Rolling motion on horizontal Surface Questions in English

(C) Let $m$ be the mass and $R$ be the radius of the solid sphere. The force $F$ is applied at the top point.
$1$. Equation of linear motion: $F + f = ma$,where $f$ is the friction force acting at the point of contact $A$ in the direction of motion.
$2$. Equation of rotational motion about the center of mass: $F \times R - f \times R = I \alpha$,where $I = \frac{2}{5} mR^2$ is the moment of inertia and $\alpha = \frac{a}{R}$ is the angular acceleration.
Substituting $I$ and $\alpha$: $F R - f R = (\frac{2}{5} mR^2) (\frac{a}{R}) \Rightarrow F - f = \frac{2}{5} ma$.
$3$. Adding the two equations: $(F + f) + (F - f) = ma + \frac{2}{5} ma \Rightarrow 2F = \frac{7}{5} ma \Rightarrow F = \frac{7}{10} ma$.

(C) The total kinetic energy of the system is the sum of the kinetic energy of the ring and the kinetic energy of the rod.
For rolling without slipping,$\omega = \frac{v}{R}$.
Kinetic energy of the ring $(KE_{ring})$: $KE_{ring} = \frac{1}{2}mv^2 + \frac{1}{2}I_{ring}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Kinetic energy of the rod $(KE_{rod})$: The rod rotates about the center of the ring. The moment of inertia of a rod of length $2R$ about its center is $I_{rod} = \frac{m(2R)^2}{12} = \frac{4mR^2}{12} = \frac{1}{3}mR^2$.
$KE_{rod} = \frac{1}{2}mv^2 + \frac{1}{2}I_{rod}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{3}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{6}mv^2 = \frac{3+1}{6}mv^2 = \frac{4}{6}mv^2 = \frac{2}{3}mv^2$.
Total kinetic energy = $KE_{ring} + KE_{rod} = mv^2 + \frac{2}{3}mv^2 = \frac{5}{3}mv^2$.

(C) The velocity of a point $P$ on the circumference of a rolling disc is given by $v_p = 2v \sin(\theta/2)$,where $v$ is the velocity of the center of mass and $\theta = \omega t$ is the angle rotated.
Since $v = R\omega$,we have $v_p = 2R\omega \sin(\omega t / 2)$.
The distance $s$ moved by point $P$ in one full rotation (time $T = 2\pi/\omega$) is the integral of the speed:
$s = \int_{0}^{T} v_p dt = \int_{0}^{2\pi/\omega} 2R\omega \sin(\omega t / 2) dt$
$s = 2R\omega \left[ -\frac{2}{\omega} \cos(\omega t / 2) \right]_{0}^{2\pi/\omega}$
$s = -4R [\cos(\pi) - \cos(0)] = -4R [-1 - 1] = 8R$.

(D) The angular momentum of the sphere about the point of contact with the ground remains conserved because the frictional force acts through the point of contact,exerting no torque about it.
Initial angular momentum about the point of contact:
$L_i = M V_0 r + I_{cm} \omega = M V_0 r + (\frac{2}{5} M r^2) (\frac{V_0}{2r}) = M V_0 r + \frac{1}{5} M V_0 r = \frac{6}{5} M V_0 r$
Final angular momentum when pure rolling starts (let final velocity be $v'$ and angular velocity be $\omega' = \frac{v'}{r}$):
$L_f = M v' r + I_{cm} \omega' = M v' r + (\frac{2}{5} M r^2) (\frac{v'}{r}) = M v' r + \frac{2}{5} M v' r = \frac{7}{5} M v' r$
Equating $L_i = L_f$:
$\frac{6}{5} M V_0 r = \frac{7}{5} M v' r$
$v' = \frac{6}{7} V_0$

(A) According to the principle of conservation of energy,the potential energy stored in the spring is converted into the total kinetic energy of the rolling cylinder (translational + rotational).
$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$
Since the cylinder rolls without slipping,$\omega = \frac{v}{R}$. For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$.
Substituting these into the energy equation:
$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$
$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$
Solving for $v$:
$v = \sqrt{\frac{2kx^2}{3M}} = x \sqrt{\frac{2k}{3M}}$
Given $M = 3 \ kg$,$k = 8 \ N/m$,and $x = 1 \ m$:
$v = 1 \times \sqrt{\frac{2 \times 8}{3 \times 3}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \approx 1.33 \ m/s$.

(C) Let the disc roll without slipping with velocity $v$ and angular velocity $\omega = v/R$. The position of point $P$ at time $t$ relative to the point of contact is given by the cycloid path. The velocity of point $P$ is $v_p = 2v \sin(\theta/2)$,where $\theta = \omega t$.
In one full rotation,$t$ goes from $0$ to $T = 2\pi/\omega$.
The distance $s$ traveled by point $P$ is the arc length of the cycloid:
$s = \int_{0}^{T} v_p dt = \int_{0}^{2\pi/\omega} 2v \sin(\omega t / 2) dt$.
$s = 2v [ -\frac{2}{\omega} \cos(\omega t / 2) ]_{0}^{2\pi/\omega}$.
$s = -\frac{4v}{\omega} [ \cos(\pi) - \cos(0) ] = -\frac{4v}{\omega} [ -1 - 1 ] = \frac{8v}{\omega}$.
Since $v = \omega R$,we have $s = 8R$.

