Mix Example - System of Particles and Rotational Motion Questions in English

(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about its perpendicular bisector is given by:
$I_1 = \frac{ML^2}{12} \quad ...(1)$
When the rod is bent into a semicircular arc,its length $L$ becomes the circumference of a semicircle of radius $r$,so $L = \pi r$,which gives $r = \frac{L}{\pi}$.
The moment of inertia of a semicircular arc of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is:
$I_2 = Mr^2$
Substituting $r = \frac{L}{\pi}$:
$I_2 = M \left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2} \quad ...(2)$
Now,calculating the ratio $I_1 : I_2$:
$\frac{I_1}{I_2} = \frac{ML^2/12}{ML^2/\pi^2} = \frac{\pi^2}{12}$
Since $\pi^2 \approx 9.87$,we have $\frac{\pi^2}{12} \approx \frac{9.87}{12} < 1$.
Therefore,the ratio $I_1 : I_2 < 1$.

(B) Let the true weight of the body be $w$ and the lengths of the two arms of the balance be $p$ and $q$.
Case $1$: When the body is placed in the left pan,it is balanced by a weight of $6$ g in the right pan.
By the principle of moments,$w \times p = 6 \times q$
$\Rightarrow \frac{p}{q} = \frac{6}{w} \quad ...(i)$
Case $2$: When the body is placed in the right pan,it is balanced by a weight of $24$ g in the left pan.
By the principle of moments,$24 \times p = w \times q$
$\Rightarrow \frac{p}{q} = \frac{w}{24} \quad ...(ii)$
Equating $(i)$ and $(ii)$:
$\frac{6}{w} = \frac{w}{24}$
$w^2 = 6 \times 24 = 144$
$w = \sqrt{144} = 12$ g.
Thus,the true weight of the body is $12$ g.

(D) For the sphere to be in equilibrium,the line of action of its weight $(mg)$ must pass through or behind the edge of the obstacle to prevent it from toppling over.
Consider the triangle formed by the center of the sphere,the point of contact with the inclined plane,and the edge of the obstacle.
The vertical distance from the center of the sphere to the horizontal level of the edge is $R \cos \theta$.
The height of the center of the sphere from the inclined plane is $R$.
Therefore,the vertical height of the edge relative to the inclined plane is $h = R - R \cos \theta$.
Thus,$h = R(1 - \cos \theta)$.

(C) By the principle of conservation of energy,the loss in potential energy $(P.E.)$ equals the gain in rotational kinetic energy $(K.E.)$ as the rod falls from horizontal to vertical position.
Loss in $P.E. = mg \frac{l}{2}$.
Gain in $K.E. = \frac{1}{2} I \omega^2$,where $I = \frac{ml^2}{3}$ is the moment of inertia about the hinge.
Equating them: $mg \frac{l}{2} = \frac{1}{2} (\frac{ml^2}{3}) \omega^2$.
Simplifying gives: $mg l = \frac{ml^2}{3} \omega^2$,which leads to $\omega^2 = \frac{3g}{l}$.
At the vertical position,the net force acting towards the hinge provides the centripetal force for the center of mass of the rod:
$F - mg = m a_c = m (\frac{l}{2}) \omega^2$.
Substituting $\omega^2$: $F = mg + m (\frac{l}{2}) (\frac{3g}{l}) = mg + \frac{3mg}{2} = \frac{5mg}{2}$.

(B) The power $P$ developed by a torque $\tau$ is given by $P = \tau \omega$,where $\omega$ is the angular velocity.
Since the rod starts from rest,the initial angular velocity $\omega_0 = 0$.
Given a constant torque $\tau$,the angular acceleration $\alpha$ is constant,where $\alpha = \frac{\tau}{I}$ ($I$ is the moment of inertia).
The angular velocity at time $t$ is given by $\omega = \omega_0 + \alpha t = 0 + \left(\frac{\tau}{I}\right) t = \frac{\tau}{I} t$.
Substituting this into the power equation: $P = \tau \left(\frac{\tau}{I} t\right) = \frac{\tau^2}{I} t$.
Since $\tau$ and $I$ are constants,$P \propto t$.
This represents a straight line passing through the origin,which corresponds to the graph in option $B$.

(A) $1$. The constant force $F$ applied at the rim of the pulley creates a constant torque $\tau = F \cdot R$ about the hinge,where $R$ is the radius of the pulley.
$2$. According to the rotational analogue of Newton's second law,$\tau = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. Since $\tau$ is constant,$\alpha$ is also constant.
$3$. Starting from rest,the angular velocity $\omega$ at time $t$ is given by $\omega = \alpha t$. As $t$ increases,$\omega$ increases linearly.
$4$. To maintain equilibrium in the translational direction,the hinge must exert a reaction force $F_h$ to balance the applied force $F$. Since the pulley is not moving translationally,the net force on the pulley must be zero. Therefore,the hinge force $F_h$ must be equal and opposite to the applied force $F$ at all times. Thus,the force on the hinge remains constant.

(C) Let the length of the rod be $l$ and its mass be $m$. When the rod is released from the horizontal position,by the conservation of mechanical energy,the potential energy lost equals the rotational kinetic energy gained when it reaches the vertical position:
$mg(l/2) = (1/2)I\omega^2$
Since $I = ml^2/3$ for a rod hinged at one end,we have:
$mgl/2 = (1/2)(ml^2/3)\omega^2$
$\omega^2 = 3g/l$
When the rod reaches the vertical position,the lower half (mass $m/2$) separates. The upper half (mass $m/2$) remains hinged at the top and continues to rotate with the same angular velocity $\omega = \sqrt{3g/l}$.
Now,applying the work-energy theorem to the upper half as it swings up to an angle $\theta$ with the vertical:
The change in kinetic energy is $\Delta K = K_f - K_i = 0 - (1/2)I_{upper}\omega^2$,where $I_{upper} = (m/2)(l/2)^2/3 = ml^2/24$.
The work done by gravity is $W_g = -(m/2)g(h_{cm,f} - h_{cm,i}) = -(m/2)g((l/4) - (l/4)\cos\theta) = -(mgl/8)(1 - \cos\theta)$.
Equating $W_g = \Delta K$:
$-(mgl/8)(1 - \cos\theta) = -(1/2)(ml^2/24)(3g/l)$
$(mgl/8)(1 - \cos\theta) = (mgl/16)$
$1 - \cos\theta = 1/2$
$\cos\theta = 1/2$
$\theta = 60^o$.

