Rolling motion on horizontal Surface Questions in English

(B) In pure rolling,the velocity of any point on the ring is the vector sum of the translational velocity of the center of mass $(v_{cm})$ and the tangential velocity due to rotation $(v_{rot} = \omega R = v_{cm})$.
For point $A$,the translational velocity is directed horizontally to the right $(v_{cm})$,and the rotational velocity is directed vertically upwards $(v_{rot} = v_{cm})$.
Since these two velocity vectors are perpendicular to each other,the net velocity $v_{net}$ is given by:
$v_{net} = \sqrt{v_{cm}^2 + v_{rot}^2} = \sqrt{v_{cm}^2 + v_{cm}^2} = \sqrt{2 v_{cm}^2} = \sqrt{2} v_{cm}$

(A) The work done by a force is given by $W = \vec{F} \cdot \vec{d}$,where $\vec{F}$ is the force and $\vec{d}$ is the displacement of the point of application.
In the case of pure rolling (rolling without slipping) on a stationary horizontal surface,the point of contact of the body with the surface is instantaneously at rest.
Since the velocity of the point of contact is $0$,the displacement of the point of contact during the time it is in contact with the surface is also $0$.
Because the displacement of the point where the friction force acts is $0$,the work done by the static friction force is $0$.

(B) For pure rolling,the linear acceleration $a$ and angular acceleration $\alpha$ are related by $a = r\alpha$.
Applying Newton's second law for translational motion:
$F - f = M a \quad \dots (1)$
Applying torque equation about the centre of mass:
$\tau = I \alpha$
$f r = (\frac{1}{2} M r^2) (\frac{a}{r})$
$f = \frac{1}{2} M a \quad \dots (2)$
Substituting $(2)$ into $(1)$:
$F - \frac{1}{2} M a = M a$
$F = \frac{3}{2} M a \Rightarrow a = \frac{2 F}{3 M}$
Now,substitute $a$ back into $(2)$ to find the friction force $f$:
$f = \frac{1}{2} M (\frac{2 F}{3 M}) = \frac{F}{3}$
For pure rolling to occur,the friction force must satisfy $f \le \mu N$,where $N = Mg$.
$\frac{F}{3} \le \mu M g$
$\mu \ge \frac{F}{3 M g}$
Thus,the minimum value of the coefficient of friction is $\mu = \frac{F}{3 M g}$.

(A) The correct answer is $A$.
When a body is rolling on a surface,it possesses both translational velocity $v$ and angular velocity $\omega$.
On a smooth horizontal surface,there is no friction to exert a torque or a force that would change the linear or angular momentum of the body.
Since there is no external torque acting on the body,the angular momentum remains conserved.
Therefore,the body will continue to roll with its constant translational velocity $v$ and constant angular velocity $\omega$ without any change in its state of motion.

(C) When the sphere is thrown,it experiences a frictional force that opposes its motion,causing it to decelerate linearly and accelerate rotationally.
Since the frictional force acts through the point of contact,the angular momentum about the point of contact on the ground remains conserved.
Initial angular momentum about the point of contact: $L_i = m u r$.
Final angular momentum when pure rolling starts (at velocity $v$ and angular velocity $\omega = v/r$): $L_f = m v r + I \omega = m v r + (\frac{2}{5} m r^2)(\frac{v}{r}) = m v r + \frac{2}{5} m v r = \frac{7}{5} m v r$.
Equating $L_i = L_f$:
$m u r = \frac{7}{5} m v r$.
Solving for $v$:
$u = \frac{7}{5} v \implies v = \frac{5u}{7}$.

(B) For a wheel rolling with a uniform velocity $v$ of its center of mass,the acceleration of the center of mass is $a_{CM} = 0$.
Each point on the rim of the wheel has a centripetal acceleration directed towards the center,given by $a_c = \frac{v^2}{R}$.
Let the topmost point be $A$ and the bottommost point be $B$.
The acceleration of point $A$ is $\vec{a}_A = \frac{v^2}{R}$ (directed downwards towards the center).
The acceleration of point $B$ is $\vec{a}_B = \frac{v^2}{R}$ (directed upwards towards the center).
The relative acceleration of $A$ with respect to $B$ is $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$.
Taking the downward direction as positive,$\vec{a}_A = \frac{v^2}{R}$ and $\vec{a}_B = -\frac{v^2}{R}$.
Thus,$\vec{a}_{AB} = \frac{v^2}{R} - (-\frac{v^2}{R}) = \frac{2v^2}{R}$.
Therefore,the magnitude of the relative acceleration is $\frac{2v^2}{R}$.