(B) Let $a$ be the linear acceleration of the center of mass and $\alpha$ be the angular acceleration of the spool.
$1$. Force equation for translational motion:
$T - f = ma$ --- $(i)$
$2$. Torque equation about the center of mass:
$f R_2 - T R_1 = I \alpha$ --- $(ii)$
$3$. Condition for rolling without slipping:
$a = \alpha R_2 \implies \alpha = \frac{a}{R_2}$ --- $(iii)$
Substituting $(iii)$ into $(ii)$:
$f R_2 - T R_1 = I \left( \frac{a}{R_2} \right)$
$f R_2^2 - T R_1 R_2 = Ia$
$a = \frac{f R_2^2 - T R_1 R_2}{I}$ --- $(iv)$
Substitute $(iv)$ into $(i)$:
$T - f = m \left( \frac{f R_2^2 - T R_1 R_2}{I} \right)$
$I(T - f) = m f R_2^2 - m T R_1 R_2$
$IT - If = m f R_2^2 - m T R_1 R_2$
$IT + m T R_1 R_2 = If + m f R_2^2$
$T(I + m R_1 R_2) = f(I + m R_2^2)$
$f = \left( \frac{I + m R_1 R_2}{I + m R_2^2} \right) T$
Since the result is positive,the assumed direction of friction (towards the left) is correct.

(A) For a body rolling without slipping,the velocity of the point of contact with the ground is zero,but its acceleration is not zero. The acceleration of the point of contact is directed towards the center of the sphere and is given by $a = \frac{v^2}{R}$,where $v$ is the velocity of the center of mass and $R$ is the radius. Thus,the statement that the acceleration of the point of contact is zero is incorrect.
Regarding the other options:
$1$. The acceleration of the $C.O.M.$ depends on external forces; if there is no external horizontal force,$a_{cm} = 0$.
$2$. Friction force is zero if the body rolls at a constant velocity on a horizontal surface without any external force,but it is non-zero if there is an external force or acceleration.
$3$. Since the point of contact is instantaneously at rest,the work done by static friction on a body rolling without slipping is always zero.

(B) The force $F$ acts at the top of the sphere,which is at a distance $2R$ from the point of contact. When the centre of mass $(CM)$ moves by a distance $S$,the point of application of force moves by $2S$.
Thus,the work done by force $F$ is $W_F = F(2S) = 2FS$.
Since the sphere rolls without slipping,the work done by static friction is zero.
Using the work-energy theorem: $W_F + W_{friction} = \Delta K$.
$2FS = \frac{1}{2}Mv^2 + \frac{1}{2}I_{cm}\omega^2$.
For a solid sphere,$I_{cm} = \frac{2}{5}MR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$.
$2FS = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$.
$2FS = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$.
$v^2 = \frac{20FS}{7M} \implies v = \sqrt{\frac{20FS}{7M}}$.
Therefore,option $B$ is correct.

(B) Let $\vec{v}_B = \vec{v}$ be the velocity of the centre of the sphere.
For pure rolling on a fixed horizontal surface,the velocity of the point of contact $A$ is $\vec{v}_A = 0$.
The velocity of the topmost point $C$ is $\vec{v}_C = 2\vec{v}$.
Now,calculating the differences:
$\vec{v}_C - \vec{v}_B = 2\vec{v} - \vec{v} = \vec{v}$
$\vec{v}_B - \vec{v}_A = \vec{v} - 0 = \vec{v}$
Therefore,$\vec{v}_C - \vec{v}_B = \vec{v}_B - \vec{v}_A$.

(C) Let the mass of the ball be $m$ and its radius be $R$. The moment of inertia of the ball about its center of mass is $I = \frac{2}{5} mR^2$.
Initially,the ball has only translational velocity $v_0$ and no angular velocity.
As the ball slides,friction acts on it,creating a torque about the point of contact $O$ on the ground. Since the torque due to friction about point $O$ is zero,the angular momentum of the ball about point $O$ is conserved.
Initial angular momentum about $O$: $L_i = m v_0 R$.
When the ball starts rolling without slipping,its velocity is $v$ and its angular velocity is $\omega = \frac{v}{R}$.
Final angular momentum about $O$ (using the parallel axis theorem): $L_f = I_{cm} \omega + m v R = (\frac{2}{5} mR^2) \frac{v}{R} + m v R = \frac{2}{5} m v R + m v R = \frac{7}{5} m v R$.
Equating initial and final angular momentum: $m v_0 R = \frac{7}{5} m v R$.
Solving for $v$: $v = \frac{5}{7} v_0$.

(B) rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground. Hence,the instantaneous speed of the point of contact is zero.
Thus,statement $(a)$ is correct.
As the body is rotating,its instantaneous acceleration is not zero because the point of contact has a centripetal acceleration directed towards the center of the body.
Hence,statement $(b)$ is incorrect.
Once perfect rolling begins,the force of static friction does no work because the point of contact is at rest relative to the surface. Hence,work done against friction is zero.
Thus,statement $(c)$ is correct.
Rolling cannot take place in the absence of friction because it is the frictional force that provides the necessary torque which makes the body roll on a surface. When the inclined plane is perfectly smooth,the wheel will simply slip under the effect of its weight.
Hence,statement $(d)$ is correct.