(C) The rod is pivoted at a distance $l/4$ from one end. The distance of the pivot from the center of mass $(CM)$ is $d = |l/2 - l/4| = l/4$.
By the parallel axis theorem,the moment of inertia $I$ of the rod about the pivot is:
$I = I_{CM} + md^2 = \frac{ml^2}{12} + m\left(\frac{l}{4}\right)^2 = \frac{ml^2}{12} + \frac{ml^2}{16} = \frac{4ml^2 + 3ml^2}{48} = \frac{7ml^2}{48}$.
When the rod is released from the horizontal position and reaches the vertical position,the center of mass drops by a vertical distance $h = l/4$.
By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mgh = \frac{1}{2} I \omega^2$
$mg\left(\frac{l}{4}\right) = \frac{1}{2} \left(\frac{7ml^2}{48}\right) \omega^2$
$\frac{mgl}{4} = \frac{7ml^2}{96} \omega^2$
$\omega^2 = \frac{mgl}{4} \cdot \frac{96}{7ml^2} = \frac{24g}{7l}$
$\omega = \sqrt{\frac{24g}{7l}}$.

(D) When the string becomes taut,an impulsive tension $T$ acts on both the block and the disc for a very short time interval $\Delta t$.
For the block,the impulse-momentum theorem gives: $-T \Delta t = m v - m v_0$,where $v_0 = 5\, m/s$ is the initial velocity and $v$ is the final velocity.
For the disc,the angular impulse-angular momentum theorem about the hinge gives: $(T \cdot r) \Delta t = I \omega$,where $I = \frac{1}{2} m r^2$ is the moment of inertia of the disc.
Since the string is taut,the tangential velocity of the disc at the rim must equal the velocity of the block,so $v = \omega r$,or $\omega = v/r$.
Substituting $\omega$ into the disc equation: $T \Delta t = \frac{I \omega}{r} = \frac{(\frac{1}{2} m r^2) (v/r)}{r} = \frac{1}{2} m v$.
Substituting $T \Delta t$ into the block equation: $-(\frac{1}{2} m v) = m v - m v_0$.
Rearranging the terms: $m v_0 = m v + \frac{1}{2} m v = \frac{3}{2} m v$.
Solving for $v$: $v = \frac{2}{3} v_0 = \frac{2}{3} \times 5 = \frac{10}{3}\, m/s$.

(A) The system consists of the table and the man. Since the table is on a smooth surface and there are no external torques acting on the system about the axis of rotation, the angular momentum $L$ of the system is conserved.
Initially, the man is at the edge of the table at a distance $R$ from the center. The initial moment of inertia of the system is $I_{initial} = I + MR^2$.
When the man moves to the center, his distance from the axis of rotation becomes zero. The final moment of inertia of the system is $I_{final} = I + M(0)^2 = I$.
Since $I_{final} < I_{initial}$, and the angular momentum $L = I\omega$ is conserved, we have $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Therefore, $\omega_{final} = \frac{I_{initial}}{I_{final}} \omega_{initial}$.
Since $I_{initial} > I_{final}$, it follows that $\omega_{final} > \omega_{initial}$.
Thus, the angular velocity of the table must increase.

(B) The angular momentum vector $\vec{L}$ of a rotating body is given by $\vec{L} = I\vec{\omega}$.
For a rod rotating about an axis passing through one end at an angle $\theta = 20^o$ with the vertical,the angular momentum vector $\vec{L}$ does not lie along the axis of rotation.
Instead,it is perpendicular to the rod. Since the rod makes an angle of $20^o$ with the vertical,the angle it makes with the horizontal is $90^o - 20^o = 70^o$.
The angular momentum vector $\vec{L}$ is perpendicular to the rod,so it makes an angle of $20^o$ with the horizontal plane.
As the rod rotates,the angular momentum vector $\vec{L}$ points downwards at an angle of $20^o$ to the horizontal,as shown in the diagram.

(B) The angular momentum vector $\vec{L}$ of the rotating rod is not aligned with the axis of rotation. As the rod rotates,the direction of the angular momentum vector changes continuously,tracing out a cone.
Since the angular momentum vector $\vec{L}$ is changing with time,there must be an external torque $\vec{\tau} = \frac{d\vec{L}}{dt}$ acting on the rod.
The change in angular momentum occurs in the horizontal plane as the rod sweeps around the vertical axis.
Therefore,the torque required to maintain this motion must be horizontal.

(C) The particle starts at $(0, 8)$ and moves with velocity $\vec{v} = 3\hat{i}\,m/s$.
After time $t = 5\,s$,the position of the particle is:
$x = x_0 + v_x t = 0 + 3 \times 5 = 15\,m$
$y = y_0 + v_y t = 8 + 0 \times 5 = 8\,m$
The position vector is $\vec{r} = 15\hat{i} + 8\hat{j}$.
The distance from the origin is $r = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\,m$.
The angular velocity $\omega$ is given by $\omega = \frac{v_{\perp}}{r}$,where $v_{\perp}$ is the component of velocity perpendicular to the position vector.
From the geometry,$\sin \theta = \frac{y}{r} = \frac{8}{17}$.
The component of velocity perpendicular to $\vec{r}$ is $v_{\perp} = v \sin \theta = 3 \times \frac{8}{17} = \frac{24}{17}\,m/s$.
Thus,$\omega = \frac{v_{\perp}}{r} = \frac{24/17}{17} = \frac{24}{289}\,rad/s$.

(D) $1)$ We know that the diagonal of a rhombus is proportional to its side length.
$\Rightarrow$ The corresponding ratio of the diagonals is $5:3:2$.
Let the lengths of the diagonals be $5y, 3y,$ and $2y$ respectively,as shown in the figure.
$2)$ Now,the velocity of $A_3$ relative to $A_0$ is given by the rate of change of the total length $A_0 A_3$:
$V_{A_3} - V_{A_0} = \frac{d(A_0 A_3)}{dt}$
Since $A_0$ is fixed,$V_{A_0} = 0$. The total length $A_0 A_3 = 5y + 3y + 2y = 10y$.
$V - 0 = \frac{d(10y)}{dt}$
$V = 10 \frac{dy}{dt} \dots (1)$
$3)$ Similarly,the velocity of $A_3$ relative to $A_2$ is the rate of change of the length $A_2 A_3$:
$V_{A_3} - V_{A_2} = \frac{d(A_2 A_3)}{dt}$
$V - V_{A_2} = \frac{d(2y)}{dt}$
$V - V_{A_2} = 2 \frac{dy}{dt} \dots (2)$
$4)$ From equation $(1)$,we have $\frac{dy}{dt} = \frac{V}{10}$.
Substituting this into equation $(2)$:
$V - V_{A_2} = 2 \left( \frac{V}{10} \right)$
$V - V_{A_2} = 0.2V$
$V_{A_2} = V - 0.2V = 0.8V$.
Thus,the velocity of $A_2$ is $0.8V$.

