In the List-$I$ below, four different paths of a particle are given as functions of time. In these functions, $\alpha$ and $\beta$ are positive constants of appropriate dimensions and $\alpha \neq \beta$. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: $\overrightarrow{ p }$ is the linear momentum, $\bar{L}$ is the angular momentum about the origin, $K$ is the kinetic energy, $U$ is the potential energy and $E$ is the total energy. Match each path in List-$I$ with those quantities in List-$II$, which are conserved for that path.
List-$I$ | List-$II$ |
$P$ $\dot{r}(t)=\alpha t \hat{t}+\beta t \hat{j}$ | $1$ $\overrightarrow{ p }$ |
$Q$ $\dot{r}(t)=\alpha \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$ | $2$ $\overrightarrow{ L }$ |
$R$ $\dot{r}(t)=\alpha(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ | $3$ $K$ |
$S$ $\dot{r}(t)=\alpha t \hat{i}+\frac{\beta}{2} t^2 \hat{j}$ | $4$ $U$ |
$5$ $E$ |
$P \rightarrow 1,2,3,4,5 ; \quad Q \rightarrow 2,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 5$
$P \rightarrow 1,2,3,4,5 ; \quad Q \rightarrow 3,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 2,5$
$P \rightarrow 2,3,4 ; \quad Q \rightarrow 5 ; \quad R \rightarrow 1,2,4 ; \quad S \rightarrow 2,5$
$P \rightarrow 1,2,3,5 ; \quad Q \rightarrow 2,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 2,5$
$A$ 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 particle of mass $20\,g$ is released with an initial velocity $5\,m/s$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be ......... $kg - m^2/s$. [Take $g = 10\,m/s^2$ ]
A solid sphere of mass $500\,g$ and radius $5\,cm$ is rotated about one of its diameter with angular speed of $10\,rad \, s ^{-1}$. If the moment of inertia of the sphere about its tangent is $x \times 10^{-2}$ times its angular momentum about the diameter. Then the value of $x$ will be ..............
A particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then