(C) The angular momentum of a body about a point $P$ is given by $\vec{L}_P = \vec{L}_{cm} + \vec{r} \times \vec{p}_{cm}$,where $\vec{L}_{cm} = I_{cm} \vec{\omega}$ is the angular momentum about the center of mass and $\vec{r} \times \vec{p}_{cm}$ is the angular momentum due to the motion of the center of mass.
For a solid sphere,the moment of inertia about the center of mass is $I_{cm} = \frac{2}{5} mR^2$.
In pure rolling motion,the angular velocity is $\omega = \frac{v_0}{R}$.
The angular momentum about the center of mass is $L_{cm} = I_{cm} \omega = (\frac{2}{5} mR^2) (\frac{v_0}{R}) = \frac{2}{5} mv_0R$.
The angular momentum due to the motion of the center of mass about the point of contact $P$ is $L_{trans} = m v_0 R$.
Since both vectors point in the same direction (into the plane),the total angular momentum is $L_P = L_{cm} + L_{trans} = \frac{2}{5} mv_0R + mv_0R = \frac{7}{5} mv_0R$.

(D) For a disc rolling without slipping on a horizontal surface, the velocity of any point on the disc is the vector sum of the velocity of the centre of mass $(v_{cm})$ and the tangential velocity due to rotation $(v_{rot} = R\omega)$.
Since the disc is rolling without slipping, the velocity of the point of contact with the ground is zero, which implies $v_{cm} = R\omega$.
At the highest point, the velocity is $v_{top} = v_{cm} + R\omega = v_{cm} + v_{cm} = 2v_{cm}$.
At the point of contact, the velocity is $v_{contact} = v_{cm} - R\omega = v_{cm} - v_{cm} = 0$.
Therefore, the velocity of the highest point is $2v_{cm}$ and the velocity of the point of contact is zero.

(B) Let the velocity of the centre of mass $B$ be $\vec{V}_B = V\hat{i}$.
For pure rolling,the velocity of the point of contact $A$ is $\vec{V}_A = 0$.
The velocity of the topmost point $C$ is $\vec{V}_C = \vec{V}_B + \vec{\omega} \times \vec{r}_{BC} = V\hat{i} + V\hat{i} = 2V\hat{i}$.
Checking the statements:
$(i) \vec{V}_C - \vec{V}_A = 2V\hat{i} - 0 = 2V\hat{i}$.
$2(\vec{V}_B - \vec{V}_C) = 2(V\hat{i} - 2V\hat{i}) = -2V\hat{i}$.
Since $2V\hat{i} \neq -2V\hat{i}$,statement $(i)$ is incorrect.
$(ii) \vec{V}_C - \vec{V}_B = 2V\hat{i} - V\hat{i} = V\hat{i}$.
$\vec{V}_B - \vec{V}_A = V\hat{i} - 0 = V\hat{i}$.
Since $V\hat{i} = V\hat{i}$,statement $(ii)$ is correct.
$(iii) |\vec{V}_C - \vec{V}_A| = |2V\hat{i} - 0| = 2V$.
$2|\vec{V}_B - \vec{V}_C| = 2|V\hat{i} - 2V\hat{i}| = 2|-V\hat{i}| = 2V$.
Since $2V = 2V$,statement $(iii)$ is correct.
$(iv) |\vec{V}_C - \vec{V}_A| = 2V$.
$4|\vec{V}_B| = 4V$.
Since $2V \neq 4V$,statement $(iv)$ is incorrect.
Thus,statements $(ii)$ and $(iii)$ are correct.

(C) The total kinetic energy $(KE)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.
$KE = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and $\omega = \frac{v}{R}$.
Substituting these,$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv^2(\frac{7}{5})$.
Given: $m = 50\,g$,$v = 5\,cm/s$.
$KE = \frac{1}{2} \times 50 \times (5)^2 \times \frac{7}{5} = 25 \times 25 \times 1.4 = 875\,ergs$.

(C) From the free body diagram of the cylinder:
$1$. The equation of translational motion is:
$ma = F - f$ ---$(i)$
where $f$ is the frictional force acting at the point of contact.
$2$. The equation of rotational motion about the centre of mass is:
$\tau = I\alpha$
$fR = \left(\frac{1}{2}mR^2\right)\alpha$
Since the cylinder is rolling without slipping,the condition $a = R\alpha$ holds,so $\alpha = \frac{a}{R}$.
Substituting this into the torque equation:
$fR = \left(\frac{1}{2}mR^2\right)\left(\frac{a}{R}\right)$
$f = \frac{1}{2}ma$ ---(ii)
$3$. Substituting the value of $f$ from equation (ii) into equation $(i)$:
$ma = F - \frac{1}{2}ma$
$F = ma + \frac{1}{2}ma = \frac{3}{2}ma$

(D) Let $F$ be the friction force acting at point $P$. The acceleration of the center of mass of the cylinder is $a_{cm} = \frac{F}{M}$.
The angular acceleration $\alpha$ of the cylinder about its center is given by $\tau = I\alpha$,where $\tau = F \cdot r$ and $I = \frac{Mr^2}{2}$.
Thus,$F \cdot r = \frac{Mr^2}{2} \alpha \Rightarrow \alpha = \frac{2F}{Mr}$.
Since the cylinder does not slip on the rug,the acceleration of the point $P$ on the cylinder must be equal to the acceleration of the rug,$a$.
The acceleration of point $P$ is $a_P = a_{cm} + \alpha r$ (in the direction of $a$).
Therefore,$a = \frac{F}{M} + \left(\frac{2F}{Mr}\right)r = \frac{F}{M} + \frac{2F}{M} = \frac{3F}{M}$.
Solving for $F$,we get $F = \frac{Ma}{3}$.