(C) The center of mass of the rod is at a distance $r = L/2$ from the bottom end (pivot point on the floor).
When the rod rotates with an angular acceleration $\alpha$,the tangential acceleration of the center of mass is $a_t = r \alpha = (L/2) \alpha$.
This tangential acceleration is perpendicular to the rod. The component of this acceleration in the vertical direction (which is the direction of the center of mass acceleration $A$ as the rod falls) is given by $A = a_t \sin \theta$.
Substituting the value of $a_t$,we get $A = (L/2) \alpha \sin \theta$.
Multiplying both sides by $2$,we obtain $2A = (L\alpha) \sin \theta$.

(B) For a disc rolling purely with constant angular velocity $\omega$,the velocity of the center of mass is $v_{cm} = R\omega$.
For point $P$ at the same horizontal level as the center $C$,the velocity vector $\vec{v}_P$ is the vector sum of the translational velocity $\vec{v}_{cm}$ and the tangential velocity $\vec{v}_{rot} = \vec{\omega} \times \vec{r}$.
Since $\vec{v}_{cm}$ is horizontal and $\vec{v}_{rot}$ is vertical at point $P$,the magnitude is $v_P = \sqrt{v_{cm}^2 + (R\omega)^2} = \sqrt{2} v_{cm}$. The velocity vector makes an angle of $45^o$ with the horizontal.
The acceleration of point $P$ is purely centripetal,directed towards the center $C$,which is horizontal.
Thus,the angle between the velocity vector (at $45^o$ to the horizontal) and the acceleration vector (horizontal) is $45^o$.

(C) For a ring rolling without slipping,the velocity of the center of mass is $u$ and the angular velocity is $\omega = u/R$.
The velocity of the highest point $B$ is $v_B = u + R\omega = u + u = 2u$.
The acceleration of the highest point $B$ is purely centripetal,directed towards the center of the ring,given by $a_c = R\omega^2 = R(u/R)^2 = u^2/R$.
The radius of curvature $R'$ of the path at any point is given by the formula $R' = v^2 / a_n$,where $v$ is the velocity of the particle and $a_n$ is the normal (centripetal) acceleration.
Substituting the values for the highest point $B$:
$R' = \frac{(2u)^2}{u^2/R} = \frac{4u^2}{u^2/R} = 4R$.

(C) For pure rolling,the velocity of the point of contact with the ground must be zero.
When a force $F$ is applied to the thread wound on the inner radius $r$ as shown,it creates a torque about the center of mass that tends to rotate the spool clockwise.
As the spool rotates clockwise,the point of contact with the ground has a tendency to slip to the left. To oppose this,the static friction $f$ acts towards the right.
However,let's analyze the torque about the point of contact $P$. The torque due to force $F$ is $\tau_P = F(R-r)$ in the clockwise direction. Since this torque is non-zero,the spool will have an angular acceleration $\alpha$ in the clockwise direction.
For pure rolling,the spool must move to the right with acceleration $a_{cm} = \alpha R$.
Considering the forces: $F + f = m a_{cm}$ (where $f$ is friction acting to the right).
Considering torque about the center of mass: $F r + f R = I \alpha$.
Since the spool moves to the right and rotates clockwise,the thread winds further. Thus,the spool moves to the right and friction acts to the right.

(B) For a solid cylinder rolling without slipping,the total energy is conserved. Let the vertical height difference between $A$ and $B$ be $H$.
At point $B$,the cylinder has reached the bottom of the rough incline. The translational kinetic energy $K_T$ and rotational kinetic energy $K_R$ are related by $K_T = 2K_R$ for a solid cylinder $(I = \frac{1}{2}mR^2)$.
From energy conservation,$mgH = K_T + K_R = 2K_R + K_R = 3K_R$. Thus,$K_R = \frac{1}{3}mgH$ and $K_T = \frac{2}{3}mgH$.
From $B$ to $C$,the surface is smooth,so there is no friction to exert a torque. Thus,the angular velocity $\omega$ and rotational kinetic energy $K_R$ remain constant.
The potential energy lost from $B$ to $C$ is converted entirely into translational kinetic energy. If the vertical height difference between $B$ and $C$ is also $H$,the additional translational kinetic energy gained is $mgH$.
Total translational kinetic energy at $C$ is $K_{T,C} = K_{T,B} + mgH = \frac{2}{3}mgH + mgH = \frac{5}{3}mgH$.
The rotational kinetic energy at $C$ is $K_{R,C} = K_{R,B} = \frac{1}{3}mgH$.
The ratio is $\frac{K_{T,C}}{K_{R,C}} = \frac{\frac{5}{3}mgH}{\frac{1}{3}mgH} = 5$.

(D) Let the friction force acting on the sphere be $f$ directed up the plane.
For the plank of mass $M$,the equation of motion along the plane is:
$M a_1 = M g \sin \theta + f$ --- $(1)$
where $a_1$ is the acceleration of the plank down the plane.
For the sphere of mass $m$,let $a_2$ be its acceleration relative to the plank down the plane. The torque equation about the center of the sphere is:
$f \cdot r = I \alpha = (\frac{1}{2} m r^2) \alpha$
Since the sphere rolls without slipping on the plank,$a_2 = \alpha r$,so $f = \frac{1}{2} m a_2$,which gives $a_2 = \frac{2f}{m}$.
The equation of motion for the sphere relative to the plank is:
$m a_2 = m g \sin \theta - f - m a_1$ --- $(2)$
Substituting $a_1 = g \sin \theta + \frac{f}{M}$ from $(1)$ into $(2)$:
$m (\frac{2f}{m}) = m g \sin \theta - f - m (g \sin \theta + \frac{f}{M})$
$2f = m g \sin \theta - f - m g \sin \theta - \frac{m f}{M}$
$3f = - \frac{m f}{M}$
$f (3 + \frac{m}{M}) = 0$
Since $(3 + \frac{m}{M}) \neq 0$,we must have $f = 0$.

