$A$ wheel of radius $0.5 \ m$ and a moment of inertia of $10 \ kg \cdot m^2$ is rotating freely at an angular speed of $70 \ rev/min$. The wheel can be stopped in $5.0 \ s$ by pressing a wet cloth against the rim and exerting a radially inward force of $88 \ N$. The coefficient of kinetic friction between the wheel and wet cloth is:

Consider a sphere of mass $m$ and radius $R$ performing pure rolling motion on a rough surface with velocity $v_0$ as shown in the figure. It makes an elastic impact with a smooth wall,moves back,and eventually starts pure rolling again.

One ice skater of mass $m$ moves with speed $2v$ to the right,while another of the same mass $m$ moves with speed $v$ toward the left,as shown in figure $I$. Their paths are separated by a distance $b$. At $t = 0$,when they are both at $x = 0$,they grasp a pole of length $b$ and negligible mass. For $t > 0$,consider the system as a rigid body of two masses $m$ separated by distance $b$,as shown in figure $II$. Which of the following is the correct formula for the motion after $t = 0$ of the skater initially at $y = b/2$?

$A$ pendulum consists of a bob of mass $m=0.1 \ kg$ and a massless inextensible string of length $L=1.0 \ m$. It is suspended from a fixed point at height $H=0.9 \ m$ above a frictionless horizontal floor. Initially,the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. $A$ horizontal impulse $P=0.2 \ kg \cdot m/s$ is imparted to the bob at some instant. After the bob slides for some distance,the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \ kg \cdot m^2/s$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. $(1)$ The value of $J$ is. . . . . . $(2)$ The value of $K$ is. . . . . Give the answers of the questions $(1)$ and $(2)$.

The position vectors of two $1 \ kg$ particles,$(A)$ and $(B),$ are given by $\overrightarrow{r}_{A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 \hat{k}) \ m$ and $\vec{r}_B = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k}) \ m$,respectively. Given $\alpha_1 = 1 \ m/s^2, \alpha_2 = 3n \ m/s, \alpha_3 = 2 \ m, \beta_1 = 2 \ m/s, \beta_2 = -1 \ m/s^2, \beta_3 = 4p \ m/s$,where $t$ is time,$n$ and $p$ are constants. At $t = 1 \ s$,$|\overrightarrow{V}_{A}| = |\overrightarrow{V}_{B}|$ and the velocities $\overrightarrow{V}_{A}$ and $\overrightarrow{V}_{B}$ are orthogonal. At $t = 1 \ s$,the magnitude of angular momentum of particle $(A)$ with respect to particle $(B)$ is $\sqrt{L} \ kg \ m^2/s$. The value of $L$ is:

$A$ rigid uniform bar $AB$ of length $L$ is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time,the angle made by the bar with the vertical is $\theta$. Which of the following statements about its motion is/are correct?
$[A]$ The midpoint of the bar will fall vertically downward
$[B]$ The trajectory of the point $A$ is a parabola
$[C]$ Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$
$[D]$ When the bar makes an angle $\theta$ with the vertical,the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$