(B) Given: Mass $M = 5\, kg$,Radius $R = 0.5\, m$,Force $F = 40\, N$. For a hollow cylinder,the moment of inertia about its central axis is $I = MR^2$.
Let $f$ be the friction force acting at the point of contact in the backward direction.
The equation for linear motion is: $F + f = Ma$,where $a = R\alpha$ for rolling without slipping.
So,$40 + f = M(R\alpha) \quad (i)$
The equation for rotational motion about the center of mass is: $(F \times R) - (f \times R) = I\alpha$.
Substituting $I = MR^2$: $(40 - f)R = (MR^2)\alpha \implies 40 - f = MR\alpha \quad (ii)$
Adding equations $(i)$ and $(ii)$:
$(40 + f) + (40 - f) = MR\alpha + MR\alpha$
$80 = 2MR\alpha$
$80 = 2 \times 5 \times 0.5 \times \alpha$
$80 = 5\alpha$
$\alpha = 16\, rad/s^2$.

(A) When a round body is given only translational velocity $v$ on a rough surface,the velocity of the point of contact is $v$ in the forward direction.
Since there is friction,it acts in the backward direction to oppose this relative motion.
This friction force creates a torque about the center of mass,which causes the body to start rotating with angular velocity $\omega$.
As $\omega$ increases,the velocity of the point of contact $v_{contact} = v - \omega R$ decreases.
The body continues to move forward while its rotational speed increases until the condition for pure rolling,$v = \omega R$,is satisfied.
Therefore,the body moves forward during the entire process until pure rolling is achieved.

(A) The total kinetic energy $(KE_{total})$ of a rolling body is given by the sum of translational and rotational kinetic energy: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 (1 + \frac{K^2}{R^2})$.
For a circular disc,the radius of gyration $K$ is given by $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
Substituting this into the formula: $KE_{total} = \frac{1}{2}mv^2 (1 + \frac{1}{2}) = \frac{1}{2}mv^2 (\frac{3}{2}) = \frac{3}{4}mv^2$.
We know that translational kinetic energy is $KE_{trans} = \frac{1}{2}mv^2$.
Therefore,$KE_{trans} = \frac{2}{3} KE_{total}$.
Given $KE_{total} = 300 \, J$,we get $KE_{trans} = \frac{2}{3} \times 300 \, J = 200 \, J$.

(C) In pure translational motion,at any instant of time,all particles of the body have the same velocity.
However,in rolling motion (which is a combination of translational and rotational motion),the velocity of any point on the body is the vector sum of its translational velocity and its rotational velocity.
Since the rotational velocity depends on the distance of the point from the axis of rotation,different points on the body have different velocities at any given instant.
Therefore,the magnitudes of the velocities of points $P_1$ and $P_2$ are not equal,i.e.,$\left| {{{\vec v}_1}} \right| \ne \left| {{{\vec v}_2}} \right|$.

(D) The frictional force always acts in the direction opposite to the tendency of motion of the contact point on the sphere relative to the surface.
$1$. For sphere $P$: It is given a clockwise spin. The contact point on the sphere has a tendency to move towards the left. Therefore,the friction force acts towards the right.
$2$. For sphere $Q$: It is given a forward linear velocity $v$. The contact point on the sphere has a tendency to move towards the right. Therefore,the friction force acts towards the left.
$3$. For sphere $R$: It is given a forward velocity $v$ and a counter-clockwise spin $\omega$. The velocity of the contact point is $v_{cp} = v - R\omega$. Since the sphere is rotating such that the bottom point moves left due to rotation and right due to translation,if $v > R\omega$,the net tendency of the contact point is to the right,so friction acts to the left. However,looking at the figure,the rotation is counter-clockwise,meaning the bottom point moves to the left due to rotation. If $v < R\omega$,the net tendency is to the left,so friction acts to the right. Given the standard representation of such problems,the net tendency for $R$ is to the right,resulting in friction to the left. Wait,let's re-evaluate: For $R$,$v$ is right,$\omega$ is counter-clockwise. The velocity of the contact point is $v_{cp} = v - R\omega$. If $v > R\omega$,friction is left. If $v < R\omega$,friction is right. Based on the standard interpretation of this specific problem,the correct sequence is Right,Left,Left.

(C) When a body rolls without slipping,its total kinetic energy $(K.E.)$ is the sum of its translational kinetic energy and rotational kinetic energy.
$K.E. = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$. Since it rolls without slipping,$\omega = \frac{v}{r}$.
Substituting these into the equation:
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$
$K.E. = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Given $m = 1\, kg$ and $v = 1\, m/s$:
$K.E. = \frac{7}{10} \times 1 \times (1)^2 = 0.7\, J$.