(D) Let $M$ be the mass of the plank and $m$ be the mass of the sphere. Let $a_p$ be the acceleration of the plank and $a_s$ be the acceleration of the sphere. Since the sphere does not slip,the acceleration of the point of contact on the sphere relative to the plank must be zero. The friction $f$ acts on the sphere in the forward direction and on the plank in the backward direction. For the sphere,$f = m a_s$ and $\tau = f R = I \alpha = (\frac{2}{5} m R^2) \alpha$. Since there is no slipping,$a_s = \alpha R$,so $f = \frac{2}{5} m a_s$. This implies $a_s = \frac{f}{m}$. The acceleration of the plank is $a_p = \frac{F - f}{M}$. Since $a_s < a_p$ is generally true for such systems,statement $A$ is correct. The work done by friction on the sphere is $W_f = \Delta K_{sphere} = K_{trans} + K_{rot} = \frac{1}{2} m v_s^2 + \frac{1}{2} I \omega^2$. Since $v_s = \omega R$,$W_f = \frac{1}{2} m v_s^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v_s}{R})^2 = \frac{1}{2} m v_s^2 + \frac{1}{5} m v_s^2 = \frac{7}{10} m v_s^2$,which is the total kinetic energy of the sphere. Thus,statement $B$ is correct. According to the work-energy theorem,the total work done by all forces on the system (plank + sphere) equals the change in total kinetic energy. The only external force doing work is $F$. Friction is an internal force,so its net work on the system is zero. Thus,$W_F = \Delta K_{total}$. Statement $C$ is also correct. Since all statements $A, B,$ and $C$ are correct,the incorrect statement is none of these.

(A) The system is rolling without slipping on a horizontal surface. The point of contact with the ground is the instantaneous axis of rotation $(IAR)$.
For pure rolling,the velocity of the centre of the ring is $v_0 = R\omega$,where $\omega$ is the angular velocity.
The kinetic energy of the system is given by $KE = \frac{1}{2} I_{IAR} \omega^2$.
First,we calculate the moment of inertia $I_{IAR}$ about the point of contact.
The ring has mass $m$ and radius $R$. The particles have masses $m$,$2m$,and $m$ at positions $(0, R)$,$(-R, 0)$,and $(R, 0)$ relative to the centre,respectively.
The point of contact is at $(0, -R)$.
The squared distances of the particles from the point of contact are:
For particle of mass $m$ at $(0, R)$: $r_1^2 = 0^2 + (R - (-R))^2 = (2R)^2 = 4R^2$.
For particle of mass $2m$ at $(-R, 0)$: $r_2^2 = (-R - 0)^2 + (0 - (-R))^2 = R^2 + R^2 = 2R^2$.
For particle of mass $m$ at $(R, 0)$: $r_3^2 = (R - 0)^2 + (0 - (-R))^2 = R^2 + R^2 = 2R^2$.
For the ring of mass $m$,$I_{ring, IAR} = I_{cm} + mR^2 = mR^2 + mR^2 = 2mR^2$.
Total $I_{IAR} = m(4R^2) + 2m(2R^2) + m(2R^2) + 2mR^2 = 4mR^2 + 4mR^2 + 2mR^2 + 2mR^2 = 12mR^2$.
Substituting $I_{IAR}$ and $\omega = v_0/R$ into the kinetic energy formula:
$KE = \frac{1}{2} (12mR^2) (v_0/R)^2 = 6mR^2 (v_0^2/R^2) = 6mv_0^2$.

(C) For the sphere to come to rest,the net impulse due to friction must stop both the linear and angular motion.
Let the mass of the sphere be $m$ and its moment of inertia about the center be $I = \frac{2}{5}mR^2$.
The friction force $f$ acts in the backward direction to oppose the linear motion,and the torque $\tau = fR$ acts to oppose the angular motion.
Let $t$ be the time taken for the sphere to come to rest.
From the impulse-momentum theorem for linear motion: $f t = m v_0$.
From the angular impulse-momentum theorem: $\tau t = I \omega_0$,which gives $(fR) t = (\frac{2}{5}mR^2) \omega_0$.
Substituting $ft = m v_0$ into the angular equation: $(m v_0) R = \frac{2}{5} m R^2 \omega_0$.
Simplifying this,we get $v_0 = \frac{2}{5} \omega_0 R$,or $5v_0 = 2\omega_0 R$.

(D) Let at the bottom position,$v$ be the linear speed of the center of mass and $\omega$ be the angular speed of rotation about its center.
From the conservation of mechanical energy,the potential energy lost equals the kinetic energy gained:
$Mgl = \frac{1}{2} M v^2 + \frac{1}{2} I_{cm} \omega^2$
Since the sphere rolls without slipping,$v = R\omega$ and $I_{cm} = \frac{2}{5}MR^2$.
Substituting these values:
$Mgl = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{v}{R})^2$
$Mgl = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2$
$v = \sqrt{\frac{10gl}{7}}$
The angular momentum $L$ about point $P$ is the sum of the angular momentum about the center of mass and the angular momentum of the center of mass about $P$:
$L = I_{cm} \omega + Mvl$
$L = (\frac{2}{5} M R^2) (\frac{v}{R}) + Mvl = \frac{2}{5} MRv + Mvl$
$L = Mv (l + \frac{2}{5}R)$
Substituting $v = \sqrt{\frac{10gl}{7}}$:
$L = M \sqrt{\frac{10gl}{7}} [l + \frac{2}{5}R]$

(A) For translational motion of the center of mass $(C.M.)$:
$F + f = ma$ (where $f$ is the friction force)
For rotational motion about the $C.M.$:
$R(F - f) = I\alpha$
For pure rolling,the condition is $a = R\alpha$,which implies $\alpha = a/R$.
Substituting $\alpha$ in the rotational equation:
$R(F - f) = I(a/R) \implies F - f = \frac{I}{R^2} a$
From the translational equation,$a = \frac{F + f}{m}$. Substituting this:
$F - f = \frac{I}{R^2} \left( \frac{F + f}{m} \right)$
$mR^2(F - f) = I(F + f)$
$F(mR^2 - I) = f(mR^2 + I)$
$f = F \frac{mR^2 - I}{mR^2 + I}$
For pure rolling to occur on a smooth surface without needing external friction,we require $f = 0$.
This implies $mR^2 - I = 0$,or $I = mR^2$.
Among the given options,a thin pipe (or ring) has a moment of inertia $I = mR^2$ about its central axis.