(D) The correct option is $(d)$.
As the disc is in combined rotation and translation,each point on the disc has a translational velocity $v$ (in the forward direction) and a tangential velocity $v = R\omega$ (due to rotation).
For the lowermost point of the disc:
$1$. The net velocity is $v_{\text{net}} = v - R\omega$. Since it is pure rolling,$v = R\omega$,therefore $v_{\text{net}} = v - v = 0$.
$2$. The acceleration of any point on a rotating body has two components: tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$.
- Since the angular velocity $\omega$ is constant,the tangential acceleration $a_t = R\alpha = 0$.
- The centripetal acceleration is $a_c = v^2/R$ directed towards the center of the disc.
- Thus,the net acceleration of the lowermost point is $a = a_c + a_t = v^2/R + 0 = v^2/R$,directed towards the center of the disc.

(A) The total kinetic energy $(K)$ of a rolling sphere 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 solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$.
Since the sphere rolls without slipping,$v = R\omega$,which implies $\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Given: $m = 500\,g = 0.5\,kg$ and $v = 0.02\,m/s$.
$K = \frac{7}{10} \times 0.5 \times (0.02)^2$
$K = 0.7 \times 0.5 \times 0.0004 = 0.35 \times 0.0004 = 0.00014\,J = 1.4 \times 10^{-4}\,J$.

(A) The total kinetic energy $(KE)$ of a body in pure rolling motion is the sum of its translational kinetic energy and rotational kinetic energy.
$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia about its centre of mass is $I = \frac{2}{5}mR^2$.
For pure rolling,the relation between linear velocity $(v)$ and angular velocity $(\omega)$ is $v = R\omega$,or $\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$
$KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Given $m = 2\,kg$ and $KE = 2240\,J$:
$2240 = \frac{7}{10} \times 2 \times v^2$
$2240 = \frac{7}{5}v^2$
$v^2 = \frac{2240 \times 5}{7} = 320 \times 5 = 1600$
$v = \sqrt{1600} = 40\,m/s$.

(C) Given: mass $m = 0.5\,kg$,initial velocity $u = 18\,m/s$,coefficient of friction $\mu = 0.3$,$g = 10\,m/s^2$,time $t = 2\,s$.
The acceleration of the disc due to friction is $a = -\mu g = -0.3 \times 10 = -3\,m/s^2$.
The velocity of the center of mass at $t = 2\,s$ is $v = u + at = 18 - 3 \times 2 = 12\,m/s$.
For pure rolling,the condition is $v = \omega r$,so $\omega = v/r$.
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 $I = \frac{1}{2}mr^2$ for a disc,we have:
$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Substituting the values:
$KE = \frac{3}{4} \times 0.5 \times (12)^2 = \frac{3}{4} \times 0.5 \times 144 = 3 \times 0.5 \times 36 = 54\,J$.

(A) The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5} MR^2$.
Since the sphere rolls without slipping,$\omega = \frac{v}{R}$.
Substituting these into the energy equation: $K = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$.
Given $K = 7 \times 10^{-3}\,J$ and $M = 1\,kg$,we have $\frac{7}{10} (1) v^2 = 7 \times 10^{-3}$.
$v^2 = 10^{-2} \implies v = 0.1\,m/s$.
Converting to $cm/s$: $v = 0.1 \times 100 = 10\,cm/s$.

(D) The rotational kinetic energy of a body is given by $K_{\text{rot}} = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a spherical shell,the moment of inertia about its center is $I = \frac{2}{3} mR^2$.
For pure rolling,$\omega = \frac{v}{R}$,so $K_{\text{rot}} = \frac{1}{2} (\frac{2}{3} mR^2) (\frac{v}{R})^2 = \frac{1}{3} mv^2$.
The total kinetic energy is $K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} mv^2}{\frac{5}{6} mv^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$.
Given that the ratio is $\frac{x}{5}$,we have $\frac{x}{5} = \frac{2}{5}$,which implies $x = 2$.

(A) When the disc rolls without slipping,the center of the disc moves forward by a distance equal to the arc length covered. For half a rotation,the center moves by a horizontal distance of $\pi R$.
The point $A$,which was initially at the top,moves to the bottom of the disc after half a rotation. The vertical displacement of point $A$ is equal to the diameter of the disc,which is $2R$.
The horizontal displacement of point $A$ is equal to the distance moved by the center,which is $\pi R$.
The total displacement $d$ is the vector sum of the horizontal and vertical displacements:
$d = \sqrt{(\text{horizontal displacement})^2 + (\text{vertical displacement})^2}$
$d = \sqrt{(\pi R)^2 + (2R)^2}$
$d = \sqrt{\pi^2 R^2 + 4R^2}$
$d = R \sqrt{\pi^2 + 4}$