(A) For a disc rolling without slipping,the translational kinetic energy is $K_t = \frac{1}{2}mv^2$.
The rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2$. Since $I = \frac{1}{2}mR^2$ and $\omega = v/R$,we have $K_r = \frac{1}{2}(\frac{1}{2}mR^2)(v/R)^2 = \frac{1}{4}mv^2$.
The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
The ratio of translational kinetic energy to total kinetic energy is $\frac{K_t}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{3}{4}mv^2} = \frac{1/2}{3/4} = \frac{2}{3}$.

(A) rolling motion can be considered as a combination of pure translation of the centre of mass $C$ with velocity $V_C = R\omega$ and pure rotation about the centre $C$ with angular velocity $\omega$.
For any point at a distance $r$ from the centre $C$,the velocity due to rotation is $v_{rot} = r\omega$.
The velocity of any point is the vector sum of the translational velocity $\vec{V}_C$ and the rotational velocity $\vec{v}_{rot}$.
For points $P$ and $Q$ on the horizontal diameter:
$1$. At point $Q$,the rotational velocity is in the same direction as the translational velocity. Thus,$V_Q = V_C + r\omega = R\omega + r\omega = (R+r)\omega$.
$2$. At point $P$,the rotational velocity is in the opposite direction to the translational velocity. Thus,$V_P = |V_C - r\omega| = |R\omega - r\omega| = (R-r)\omega$.
Since $R > r$,we have $V_Q > V_C > V_P$.

(B) For a wheel rolling without slipping,the velocity of any point $P$ on the rim is the vector sum of the translational velocity of the center of mass $(v)$ and the tangential velocity due to rotation ($v$ relative to the center).
The translational velocity vector is $\vec{v}_{cm} = v\hat{i}$.
The rotational velocity vector at point $P$ is $\vec{v}_{rot} = v\sin\theta\hat{i} + v\cos\theta\hat{j}$ (based on the geometry of the angle $\theta$ with the vertical).
The resultant velocity $\vec{v}_P = \vec{v}_{cm} + \vec{v}_{rot} = (v + v\sin\theta)\hat{i} + v\cos\theta\hat{j}$.
The magnitude is $V_P = \sqrt{(v + v\sin\theta)^2 + (v\cos\theta)^2} = \sqrt{v^2(1 + \sin^2\theta + 2\sin\theta + \cos^2\theta)} = \sqrt{v^2(2 + 2\sin\theta)} = v\sqrt{2(1 + \sin\theta)}$.
However,if $\theta$ is defined such that the velocity components result in the standard form $2v\cos(\theta/2)$,we use the vector addition of two velocities of magnitude $v$ with an angle $\theta$ between them: $V_P = \sqrt{v^2 + v^2 + 2v^2\cos\theta} = \sqrt{2v^2(1 + \cos\theta)} = \sqrt{2v^2(2\cos^2(\theta/2))} = 2v\cos(\theta/2)$.

(D) For a disc rolling without slipping on a horizontal rough surface with uniform angular velocity $\omega$,the velocity of the point of contact is zero.
However,the acceleration of the lowest point is not zero.
The acceleration of any point on the rim of the disc consists of two components: centripetal acceleration $(a_c = \omega^2 R)$ directed towards the center and tangential acceleration $(a_t = \alpha R)$.
Since the angular velocity is uniform,$\alpha = 0$,so $a_t = 0$.
The centripetal acceleration $a_c = \omega^2 R$ is directed vertically upwards towards the center of the disc.
Therefore,the net acceleration of the lowest point is $\omega^2 R$ (upwards),which is non-zero.
Thus,the $Assertion$ is incorrect and the $Reason$ is correct.

(A) The work required to stop the disc is equal to the change in its total kinetic energy.
Total kinetic energy of a rolling disc is given by $KE = KE_{translational} + KE_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a disc,the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
Substituting these,$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Given: $m = 100\; kg$,$v = 20\; cm/s = 0.2\; m/s$.
$KE = \frac{3}{4} \times 100 \times (0.2)^2 = 75 \times 0.04 = 3\; J$.
Since the work done to stop the object is $W = \Delta KE = 0 - KE_{initial} = -3\; J$,the magnitude of work needed is $3\; J$.

(A) Given: Mass $m = 500 \; g = 0.5 \; kg$,velocity $v = 5.00 \; cm/s = 0.05 \; m/s$.
For a solid sphere rolling without slipping,the moment of inertia about the center of mass is $I = \frac{2}{5} mR^2$.
The total kinetic energy $KE$ is the sum of translational and rotational kinetic energy:
$KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$.
Since $v = R\omega$,we have $\omega = v/R$.
$KE = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mR^2) (\frac{v}{R})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$.
Substituting the values:
$KE = 0.7 \times 0.5 \; kg \times (0.05 \; m/s)^2 = 0.35 \times 0.0025 \; J = 8.75 \times 10^{-4} \; J$.

(4 J) Radius of the hoop,$r = 2 \; m$.
Mass of the hoop,$m = 100 \; kg$.
Velocity of the centre of mass,$v = 20 \; cm/s = 0.2 \; m/s$.
The total kinetic energy $(K)$ of a rolling hoop is the sum of its translational kinetic energy and rotational kinetic energy.
$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a hoop,the moment of inertia about its centre is $I = mr^2$.
Since the hoop is rolling without slipping,$v = r\omega$,which implies $\omega = v/r$.
Substituting $I$ and $\omega$ into the energy equation:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
The work required to stop the hoop is equal to its total kinetic energy.
$W = K = mv^2 = 100 \; kg \times (0.2 \; m/s)^2 = 100 \times 0.04 = 4 \; J$.