(A) When the disc is placed on the rough surface,slipping occurs. To oppose the relative motion at the point of contact,kinetic friction $f$ acts in the direction opposite to the velocity $v_0$.
For linear motion,the force of friction $f$ provides retardation $a = \frac{f}{m}$.
For rotational motion,the torque due to friction about the center of mass is $\tau = f \cdot r$. Since the torque acts to oppose the initial angular velocity $\omega_0$,it provides angular retardation $\alpha = \frac{\tau}{I} = \frac{f \cdot r}{\frac{1}{2} m r^2} = \frac{2f}{mr}$.
If the disc comes to rest after time $t$,both linear velocity and angular velocity become zero simultaneously:
$v_0 - at = 0 \implies t = \frac{v_0}{a} = \frac{v_0 m}{f}$
$\omega_0 - \alpha t = 0 \implies t = \frac{\omega_0}{\alpha} = \frac{\omega_0 mr}{2f}$
Equating the two expressions for time $t$:
$\frac{v_0 m}{f} = \frac{\omega_0 mr}{2f}$
$\frac{v_0}{\omega_0 r} = \frac{1}{2}$

(A) Let $M$ be the mass and $R$ be the radius of the solid sphere. The moment of inertia about the center of mass is $I = \frac{2}{5}MR^2$.
Applying Newton's second law for linear motion: $F + f = Ma$,where $f$ is the friction force.
Applying torque equation about the center of mass: $F(R/2) - fR = I\alpha = (\frac{2}{5}MR^2)\alpha$.
Since there is no slipping at the point of contact,the acceleration of the center of mass $a = R\alpha$,so $\alpha = a/R$.
Substituting $\alpha$: $F/2 - f = \frac{2}{5}Ma \implies f = F/2 - \frac{2}{5}Ma$.
Substituting $f$ into the linear equation: $F + F/2 - \frac{2}{5}Ma = Ma \implies \frac{3F}{2} = \frac{7}{5}Ma \implies a = \frac{15F}{14M}$.
The angular acceleration is $\alpha = a/R = \frac{15F}{14MR}$.
The acceleration of the top point is $a_{top} = a + R\alpha = a + a = 2a$.
Therefore,$a_{top} = 2 \times \frac{15F}{14M} = \frac{15F}{7M}$.

(B) For pure rolling,the point of contact $P$ must be at rest relative to the surface.
Let $a_{cm}$ be the acceleration of the center of mass and $\alpha$ be the angular acceleration.
The condition for pure rolling is $a_{cm} = R\alpha$.
Applying Newton's second law for linear motion: $F - f = M a_{cm} \Rightarrow a_{cm} = \frac{F - f}{M}$.
Applying Newton's second law for rotational motion about the center of mass: $\tau = I\alpha \Rightarrow fR = I\alpha \Rightarrow \alpha = \frac{fR}{I}$.
Substituting these into the rolling condition: $\frac{F - f}{M} = R \left( \frac{fR}{I} \right) = \frac{fR^2}{I}$.
$F - f = \frac{M R^2}{I} f \Rightarrow F = f \left( 1 + \frac{M R^2}{I} \right)$.
Since $F$ is positive,$f$ must be positive,meaning it acts in the same direction as $F$ (forward direction).

(D) Due to translational motion only,the friction force $f^{\prime}$ should act backwards to oppose the sliding tendency.
Due to rotational motion only,the torque produced by the force $F$ (applied above the centre of mass $C$) causes a tendency to rotate,which creates a friction force $f^{\prime \prime}$ acting forwards at the point of contact.
Thus,the direction of the net friction force depends on the relative magnitudes of $f^{\prime}$ and $f^{\prime \prime}$.
Since the exact height of the point of application of force $F$ above the centre of mass $C$ is not specified,the relative magnitudes of these opposing friction components cannot be determined.
Therefore,the direction of the net friction force cannot be interpreted.

(B) When a force $F$ is applied vertically upwards on the inner radius $r$ of the spool (with outer radius $R$),it creates a torque about the center of mass of the spool.
Since the force is applied to the right of the center,it produces a clockwise torque about the center of mass.
This torque tends to rotate the spool in the clockwise direction.
Due to this clockwise rotation,the point of contact of the spool with the ground tends to move towards the left.
Therefore,the static friction force $f$ acts in the opposite direction,which is towards the right,to oppose this tendency of motion.

(C) From the conservation of linear momentum,the velocity of the center of mass $(v_{cm})$ is given by:
$m(2v) - m(v) = (2m)v_{cm} \Rightarrow v_{cm} = v/2$ towards the right.
From the conservation of angular momentum about the center of mass:
$L_i = L_f$
$m(2v)(b/2) + m(v)(b/2) = I\omega$
where $I = m(b/2)^2 + m(b/2)^2 = mb^2/2$.
$(3/2)mvb = (mb^2/2)\omega \Rightarrow \omega = 3v/b$.
The motion of the skater initially at $y = b/2$ is a combination of translation of the center of mass and rotation about it:
$x(t) = v_{cm}t + (b/2)\sin(\omega t) = 0.5vt + 0.5b\sin(3vt/b)$
$y(t) = (b/2)\cos(\omega t) = 0.5b\cos(3vt/b)$
Thus,the correct option is $C$.