(C) For a solid sphere rolling without slipping,the angular momentum $L$ about the axis of rotation is $L = I_{com}\omega = (\frac{2}{5}MR^2)\omega$.
Since $V_{com} = R\omega$,we have $L = \frac{2}{5}MR^2(\frac{V_{com}}{R}) = \frac{2}{5}MRV_{com}$.
The total kinetic energy $K$ is $K = \frac{1}{2}I_{com}\omega^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{2}(\frac{2}{5}MR^2)(\frac{V_{com}}{R})^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{5}MV_{com}^2 + \frac{1}{2}MV_{com}^2 = \frac{7}{10}MV_{com}^2$.
The ratio $\frac{L}{K} = \frac{\frac{2}{5}MRV_{com}}{\frac{7}{10}MV_{com}^2} = \frac{2}{5} \times \frac{10}{7} \times \frac{R}{V_{com}} = \frac{4}{7} \times \frac{R}{R\omega} = \frac{4}{7\omega}$.
Given $\frac{L}{K} = \frac{\pi}{22}$,we have $\frac{4}{7\omega} = \frac{\pi}{22}$. Assuming $\pi \approx \frac{22}{7}$,we get $\frac{4}{7\omega} = \frac{22/7}{22} = \frac{1}{7}$.
Therefore,$7\omega = 28$,which gives $\omega = 4\,rad/s$.

(D) According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy $\Delta KE$.
Since the disc is rolling, its total kinetic energy $KE$ is the sum of translational and rotational kinetic energy: $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid disc, the moment of inertia $I = \frac{1}{2}mR^2$ and for rolling without slipping, $\omega = \frac{v}{R}$.
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 $m = 50 \,kg$ and $v = 0.4 \,m/s$, the initial kinetic energy is $KE_i = \frac{3}{4} \times 50 \times (0.4)^2 = \frac{3}{4} \times 50 \times 0.16 = 0.75 \times 8 = 6 \,J$.
To stop the disc, the final kinetic energy $KE_f = 0$.
Thus, $W = KE_f - KE_i = 0 - 6 = -6 \,J$.
The absolute value of the work done is $|W| = 6 \,J$.

(A) For a hollow sphere rolling without slipping,the moment of inertia about its axis of symmetry is $I = \frac{2}{3} mR^2$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} mR^2) \omega^2 = \frac{1}{3} mR^2 \omega^2$.
The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$.
Since $v = R\omega$,we have $K_{total} = \frac{1}{2} (\frac{2}{3} mR^2) \omega^2 + \frac{1}{2} m(R\omega)^2 = \frac{1}{3} mR^2 \omega^2 + \frac{1}{2} mR^2 \omega^2 = (\frac{2+3}{6}) mR^2 \omega^2 = \frac{5}{6} mR^2 \omega^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3} mR^2 \omega^2}{\frac{5}{6} mR^2 \omega^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$.
Comparing this with $\frac{x}{5}$,we get $x = 2$.

(A) In the case of pure rolling motion,the velocity of any point on the rim of the wheel is the vector sum of the translational velocity of the center of mass $(v)$ and the tangential velocity due to rotation $(v = r\omega)$.
For the topmost point $P$,the translational velocity and the rotational velocity are in the same direction. Thus,the resultant velocity is $v + v = 2v$.
For the lowest point $Q$,the translational velocity is in the forward direction $(v)$ and the rotational velocity is in the backward direction $(v)$. Thus,the resultant velocity is $v - v = 0$.
Since the velocity of point $P$ is $2v$ and the velocity of point $Q$ is $0$,point $P$ moves faster than point $Q$.

(C) Let $\vec{V}_0$ be the velocity of the centre of the sphere. For a sphere rolling without slipping on a fixed horizontal surface:
$\vec{V}_A = 0$ (velocity of the point of contact)
$\vec{V}_B = \vec{V}_0$ (velocity of the centre)
$\vec{V}_C = 2\vec{V}_0$ (velocity of the topmost point)
Now,let's evaluate the given options:
For $(B)$: $\vec{V}_C - \vec{V}_B = 2\vec{V}_0 - \vec{V}_0 = \vec{V}_0$ and $\vec{V}_B - \vec{V}_A = \vec{V}_0 - 0 = \vec{V}_0$. Thus,$\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$ is correct.
For $(C)$: $|\vec{V}_C - \vec{V}_A| = |2\vec{V}_0 - 0| = 2|\vec{V}_0|$ and $2|\vec{V}_B - \vec{V}_C| = 2|\vec{V}_0 - 2\vec{V}_0| = 2|-\vec{V}_0| = 2|\vec{V}_0|$. Thus,$|\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$ is correct.
Therefore,both $(B)$ and $(C)$ are correct statements.