(N/A) The angular speed of the disc is $\omega_{o}$ and its radius is $R$. The linear velocity $v$ of a point at a distance $r$ from the center is given by $v = r \omega_{o}$.
For point $A$ (at distance $R$ from the center,at the top): $v_{A} = R \omega_{o}$,directed tangentially to the right.
For point $B$ (at distance $R$ from the center,at the bottom): $v_{B} = R \omega_{o}$,directed tangentially to the left.
For point $C$ (at distance $R/2$ from the center): $v_{C} = (R/2) \omega_{o}$,directed tangentially to the right.
The disc will not roll. Rolling motion requires the presence of friction to provide the necessary torque and to satisfy the condition $v_{cm} = R \omega$. Since the table is perfectly frictionless,there is no external force to cause translational motion of the center of mass,and thus the disc will simply continue to rotate about its stationary center.

(N/A) To make the disc roll,an external torque is required to change its angular velocity. Friction provides this necessary torque.
$(a)$ The point of contact $B$ has a velocity relative to the surface due to the initial rotation $\omega_0$. Since the disc is rotating clockwise,the point $B$ has a linear velocity directed to the left. Friction acts in the direction opposite to the relative velocity,so the frictional force at $B$ acts tangentially to the right. The torque due to this friction about the center of the disc is directed outward,perpendicular to the plane of the disc,which acts to reduce the angular velocity.
$(b)$ Perfect rolling begins when the velocity of the point of contact $B$ becomes zero relative to the surface $(v = r\omega)$. Once perfect rolling is achieved,there is no relative motion between the point of contact and the surface. Therefore,the force of kinetic friction ceases to act,and the force of friction becomes zero.

(B) Given: Radius $r = 10\; cm = 0.1\; m$,initial angular speed $\omega_0 = 10\pi\; rad\; s^{-1}$,coefficient of kinetic friction $\mu_k = 0.2$,initial linear velocity $u = 0$.
The frictional force $f = \mu_k mg$ provides linear acceleration $a = \mu_k g$. Using $v = u + at$,we get $v = \mu_k gt$.
The torque $\tau = -f r = -I\alpha$ causes angular deceleration $\alpha = \frac{\mu_k mgr}{I}$. Using $\omega = \omega_0 - \alpha t$,we get $\omega = \omega_0 - \frac{\mu_k mgr}{I}t$.
Rolling starts when $v = r\omega$. Substituting the expressions:
$\mu_k gt = r(\omega_0 - \frac{\mu_k mgr}{I}t) \implies \mu_k gt = r\omega_0 - \frac{\mu_k mgr^2}{I}t$.
For the ring $(I = mr^2)$:
$\mu_k gt_r = r\omega_0 - \mu_k gt_r \implies 2\mu_k gt_r = r\omega_0 \implies t_r = \frac{r\omega_0}{2\mu_k g} = \frac{0.1 \times 10\pi}{2 \times 0.2 \times 9.8} \approx 0.80\; s$.
For the disc $(I = \frac{1}{2}mr^2)$:
$\mu_k gt_d = r\omega_0 - 2\mu_k gt_d \implies 3\mu_k gt_d = r\omega_0 \implies t_d = \frac{r\omega_0}{3\mu_k g} = \frac{0.1 \times 10\pi}{3 \times 0.2 \times 9.8} \approx 0.53\; s$.
Since $t_d < t_r$,the disc starts rolling earlier than the ring.

(B, D, E) False: Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling,the frictional force acts in the forward direction to provide the necessary torque,while the centre of mass moves forward.
$(b)$ True: Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence,its instantaneous speed is zero.
$(c)$ False: When a body is rolling,the point of contact has a centripetal acceleration directed towards the centre of the body. Thus,its instantaneous acceleration is not zero.
$(d)$ True: In perfect rolling,there is no relative motion between the point of contact and the surface. Since the point of contact is at rest relative to the surface,the work done by friction is zero.
$(e)$ True: The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force,the body slips down the inclined plane under the effect of its own weight.

(N/A) For a rigid body (like a sphere,circular disc,or wheel) rolling without slipping,the point of contact with the surface must have zero instantaneous velocity relative to the surface.
Consider a circular disc of radius $R$ rolling on a horizontal surface. Let $v_{cm}$ be the velocity of the center of mass $C$ and $\omega$ be the angular velocity about the center.
The velocity of any point $P$ on the rim of the disc is the vector sum of the velocity of the center of mass and the tangential velocity due to rotation: $\vec{v}_P = \vec{v}_{cm} + \vec{v}_{rot}$.
At the point of contact $P_0$ with the ground,the velocity due to rotation $\vec{v}_{rot}$ is directed backwards with magnitude $R\omega$. The velocity of the center of mass $\vec{v}_{cm}$ is directed forwards.
For the condition of rolling without slipping,the velocity at the point of contact $P_0$ must be zero:
$\vec{v}_{P_0} = \vec{v}_{cm} + \vec{v}_{rot} = 0$
Since $\vec{v}_{cm}$ is forward and $\vec{v}_{rot}$ is backward at $P_0$:
$v_{cm} - R\omega = 0$
Therefore,the necessary condition is $v_{cm} = R\omega$.