(B) Let the mass of the ring $A$ be $m$ and the mass of the body $B$ be $m$. Initially,the ring is rolling without sliding,so its angular velocity is $\omega = v/R$. The total initial kinetic energy is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(v/R)^2 = mv^2$.
At the maximum height $h$,the ring and the body $B$ move together with a common horizontal velocity $v'$. By conservation of linear momentum in the horizontal direction: $mv = (m + m)v'$,which gives $v' = v/2$.
Since all surfaces are smooth,there is no friction,so the angular velocity $\omega$ of the ring remains constant throughout the motion.
Applying the conservation of mechanical energy:
$K_i = K_f + U_f$
$mv^2 = \frac{1}{2}(m+m)(v')^2 + \frac{1}{2}I\omega^2 + mgh$
$mv^2 = \frac{1}{2}(2m)(v/2)^2 + \frac{1}{2}(mR^2)(v/R)^2 + mgh$
$mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}mv^2 + mgh$
$mv^2 = \frac{3}{4}mv^2 + mgh$
$mgh = mv^2 - \frac{3}{4}mv^2 = \frac{1}{4}mv^2$
$h = \frac{v^2}{4g}$

(C) The center of mass $(CM)$ of the system is at the midpoint of the rod. The distance of each mass from the $CM$ is $r = L/2$.
The impulse $J = MV$ is applied at one end. The component of the impulse perpendicular to the rod is $J_{\perp} = J \sin 30^{\circ} = MV \sin 30^{\circ} = MV(1/2) = MV/2$.
The angular impulse about the $CM$ is $\tau_{impulse} = J_{\perp} \times r = (MV/2) \times (L/2) = MVL/4$.
The angular impulse is equal to the change in angular momentum: $\tau_{impulse} = I \omega$,where $I$ is the moment of inertia about the $CM$.
The moment of inertia of the two masses about the $CM$ is $I = M(L/2)^2 + M(L/2)^2 = 2M(L^2/4) = ML^2/2$.
Equating the angular impulse to the change in angular momentum:
$ML^2/2 \times \omega = MVL/4$
Solving for $\omega$:
$\omega = \frac{MVL/4}{ML^2/2} = \frac{MVL}{4} \times \frac{2}{ML^2} = \frac{V}{2L}$.
Thus,the angular velocity is $\frac{V}{2L}$.

(A) Given: Mass $m = 2 \, kg$,Length $L = 1 \, m$,Impulse $J = 10 \, Ns$.
The moment of inertia of the rod about its upper end is $I = \frac{1}{3} m L^2$.
Substituting the values: $I = \frac{1}{3} \times 2 \times (1)^2 = \frac{2}{3} \, kg \cdot m^2$.
Angular impulse is equal to the change in angular momentum:
$J \times L = I \omega - 0$
$10 \times 1 = \frac{2}{3} \omega$
$\omega = \frac{10 \times 3}{2} = 15 \, rad/s$.
The kinetic energy of the rod just after impact is given by:
$KE = \frac{1}{2} I \omega^2$
$KE = \frac{1}{2} \times \left(\frac{2}{3}\right) \times (15)^2$
$KE = \frac{1}{3} \times 225 = 75 \, J$.

(B) The rod will have both translational and rotational motion about its centre of mass $(C.M.)$.
From the conservation of linear momentum:
$m v_0 = M V_{CM} \implies V_{CM} = \frac{m v_0}{M} \dots(1)$
From the conservation of angular momentum about the $C.M.$:
$m v_0 x = I \omega = \left( \frac{M L^2}{12} \right) \omega \implies \omega = \frac{12 m v_0 x}{M L^2} \dots(2)$
For point $A$ to be at rest,the velocity of the $C.M.$ must be equal to the velocity of point $A$ due to rotation:
$V_{CM} = \omega \left( \frac{L}{2} \right) \dots(3)$
Substituting $(1)$ and $(2)$ into $(3)$:
$\frac{m v_0}{M} = \left( \frac{12 m v_0 x}{M L^2} \right) \left( \frac{L}{2} \right)$
$\frac{m v_0}{M} = \frac{6 m v_0 x}{M L}$
$1 = \frac{6 x}{L} \implies x = \frac{L}{6}$

(D) From the equation of linear momentum,$m v_{cm} = J$,so $v_{cm} = \frac{J}{m}$.
From the equation of angular momentum about the center of mass $(CM)$,$J \cdot \frac{l}{2} = I \omega = \frac{m l^2}{12} \cdot \omega$,which gives $\omega = \frac{6 J}{m l}$.
At time $t = \frac{\pi m l}{12 J}$,the angle rotated by the rod is $\theta = \omega t = \left( \frac{6 J}{m l} \right) \left( \frac{\pi m l}{12 J} \right) = \frac{\pi}{2}$.
The tangential velocity of particle $P$ due to rotation about the $CM$ is $v_t = r \omega = \frac{l}{6} \cdot \frac{6 J}{m l} = \frac{J}{m}$.
Since the rod has rotated by $\frac{\pi}{2}$,the velocity vector due to rotation is now perpendicular to the initial velocity vector $v_{cm}$.
The resultant velocity of particle $P$ is $v_p = \sqrt{v_{cm}^2 + v_t^2} = \sqrt{\left( \frac{J}{m} \right)^2 + \left( \frac{J}{m} \right)^2} = \sqrt{2} \frac{J}{m}$.

(A) Given: Mass of the particle $= m$,Mass of the rod $= M$,Length of the rod $= L$.
Applying the principle of conservation of angular momentum about the hinge:
$m v (L/2) = I \omega$
Since the moment of inertia of the rod about the hinge is $I = M L^2 / 3$:
$m v (L/2) = (M L^2 / 3) \omega$
$\omega = (3 m v) / (2 M L)$
Since the collision is elastic and the particle comes to rest,the initial kinetic energy of the particle is converted into the rotational kinetic energy of the rod:
$(1/2) m v^2 = (1/2) I \omega^2$
$(1/2) m v^2 = (1/2) (M L^2 / 3) [(3 m v) / (2 M L)]^2$
$m v^2 = (M L^2 / 3) [9 m^2 v^2 / (4 M^2 L^2)]$
$m v^2 = (3 m^2 v^2) / (4 M)$
$1 = (3 m) / (4 M)$
$M / m = 3 / 4$

(D) The rotational kinetic energy is given by $K = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the axis of rotation.
$1$. For rotation about the $y$-axis: The masses $m$ are on the $y$-axis,so their distance from the axis is $0$. The masses $M$ are at distance $a$ from the $y$-axis. Thus,$I_y = M(a)^2 + M(a)^2 = 2Ma^2$. The kinetic energy is $K_y = \frac{1}{2}(2Ma^2)\omega^2 = Ma^2\omega^2$. This is independent of $m$.
$2$. For rotation about the $z$-axis: The masses $M$ are at distance $a$ from the $z$-axis,and masses $m$ are at distance $b$ from the $z$-axis. Thus,$I_z = M(a)^2 + M(a)^2 + m(b)^2 + m(b)^2 = 2Ma^2 + 2mb^2$. The kinetic energy is $K_z = \frac{1}{2}(2Ma^2 + 2mb^2)\omega^2 = (Ma^2 + mb^2)\omega^2$. This depends on $m$.
$3$. Since the moment of inertia $I$ depends on the axis of rotation,the rotational kinetic energy $K = \frac{1}{2}I\omega^2$ also depends on the axis of rotation.
Therefore,all statements are correct.

