$A$ 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. What is the direction of the rod's angular momentum vector (calculated about its lower end)?

Obtain the relation between torque of a system of particles and angular moment. 

A ball of mass $1 \,kg$ is projected with a velocity of $20 \sqrt{2}\,m / s$ from the origin of an $x y$ co-ordinate axis system at an angle $45^{\circ}$ with $x$-axis (horizontal). The angular momentum [In $SI$ units] of the ball about the point of projection after $2 \,s$ of projection is [take $g=10 \,m / s ^2$ ] ( $y$-axis is taken as vertical)

$A$ paritcle falls freely near the surface of the earth. Consider $a$ fixed point $O$ (not vertically below the particle) on the ground.

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 . The angular speed of the door just after the bullet embeds into it will be

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