(C) In pure rolling motion,a circular body rotates about its center of mass while the center of mass moves linearly.
Let $v_{cm}$ be the linear velocity of the center of mass and $\omega$ be the angular velocity of the body.
The velocity of any point on the rim of the body at a distance $R$ from the center is the vector sum of the linear velocity of the center and the tangential velocity due to rotation.
For the point of contact with the ground,the linear velocity is $v_{cm}$ (forward) and the tangential velocity due to rotation is $v_{rot} = R\omega$ (backward).
In pure rolling,the condition is $v_{cm} = R\omega$.
Therefore,the net velocity of the point of contact is $v_{net} = v_{cm} - R\omega = 0$.
Since the net velocity is zero,the point of contact is instantaneously at rest relative to the surface.

(B) No,the relation $v = r\omega$ is valid only for the case of pure rolling motion where the point of contact with the surface has zero instantaneous velocity relative to the surface.
In the case of rolling with slipping,the velocity of the point of contact is not zero,meaning the translational velocity $v$ and the angular velocity $\omega$ do not satisfy the condition $v = r\omega$ at the point of contact.

(N/A) The disc was in pure rotational motion before being brought in contact with the table,hence,$u_{CM} = 0$.
$(b)$ The linear velocity of a point on the rim decreases when the disc is placed in contact with the table due to the opposing force of kinetic friction.
$(c)$ The centre of mass acquires a linear velocity in the direction of motion when the rotating disc is placed in contact with the table due to the force of friction.
$(d)$ Kinetic friction is responsible for these effects.
$(e)$ Rolling without slipping begins when the condition $v_{CM} = \omega R$ is satisfied.
$(f)$ The acceleration produced in the centre of mass due to friction is $a_{CM} = \frac{f}{m} = \frac{\mu_k N}{m} = \frac{\mu_k mg}{m} = \mu_k g$.
The angular retardation produced by the torque due to friction is $\alpha = \frac{\tau}{I} = \frac{\mu_k mgR}{I}$ (since $\tau = fR = \mu_k mgR$).
Using $v_{CM} = u_{CM} + a_{CM}t$ (with $u_{CM} = 0$),we get $v_{CM} = \mu_k gt$.
Using $\omega = \omega_0 - \alpha t$,we get $\omega = \omega_0 - \frac{\mu_k mgR}{I}t$.
For rolling without slipping,$\frac{v_{CM}}{R} = \omega$.
Substituting the expressions: $\frac{\mu_k gt}{R} = \omega_0 - \frac{\mu_k mgR}{I}t$.
Solving for $t$: $t = \frac{R\omega_0}{\mu_k g(1 + \frac{mR^2}{I})}$.

(D) Let $M$ be the mass of the disc. The forces acting on the disc are the applied force $F$ at the center and the friction force $f$ at the point of contact,acting in the opposite direction to $F$.
$1$. Translational motion equation: $F - f = Ma$ ... $(i)$
$2$. Rotational motion equation about the center of mass: $\tau = I\alpha$. Since the friction force $f$ acts at the rim,the torque is $\tau = fR$. The moment of inertia of a disc about its center is $I = \frac{1}{2}MR^2$. Thus,$fR = (\frac{1}{2}MR^2)\alpha$.
$3$. For pure rolling,the condition is $a = R\alpha$,which implies $\alpha = \frac{a}{R}$.
Substituting $\alpha$ into the torque equation: $fR = (\frac{1}{2}MR^2)(\frac{a}{R}) = \frac{1}{2}MaR$,which simplifies to $f = \frac{1}{2}Ma$ or $Ma = 2f$ ... $(ii)$.
$4$. Substitute $(ii)$ into $(i)$: $F - f = 2f$,which gives $F = 3f$.
$5$. For rolling without slipping,the friction force must satisfy $f \le \mu N$,where $N = Mg$ is the normal force. Therefore,$f \le \mu Mg$.
$6$. Since $F = 3f$,the maximum force $F$ is $F_{max} = 3f_{max} = 3\mu Mg$.

(D) For a rolling disc,the total kinetic energy $(K_{total})$ is the sum of translational kinetic energy and rotational kinetic energy.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $v = R\omega$ and for a disc $I = \frac{1}{2}mR^2$,we have $K_{rot} = \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{4}mv^2$.
Thus,$K_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
The ratio of total kinetic energy to rotational kinetic energy is $\frac{K_{total}}{K_{rot}} = \frac{\frac{3}{4}mv^2}{\frac{1}{4}mv^2} = \frac{3}{1}$.

(A) For a ring in pure rolling,the moment of inertia about the center of mass is $I_{CM} = MR^2$.
The angular velocity $\omega$ is related to the velocity of the center of mass $V_{CM}$ by the condition of pure rolling: $\omega = \frac{V_{CM}}{R}$.
Given $V_{CM} = 2\; m/s$ and $R = 4\; m$,we have $\omega = \frac{2}{4} = 0.5\; rad/s$.
The angular momentum $L$ of a body about a point (the origin) is given by the sum of the angular momentum about its center of mass and the angular momentum of the center of mass about the origin:
$L = I_{CM}\omega + M V_{CM} R$.
Substituting the given values:
$L = (MR^2) \omega + M V_{CM} R$
$L = (2 \times 4^2) \times 0.5 + (2 \times 2 \times 4)$
$L = (2 \times 16 \times 0.5) + 16$
$L = 16 + 16 = 32\; kg \cdot m^2/s$.