(C) Let $M = 2 \ kg$,$R = 0.5 \ m$,$a = 0.3 \ m/s^2$,and $N = 2 \ N$ (force applied by the stick).
For a ring,the moment of inertia about the center is $I = MR^2$.
Since it rolls without slipping,the angular acceleration is $\alpha = a/R$.
Let $f_s$ be the friction from the ground and $f_a$ be the friction from the stick.
Applying Newton's second law for translational motion: $N - f_s = Ma$.
$2 - f_s = 2 \times 0.3 = 0.6 \implies f_s = 1.4 \ N$.
Applying torque equation about the center of the ring: $(f_s - f_a)R = I\alpha$.
$(1.4 - f_a)R = (MR^2)(a/R) = MaR$.
$1.4 - f_a = Ma = 2 \times 0.3 = 0.6$.
$f_a = 1.4 - 0.6 = 0.8 \ N$.
The friction from the stick is $f_a = \mu N$,where $\mu = P/10$.
$0.8 = (P/10) \times 2$.
$0.8 = P/5 \implies P = 4$.

(C) Let $R$ be the radius of the roller and $r$ be the radius of the axle.
Given: $2R = 20 \ cm \implies R = 10 \ cm$ and $2r = 10 \ cm \implies r = 5 \ cm$.
For the roller to roll without slipping on the floor,the velocity of the center of the roller is $V_{\text{center}} = \omega R$,where $\omega$ is the angular velocity.
The velocity of the scale,which is on top of the axle,is $V_{\text{scale}} = V_{\text{center}} + \omega r$.
Substituting $\omega = \frac{V_{\text{center}}}{R}$,we get:
$V_{\text{scale}} = V_{\text{center}} + \left(\frac{V_{\text{center}}}{R}\right)r = V_{\text{center}} \left(1 + \frac{r}{R}\right)$.
Given $r = 5 \ cm$ and $R = 10 \ cm$,we have $V_{\text{scale}} = V_{\text{center}} \left(1 + \frac{5}{10}\right) = 1.5 V_{\text{center}}$.
If the roller moves a distance $d_{\text{roller}} = 50 \ cm$ in time $t$,then $V_{\text{center}} \cdot t = 50 \ cm$.
The distance moved by the scale in the same time $t$ is $d_{\text{scale}} = V_{\text{scale}} \cdot t = 1.5 V_{\text{center}} \cdot t = 1.5 \times 50 \ cm = 75 \ cm$.
Thus,the scale moves $75 \ cm$ relative to the ground,while the center of the roller moves $50 \ cm$ relative to the ground. The position of the scale relative to the center of the roller is $75 \ cm - 50 \ cm = 25 \ cm$ ahead of the center.

(C) At $t=0$,$\omega=0$. At $t=\sqrt{\pi} \text{ s}$,$\omega = \alpha t = \frac{2}{3} \sqrt{\pi} \text{ rad/s}$.
The linear velocity of the center of mass is $v_{cm} = \omega R = \frac{2}{3} \sqrt{\pi} \times 1 = \frac{2}{3} \sqrt{\pi} \text{ m/s}$.
The angle rotated by the disk is $\theta = \frac{1}{2} \alpha t^2 = \frac{1}{2} \times \frac{2}{3} \times (\sqrt{\pi})^2 = \frac{\pi}{3} = 60^{\circ}$.
At this instant,the stone is at a height $y = R - R \cos \theta = 1 - \cos 60^{\circ} = 1 - 0.5 = 0.5 \text{ m}$ from the ground.
The velocity of the stone relative to the ground is the vector sum of the velocity of the center of mass and the tangential velocity relative to the center of mass. The vertical component of this velocity is $v_y = v_{cm} \sin \theta = (\frac{2}{3} \sqrt{\pi}) \sin 60^{\circ} = \frac{2}{3} \sqrt{\pi} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3\pi}}{3} \text{ m/s}$.
The additional height gained by the stone after detaching is $h = \frac{v_y^2}{2g} = \frac{(\frac{\sqrt{3\pi}}{3})^2}{2 \times 10} = \frac{3\pi / 9}{20} = \frac{\pi}{60} \text{ m}$.
The total maximum height from the plane is $H = y + h = 0.5 + \frac{\pi}{60} = \frac{1}{2} + \frac{\pi/6}{10}$.
Comparing with $\frac{1}{2} + \frac{x}{10}$,we get $x = \frac{\pi}{6} \approx \frac{3.14}{6} \approx 0.523$. Thus,$x \approx 0.52$.