(D) The angular momentum $L$ of the particle about point $O$ is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicular distance from $O$ to the line of motion. Since the particle is falling freely,its speed $v$ increases with time,while $d$ remains constant. Therefore,the angular momentum $L$ increases.
The moment of inertia $I$ of the particle about $O$ is $I = mr^2$,where $r$ is the distance from $O$ to the particle. As the particle falls,the distance $r$ decreases,so the moment of inertia $I$ decreases.
The angular velocity $\omega$ of the particle about $O$ is given by $\omega = \frac{v_{\perp}}{r} = \frac{v \sin \theta}{r}$. As the particle falls,$v$ increases,$\sin \theta$ increases (since $\theta$ increases),and $r$ decreases. Consequently,the angular velocity $\omega$ increases.
Since all statements are correct,the correct option is $D$.

(D) In free space,there is no external torque acting on the man,so his angular momentum $(L)$ remains conserved.
By changing the shape of his body (spreading arms or curling up),the man changes his moment of inertia $(I)$.
Since $L = I \omega$,if $I$ changes,the angular velocity $(\omega)$ must also change to keep $L$ constant.
Furthermore,the rotational kinetic energy is given by $K.E. = \frac{L^2}{2I}$. Since $L$ is constant,a change in $I$ results in a change in rotational kinetic energy.
Therefore,by changing his body shape,the man can change his moment of inertia,angular velocity,and rotational kinetic energy.
Thus,the correct option is $D$.

(D) For a ring rolling without slipping,the velocity of the point of contact $A$ with the ground is zero,so $v_A = u - R\omega = 0$,which implies $u = R\omega$.
The velocity of the top point $B$ is $v_B = u + R\omega = 2u$.
Thus,the range of speeds for any point $P$ on the ring is $0 \leq v \leq 2u$.
Let $P$ be a point such that the radius $CP$ makes an angle $\theta$ with the horizontal. The velocity vector of $P$ is the vector sum of the translational velocity $\vec{u}$ and the rotational velocity $\vec{v}_{rot} = \vec{\omega} \times \vec{r}$.
The magnitude of velocity is $v_P = \sqrt{u^2 + (R\omega)^2 + 2u(R\omega)\cos(\theta + 90^\circ)}$.
Since $u = R\omega$,$v_P = \sqrt{u^2 + u^2 + 2u^2(-\sin\theta)} = u\sqrt{2(1 - \sin\theta)}$.
For option $B$: If $CP$ is horizontal,$\theta = 0^\circ$,then $v_P = u\sqrt{2(1 - 0)} = u\sqrt{2}$. This is correct.
For option $C$: If $\theta = 30^\circ$ and $P$ is below the horizontal,then $v_P = u\sqrt{2(1 - \sin 30^\circ)} = u\sqrt{2(1 - 0.5)} = u\sqrt{2(0.5)} = u$. This is also correct.
Therefore,all statements are correct.

(C) The motion of the centre of mass of a body is determined by the net external force acting on it,according to Newton's second law,$F_{ext} = Ma_{cm}$.
For the yo-yo to move to the right,the horizontal component of the applied force must be directed towards the right.
$1$. Force $F_1$ is applied horizontally to the right. Therefore,when $F_1$ is applied,the net force on the yo-yo is to the right,causing the centre of mass to move to the right.
$2$. Force $F_3$ is applied vertically upwards. It has no horizontal component,so it will not cause the centre of mass to move horizontally.
$3$. Force $F_2$ is applied at an angle such that its horizontal component is directed towards the right. Therefore,when $F_2$ is applied,the centre of mass will also move to the right.
Comparing the options,option $C$ is a correct statement.

(D) Let $F$ be the applied force,$m$ be the mass,and $R$ be the radius of the disc. Circumference $s = 2\pi R$.
Between $A$ and $B$,the disc rolls without slipping. The equations of motion are:
$F - f = ma_1$ and $fR = I\alpha = (\frac{1}{2}mR^2)\alpha$. Since $a_1 = R\alpha$,we get $f = \frac{1}{2}ma_1$. Substituting this,$F - \frac{1}{2}ma_1 = ma_1 \implies F = \frac{3}{2}ma_1 \implies a_1 = \frac{2F}{3m}$.
The distance $AB = s = 2\pi R = \frac{1}{2}a_1 T^2$. Thus,$T^2 = \frac{4\pi R}{a_1} = \frac{4\pi R}{2F/3m} = \frac{6\pi mR}{F}$.
At point $B$,the linear velocity $v_B = a_1 T$ and angular velocity $\omega_B = \alpha T = \frac{a_1}{R}T$.
To the right of $B$,the surface is smooth,so friction $f = 0$. The new linear acceleration $a_2 = F/m$. Since $F = \frac{3}{2}ma_1$,$a_2 = \frac{3}{2}a_1 > a_1$. Thus,linear acceleration increases.
In time $T$ after $B$,the distance covered is $d = v_B T + \frac{1}{2}a_2 T^2 = (a_1 T)T + \frac{1}{2}(\frac{3}{2}a_1)T^2 = a_1 T^2 + \frac{3}{4}a_1 T^2 = \frac{7}{4}a_1 T^2$. Since $s = \frac{1}{2}a_1 T^2$,$d = \frac{7}{2}s > s$.
Since there is no friction to the right of $B$,the angular velocity remains constant at $\omega_B = \frac{a_1 T}{R}$. The time taken for one rotation is $t = \frac{2\pi}{\omega_B} = \frac{2\pi R}{a_1 T} = \frac{s}{a_1 T} = \frac{\frac{1}{2}a_1 T^2}{a_1 T} = \frac{T}{2}$.
All statements are correct.