(B) For a wheel rolling without slipping,the velocity of the centre of mass is $v_{0} = \omega R$,where $\omega$ is the angular velocity and $R$ is the radius of the wheel.
For a particle on the rim at the same level as the centre,the velocity has two components:
$1$. The translational velocity $v_{0}$ directed horizontally.
$2$. The tangential velocity $v_{t} = \omega R$ directed vertically (downwards or upwards depending on the side).
Since $v_{0} = \omega R$,the magnitude of the resultant velocity $v$ is given by:
$v = \sqrt{v_{0}^{2} + v_{t}^{2}} = \sqrt{v_{0}^{2} + (\omega R)^{2}}$
$v = \sqrt{v_{0}^{2} + v_{0}^{2}} = \sqrt{2 v_{0}^{2}} = \sqrt{2} \, v_{0}$
Comparing this with the given expression $\sqrt{x} \, v_{0}$,we get $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

(B) The total kinetic energy of a rolling body is the sum of its rotational kinetic energy and translational kinetic energy.
$K_{\text{total}} = K_{\text{rotational}} + K_{\text{translational}}$
For a solid sphere,the moment of inertia about its center of mass is $I_{\text{cm}} = \frac{2}{5} mR^2$.
Rotational kinetic energy is given by $K_{\text{rot}} = \frac{1}{2} I_{\text{cm}} \omega^2$.
Substituting $I_{\text{cm}}$ and using the pure rolling condition $v = R\omega$ (or $\omega = v/R$):
$K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{5} mR^2 \right) \left( \frac{v^2}{R^2} \right) = \frac{1}{5} mv^2$.
Translational kinetic energy is $K_{\text{trans}} = \frac{1}{2} mv^2$.
Total kinetic energy is $K_{\text{total}} = \frac{1}{5} mv^2 + \frac{1}{2} mv^2 = \left( \frac{2+5}{10} \right) mv^2 = \frac{7}{10} mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is:
$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.

(C) Let $L_{O}$ be the angular momentum of the shell about the origin $O$.
Since the shell is rolling without slipping on the horizontal plane,the velocity of the center of mass is given by $V_{cm} = \omega R$.
The total angular momentum about the origin $O$ is the sum of the angular momentum due to the motion of the center of mass and the angular momentum about the center of mass:
$L_{O} = m V_{cm} R + I_{cm} \omega$
For a spherical shell,the moment of inertia about its center of mass is $I_{cm} = \frac{2}{3} m R^{2}$.
Substituting the values $(m = 1 \, kg)$:
$L_{O} = (1) (\omega R) R + (\frac{2}{3} (1) R^{2}) \omega$
$L_{O} = \omega R^{2} + \frac{2}{3} R^{2} \omega$
$L_{O} = (1 + \frac{2}{3}) R^{2} \omega = \frac{5}{3} R^{2} \omega$
Comparing this with the given expression $\frac{a}{3} R^{2} \omega$,we get:
$\frac{a}{3} = \frac{5}{3}$
Therefore,$a = 5$.

(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of contact to oppose the slipping.
The force of friction $f = \mu_k mg$ acts in the forward direction,causing linear acceleration $a_{CM} = \frac{f}{m} = \mu_k g$.
The torque due to friction about the center of mass is $\tau = -fR = -\mu_k mgR$. The angular deceleration is $\alpha = \frac{\tau}{I} = \frac{-\mu_k mgR}{\frac{2}{5}mR^2} = -\frac{5\mu_k g}{2R}$.
After time $t$,the linear velocity is $v_{CM} = a_{CM}t = \mu_k gt$ and the angular velocity is $\omega_f = \omega + \alpha t = \omega - \frac{5\mu_k g}{2R}t$.
Pure rolling begins when $v_{CM} = \omega_f R$. Substituting the expressions:
$\mu_k gt = \left(\omega - \frac{5\mu_k g}{2R}t\right)R$
$\mu_k gt = \omega R - \frac{5}{2}\mu_k gt$
$\frac{7}{2}\mu_k gt = \omega R \implies \mu_k gt = \frac{2}{7}\omega R$.
Substituting this back into the expression for $\omega_f$:
$\omega_f = \frac{v_{CM}}{R} = \frac{\mu_k gt}{R} = \frac{1}{R} \left(\frac{2}{7}\omega R\right) = \frac{2}{7}\omega$.

(B) In pure rolling motion,the point of contact of the cylinder with the horizontal surface is instantaneously at rest relative to the surface.
Since there is no relative motion between the cylinder and the surface at the point of contact,the static friction force acting at this point is zero for a cylinder rolling at a constant velocity.
Therefore,the force of friction is zero.

(B) The total kinetic energy $(K)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.
$K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a circular disc,the moment of inertia about its central axis is $I = \frac{1}{2}mR^2$.
Since it rolls without slipping,$\omega = \frac{v}{R}$.
Substituting these into the equation:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
Given: $m = 2 \,kg$,$v = 2 \,m/s$.
$K = \frac{3}{4} \times 2 \times (2)^2 = \frac{3}{4} \times 2 \times 4 = 6 \,J$.