(B) The linear kinetic energy of the centre of mass is given by $K_{linear} = \frac{1}{2} mv_{cm}^2$.
The rotational kinetic energy of the sphere is given by $K_{rotational} = \frac{1}{2} I \omega^2$.
For a solid sphere,the moment of inertia about its centre of mass is $I = \frac{2}{5} mR^2$.
Since the sphere is rolling without slipping,we have the condition $v_{cm} = \omega R$.
Substituting these into the ratio:
$\frac{K_{linear}}{K_{rotational}} = \frac{\frac{1}{2} mv_{cm}^2}{\frac{1}{2} (\frac{2}{5} mR^2) \omega^2} = \frac{mv_{cm}^2}{\frac{2}{5} m(v_{cm}^2)} = \frac{1}{2/5} = \frac{5}{2}$.
Thus,the ratio is $5/2$.

(A) Let $F = 49 \ N$ be the applied force,$m = 20 \ kg$ be the mass of the solid sphere,and $r$ be its radius.
For a solid sphere,the moment of inertia about its center of mass is $I_{cm} = \frac{2}{5} mr^2$.
The moment of inertia about the point of contact (bottom point) is $I = I_{cm} + mr^2 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} mr^2$.
The torque about the point of contact is $\tau = F \times (2r)$.
Using $\tau = I \alpha$,we have $F \times 2r = (\frac{7}{5} mr^2) \alpha$.
$49 \times 2r = \frac{7}{5} \times 20 \times r^2 \times \alpha$.
$98r = 28r^2 \alpha$.
Since the sphere rolls without slipping,the linear acceleration $a$ of the center is $a = r \alpha$,so $\alpha = \frac{a}{r}$.
Substituting this into the equation: $98r = 28r^2 (\frac{a}{r}) = 28ra$.
$98 = 28a$.
$a = \frac{98}{28} = 3.5 \ m/s^2$.

(A) For a wheel rolling on a plane surface,the velocity of the highest point $B$ is $V_B = 2v$,where $v$ is the velocity of the center of mass.
Given $V_B = 8 \ m/s$,we have $2v = 8 \ m/s$,which implies $v = 4 \ m/s$.
The point $A$ at the bottom is the instantaneous center of rotation.
The velocity of any point $P$ on the rim is given by $V_P = \omega r_P$,where $r_P$ is the distance from the instantaneous center $A$.
For a point $P$ at the same level as the center $C$,the distance $AP = \sqrt{R^2 + R^2} = R\sqrt{2}$,where $R$ is the radius of the wheel.
Since $v = \omega R$,we have $\omega = v/R$.
Thus,$V_P = (v/R) \times (R\sqrt{2}) = v\sqrt{2}$.
Substituting $v = 4 \ m/s$,we get $V_P = 4\sqrt{2} \ m/s$.

(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,the point of contact has a centripetal acceleration directed towards the center of the body,so its instantaneous acceleration is not zero.
Hence,statement $B$ is incorrect.
In perfect rolling on ground,the velocity of the point of contact is zero.
Since the point of contact does not move relative to the ground,the work done against friction is zero.
Thus,statement $C$ is correct.
Rolling cannot take place in the absence of friction on an inclined plane because it is the frictional force that provides the necessary torque which makes the body roll.
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 distance travelled by the wheel in a half revolution is given by $d = \frac{C}{2} = \frac{2 \pi r}{2} = \pi r$,where $C$ is the circumference of the wheel.
After a half revolution,the point initially in contact with the surface (point $A$) moves to the top of the wheel (point $B$).
The horizontal distance covered by the center of the wheel is $\pi r$,and the vertical displacement of the point is equal to the diameter of the wheel,which is $2r$.
Using the Pythagorean theorem for the displacement vector $AB$:
$AB = \sqrt{(\text{horizontal distance})^2 + (\text{vertical distance})^2}$
$AB = \sqrt{(\pi r)^2 + (2r)^2} = r \sqrt{\pi^2 + 4}$
Given $r = 1 \ m$,the magnitude of the displacement is $AB = \sqrt{\pi^2 + 4} \ m$.

(D) The total kinetic energy of a body rolling without slipping is given by $K.E. = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2 = \frac{1}{2} Mv^2 (1 + \frac{k^2}{R^2})$.
For a ring,the radius of gyration $k = R$,so $K.E._{\text{ring}} = \frac{1}{2} Mv^2 (1 + 1) = Mv^2$.
Given $K.E._{\text{ring}} = 6 \ J$,therefore $Mv^2 = 6 \ J$.
For a disc,the moment of inertia $I = \frac{1}{2} MR^2$,so $k^2 = \frac{1}{2} R^2$.
$K.E._{\text{disc}} = \frac{1}{2} Mv^2 (1 + \frac{1}{2}) = \frac{1}{2} Mv^2 (\frac{3}{2}) = \frac{3}{4} Mv^2$.
Substituting $Mv^2 = 6 \ J$ into the equation for the disc:
$K.E._{\text{disc}} = \frac{3}{4} \times 6 = \frac{18}{4} = 4.5 \ J = \frac{9}{2} \ J$.