(C) The total mass of the system is $M = m + m = 2m$.
Since the sphere is hollow and filled with water,the moment of inertia of the shell is $I_{shell} = \frac{2}{3} m R^2$.
Assuming the water behaves as a rigid body rotating with the shell,the total moment of inertia is $I = I_{shell} + I_{water} = \frac{2}{3} m R^2 + 0$ (assuming water is non-viscous and does not rotate,but in standard physics problems of this type,we treat the system as a rigid body with $I = \frac{2}{3} m R^2$).
For pure rolling,$\omega = \frac{v}{R}$.
The angular momentum about a point on the ground is $L = I_{cm} \omega + M v R$.
Substituting the values: $L = (\frac{2}{3} m R^2) \times (\frac{v}{R}) + (2m) v R$.
$L = \frac{2}{3} m v R + 2 m v R = \frac{8}{3} m v R$.

(D) Let $v_{cm}$ be the velocity of the centre of mass of the cylinder and $\omega$ be its angular velocity.
At the contact point between the cylinder and the plank (bottom point),the velocity of the cylinder must match the velocity of the plank because there is no slipping. Thus,$v_{cm} - R\omega = v$ (assuming $v_{cm}$ is to the right and $\omega$ is clockwise).
At the contact point between the cylinder and the top string (top point),the velocity of the cylinder must match the velocity of the string. Since the string is fixed to the wall,its velocity is $0$. Thus,$v_{cm} + R\omega = 0$.
Adding these two equations: $(v_{cm} - R\omega) + (v_{cm} + R\omega) = v + 0$,which gives $2v_{cm} = v$,so $v_{cm} = v/2$.
Subtracting the equations: $(v_{cm} - R\omega) - (v_{cm} + R\omega) = v - 0$,which gives $-2R\omega = v$,so $\omega = -v/(2R)$ (the negative sign indicates counter-clockwise rotation).
Therefore,the speed of the centre of mass of the cylinder is $v/2$.

(D) For a disc of mass $m$ and radius $R$ rolling with angular velocity $\omega$ and linear velocity $v = R\omega$:
The angular momentum about the center of mass $O$ is $L_O = I_O \omega = (\frac{1}{2}mR^2)\omega$.
The angular momentum about any point $P$ is given by $\vec{L}_P = \vec{L}_{cm} + \vec{r}_{cm/P} \times m\vec{v}_{cm}$.
For point $B$ (point of contact): $\vec{L}_B = I_O \omega + mR^2 \omega = \frac{1}{2}mR^2 \omega + mR^2 \omega = \frac{3}{2}mR^2 \omega$.
For point $A$ (topmost point): $\vec{L}_A = I_O \omega - mR^2 \omega = \frac{1}{2}mR^2 \omega - mR^2 \omega = -\frac{1}{2}mR^2 \omega$.
The magnitude $|L_B| = \frac{3}{2}mR^2 \omega$ and $|L_A| = \frac{1}{2}mR^2 \omega$. Thus,$|L_B| = 3|L_A|$.
Since $L_A$ is negative,it is anticlockwise,and $L_B$ is positive,it is clockwise. Therefore,all statements are correct.

(D) When a cylinder rolls without sliding on an inclined plane,the friction force $f$ depends on the net torque and the net force acting on the body.
For a cylinder of mass $M$ and radius $R$ on an incline of angle $\theta$,the equation of motion is $Mg \sin \theta - f = Ma$ and the torque equation about the center of mass is $fR = I\alpha = (\frac{1}{2}MR^2)(\frac{a}{R})$.
Solving these,we get $f = \frac{1}{3}Mg \sin \theta$,which is directed upward.
However,if an external force (like an applied force $F$) is present,the direction of friction can change to oppose the tendency of sliding or to provide the necessary torque for rolling.
If the external force is applied such that the required torque for rolling is already provided,the friction force can be zero.
Thus,friction can be upward,downward,or zero depending on the external force and the conditions of motion.
Therefore,statements $(B)$ and $(C)$ are correct.

(D) When the ring enters the rough surface,it is purely sliding forward with velocity $v_0$ and has no initial angular velocity $(\omega = 0)$.
$1$. Since the ring is sliding,the kinetic friction force $f = \mu N = \mu Mg$ acts on it,which is the limiting friction.
$2$. The friction force acts at the point of contact in a direction opposite to the velocity of the point of contact,which is opposite to the direction of motion of the ring.
$3$. The friction force $f$ creates a torque $\tau = f \times R$ about the centre of mass. Since the friction acts at the bottom point towards the left,the torque is in the clockwise direction,which causes an angular acceleration $\alpha$ in the clockwise sense.
Therefore,all the given statements are correct.

(C) $1$. When the ring enters the rough surface,a kinetic frictional force $f = \mu Mg$ acts on the ring in the direction opposite to its velocity $v_0$.
$2$. Since an external force (friction) acts on the ring,the linear momentum is not conserved.
$3$. The frictional force exerts a torque about the centre of mass $(CM)$ of the ring,causing it to rotate. Therefore,the angular momentum about the $CM$ is not conserved.
$4$. The frictional force acts at the point of contact with the surface. If we choose any point on the horizontal surface as the reference point,the line of action of the frictional force passes through this point. Thus,the torque due to friction about any point on the horizontal surface is zero. Consequently,the angular momentum of the ring is conserved about any point on the horizontal surface.
$5$. Since friction is a non-conservative force,it does work on the system,so the mechanical energy is not conserved.

(D) When the ring enters the rough surface,kinetic friction acts in the direction opposite to the velocity $v_0$. The force of friction is $f = \mu Mg$.
This force causes a retardation in linear motion: $a = -\frac{f}{M} = -\mu g$.
Also,this friction force provides a torque about the center of mass: $\tau = fR = \mu MgR$.
The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{\mu MgR}{MR^2} = \frac{\mu g}{R}$.
Let $t$ be the time when pure rolling starts. At this time,the velocity of the center of mass is $v = v_0 - at = v_0 - \mu gt$ and the angular velocity is $\omega = \alpha t = \frac{\mu gt}{R}$.
For pure rolling,$v = \omega R$,so $v_0 - \mu gt = (\frac{\mu gt}{R})R = \mu gt$.
This gives $2\mu gt = v_0$,or $t = \frac{v_0}{2\mu g}$.
The velocity at this time is $v = v_0 - \mu g(\frac{v_0}{2\mu g}) = v_0 - \frac{v_0}{2} = \frac{v_0}{2}$.
Pure rolling starts when the point of contact becomes stationary relative to the surface. Thus,all statements are correct.