \ m$ and its velocity is $\vec{v} = (-\sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{\pi} \hat{k}) \ m \ s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity,respectively. Ignore gravity. When $z = 0.5 \ m$,the value of $(x v_y - y v_x)$ is . . . . . $m^2 \ s^{-1}$.