(A) Let the radius of the wheel be $R = 2 \ cm$.
Initially,point $P$ is at the origin $(0, 0)$.
After half a rotation,the center of the wheel moves forward by a distance equal to half the circumference,which is $\pi R$.
The new position of the center is $(\pi R, R)$.
After half a rotation,point $P$ moves from the bottom to the top of the wheel.
The new coordinates of point $P$ will be $(\pi R, 2R)$.
The displacement $d$ is the distance between the initial position $(0, 0)$ and the final position $(\pi R, 2R)$.
Using the distance formula: $d = \sqrt{(\pi R - 0)^2 + (2R - 0)^2} = \sqrt{\pi^2 R^2 + 4R^2} = R\sqrt{\pi^2 + 4}$.
Substituting $R = 2 \ cm$: $d = 2\sqrt{\pi^2 + 4} \ cm$ or $2(\pi^2 + 4)^{1/2} \ cm$.

(C) The total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energies.
$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m r^2$ and the rolling condition is $V = r \omega$.
$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} (\frac{2}{5} m r^2) (\frac{V}{r})^2 = \frac{1}{2} m V^2 + \frac{1}{5} m V^2 = \frac{7}{10} m V^2$.
When the sphere compresses the spring by a maximum distance $x$,all its kinetic energy is converted into the potential energy of the spring $(U = \frac{1}{2} k x^2)$.
$\frac{1}{2} k x^2 = \frac{7}{10} m V^2$
$x^2 = \frac{14}{10} \frac{m V^2}{k} = \frac{1.4 \times 2 \times 6^2}{36} = \frac{1.4 \times 2 \times 36}{36} = 2.8$.
$x = \sqrt{2.8} \approx 1.67 \ m$.
Note: Given the options provided,there appears to be a discrepancy in the calculation or the provided options. Based on the standard physics approach,the result is $\sqrt{2.8} \ m$.

(A) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the objects roll without slipping,$v = R\omega$,so $\omega^2 = \frac{v^2}{R^2}$.
For a ring,the moment of inertia $I_{ring} = mR^2$. Thus,$K.E._{ring} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Given $K.E._{ring} = 4 \ J$,we have $mv^2 = 4 \ J$.
For a disc,the moment of inertia $I_{disc} = \frac{1}{2}mR^2$. Thus,$K.E._{disc} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Substituting $mv^2 = 4 \ J$,we get $K.E._{disc} = \frac{3}{4} \times 4 \ J = 3 \ J$.

(C) The total kinetic energy of an object in pure rolling motion is given by:
$E = \frac{1}{2} m v_{cm}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$
where $k$ is the radius of gyration,$R$ is the radius,and $m$ is the mass of the object.
Given $m = 1 \ kg$,the expression becomes:
$\frac{E}{v_{cm}^{2}} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right) \quad ...(i)$
From the given graph,the slope of the line is $\frac{E}{v_{cm}^{2}} = \frac{3}{4}$.
Substituting this value into Eq. $(i)$:
$\frac{3}{4} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$
$\frac{3}{2} = 1 + \frac{k^{2}}{R^{2}}$
$\frac{k^{2}}{R^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$
We know that for a disc,the moment of inertia $I = \frac{1}{2} m R^{2}$,so $k^{2} = \frac{1}{2} R^{2}$,which means $\frac{k^{2}}{R^{2}} = \frac{1}{2}$.
Thus,the object is a disc.

(C) In pure rolling motion,the total kinetic energy is the sum of translational and rotational kinetic energy.
Translational $KE = \frac{1}{2}mv^2$.
Rotational $KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.
Total $KE = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2}) = \frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})$.
The fraction of total energy associated with rotation is $\frac{Rotational \ KE}{Total \ KE} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})} = \frac{K^2}{K^2+R^2}$.

(A) For a rolling disc of mass $M$ and radius $R$ moving with velocity $v$ and angular velocity $\omega = v/R$,the translational kinetic energy is $K_t = \frac{1}{2} Mv^2$.
The rotational kinetic energy about the center of mass is $K_r = \frac{1}{2} I \omega^2$.
For a disc,the moment of inertia $I = \frac{1}{2} MR^2$.
Substituting $I$ and $\omega = v/R$,we get $K_r = \frac{1}{2} (\frac{1}{2} MR^2) (v/R)^2 = \frac{1}{4} Mv^2$.
The total kinetic energy $K = K_t + K_r = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$.
Comparing $K$ with $K_r$: $K = \frac{3/4 Mv^2}{1/4 Mv^2} K_r = 3 K_r$.
Thus,the total kinetic energy is $3$ times its rotational kinetic energy.

(C) The total kinetic energy of a body rolling without sliding is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.
For a circular disc,the moment of inertia is $I = \frac{1}{2}mR^2$,so $k^2 = \frac{1}{2}R^2$. The kinetic energy is $K_{disc} = \frac{1}{2}mv^2(1 + \frac{1}{2}) = \frac{3}{4}mv^2 = 6 \ J$.
From this,$\frac{1}{2}mv^2 = \frac{6 \times 2}{3} = 4 \ J$.
For a thin circular ring,the moment of inertia is $I = mR^2$,so $k^2 = R^2$. The kinetic energy is $K_{ring} = \frac{1}{2}mv^2(1 + 1) = mv^2$.
Since $\frac{1}{2}mv^2 = 4 \ J$,then $mv^2 = 8 \ J$.
Therefore,the total kinetic energy of the ring is $8 \ J$.

(B) For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5}MR^2$.
Since the sphere is rolling without slipping,the velocity of the center of mass is $v = R\omega$,where $\omega$ is the angular velocity.
The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} (\frac{2}{5}MR^2) \omega^2 = \frac{1}{5}M(R\omega)^2 = \frac{1}{5}Mv^2$.
The translational kinetic energy is $K_{trans} = \frac{1}{2}Mv^2$.
The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{2+5}{10}Mv^2 = \frac{7}{10}Mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.
Thus,the ratio is $2: 7$.

(C) The initial angular momentum of the hoop about the point of contact is $L_i = I_{cm} \omega_0 + m r^2 \omega_0 = m r^2 \omega_0 + m r^2 \omega_0 = 2 m r^2 \omega_0$. However,considering the angular momentum about the contact point $P$ at the instant it is placed: $L_i = I_P \omega_0 = (I_{cm} + m r^2) \omega_0 = (m r^2 + m r^2) \omega_0 = 2 m r^2 \omega_0$.
When the hoop stops slipping,it rolls without slipping,so $v = r \omega$. The final angular momentum about the contact point $P$ is $L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = m r v + m r v = 2 m r v$.
Since the frictional force acts through the contact point,the net torque about the contact point is zero. Thus,angular momentum is conserved: $L_i = L_f$.
$2 m r^2 \omega_0 = 2 m r v$.
Dividing both sides by $2 m r$,we get $v = r \omega_0$.
Wait,re-evaluating: $L_i = I_{cm} \omega_0 = m r^2 \omega_0$. The angular momentum about the contact point is $L_i = I_{cm} \omega_0 = m r^2 \omega_0$.
$L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = 2 m r v$.
Equating $L_i = L_f$: $m r^2 \omega_0 = 2 m r v$.
Therefore,$v / \omega_0 = r / 2 = 50 \ cm / 2 = 25 \ cm$.

(D) Given, mass of the circular ring, $m = 10 \text{ kg}$.
Linear speed of the center of mass, $v = 1.5 \text{ m/s}$.
The total initial kinetic energy of the rolling ring is the sum of its rotational and translational kinetic energies:
$K_i = K_{\text{rotational}} + K_{\text{translational}} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$.
For a ring, the moment of inertia $I = m R^2$ and the angular velocity $\omega = \frac{v}{R}$.
Substituting these values:
$K_i = \frac{1}{2} (m R^2) \left(\frac{v}{R}\right)^2 + \frac{1}{2} m v^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2$.
$K_i = 10 \times (1.5)^2 = 10 \times 2.25 = 22.5 \text{ J}$.
According to the work-energy theorem, the work required to stop the ring is equal to the change in kinetic energy:
$W = K_f - K_i = 0 - 22.5 = -22.5 \text{ J}$.

(D) For rolling without slipping,the velocity is $v = r \omega$.
Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$.
Substituting $v = r \omega$,we get $E = \frac{1}{2} I \omega^2 + \frac{1}{2} m r^2 \omega^2 = \frac{1}{2} (I + m r^2) \omega^2$.
Rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2$.
Given that $K_{rot} = 50 \% \text{ of } E$,so $K_{rot} = \frac{1}{2} E$.
Substituting the expressions: $\frac{1}{2} I \omega^2 = \frac{1}{2} [\frac{1}{2} (I + m r^2) \omega^2]$.
$I = \frac{1}{2} I + \frac{1}{2} m r^2$.
$\frac{1}{2} I = \frac{1}{2} m r^2$,which implies $I = m r^2$.
The moment of inertia $I = m r^2$ corresponds to a thin circular ring.

(A) The angular momentum of a body rolling on a surface with respect to a point $O$ on the surface is given by the sum of the angular momentum due to the motion of the centre of mass and the angular momentum due to rotation about the centre of mass.
$L_O = L_{\text{linear}} + L_{\text{rotational}} = MvR + I\omega$
Since the body is rolling without slipping,$v = R\omega$,so $\omega = \frac{v}{R}$.
$L_O = MvR + I\left(\frac{v}{R}\right) = vR \left(M + \frac{I}{R^2}\right)$
For a solid sphere,$I_1 = \frac{2}{5}MR^2$. Thus,$L_1 = vR \left(M + \frac{2}{5}M\right) = \frac{7}{5}MvR$.
For a solid cylinder,$I_2 = \frac{1}{2}MR^2$. Thus,$L_2 = vR \left(M + \frac{1}{2}M\right) = \frac{3}{2}MvR$.
The ratio is $\frac{L_1}{L_2} = \frac{\frac{7}{5}MvR}{\frac{3}{2}MvR} = \frac{7}{5} \times \frac{2}{3} = \frac{14}{15}$.

(B) For a solid sphere rolling without slipping,the moment of inertia about its center of mass is $I = \frac{2}{5}MR^2$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{5}MR^2\omega^2$.
Since the sphere is rolling without slipping,the velocity of the center of mass is $v = R\omega$,so $\omega = \frac{v}{R}$.
Substituting $\omega$ into the rotational kinetic energy expression: $K_{rot} = \frac{1}{5}MR^2(\frac{v}{R})^2 = \frac{1}{5}Mv^2$.
The translational kinetic energy is $K_{trans} = \frac{1}{2}Mv^2$.
The ratio of rotational to translational kinetic energy is $\frac{K_{rot}}{K_{trans}} = \frac{\frac{1}{5}Mv^2}{\frac{1}{2}Mv^2} = \frac{2}{5}$.
Thus,the ratio is $2: 5$.

(A) The total kinetic energy $(KE)$ of a rolling body is the sum of its translational and rotational kinetic energies, given by the formula:
$KE = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2$
For a solid sphere, the moment of inertia about its centre is $I = \frac{2}{5} mR^2$ and the angular velocity is $\omega = \frac{v}{R}$.
Substituting these values, we get:
$KE = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mR^2)(\frac{v}{R})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$
Given $m = 5 \,kg$ and $v = 4 \,m/s$:
$KE = \frac{7}{10} \times 5 \times (4)^2 = \frac{7}{10} \times 5 \times 16 = 7 \times 8 = 56 \,J$
Thus, the correct option is $A$.

(C) Let $v$ be the final velocity of the centre of the sphere and $\omega$ be the final angular velocity when it starts rolling without slipping.
Since the friction force acts at the point of contact,the net torque about the point of contact is zero.
Therefore,the angular momentum about the point of contact remains conserved.
Initial angular momentum about the point of contact: $L_i = m v_0 r$
Final angular momentum about the point of contact: $L_f = mvr + I_{cm}\omega$
For a solid sphere,the moment of inertia about its centre is $I_{cm} = \frac{2}{5} mr^2$.
Since it rolls without slipping,the condition is $v = r\omega$,or $\omega = \frac{v}{r}$.
Equating initial and final angular momentum:
$mv_0 r = mvr + (\frac{2}{5} mr^2)(\frac{v}{r})$
$mv_0 r = mvr + \frac{2}{5} mvr$
$v_0 = v + \frac{2}{5} v$
$v_0 = \frac{7}{5} v$
$v = \frac{5}{7} v_0$

(A, C, D) For a circular disc rolling without slipping on a horizontal floor, the velocity of any point on the rim is the vector sum of the translational velocity of the centre $(v)$ and the tangential velocity due to rotation $(\omega R)$.
Since the disc rolls without slipping, $v = \omega R$.
$1$. At the point of contact $(P)$ with the floor, the velocity is $v_P = v - \omega R = v - v = 0$.
$2$. At the top point $(S)$, the velocity is $v_S = v + \omega R = v + v = 2v$.
$3$. At any other point on the rim, the magnitude of the velocity lies between $0$ and $2v$. Specifically, at the level of the centre, the velocity is $\sqrt{v^2 + v^2} = \sqrt{2}v$.
Thus, the possible values for the velocity of a point on the rim include $0$, $v$, $\sqrt{2}v$, and $2v$. Among the given options, $v$, $2v$, and $0$ are valid.

(B) The friction force $f = \mu_K mg$ acts opposite to the direction of translational velocity $v$. The linear acceleration is $a = -\mu_K g$.
The velocity at time $t$ is given by $v(t) = v_0 - \mu_K gt$.
The torque due to friction about the center is $\tau = fR = \mu_K mgR$.
The angular acceleration is $\alpha = \tau/I = \frac{\mu_K mgR}{(1/2)mR^2} = \frac{2\mu_K g}{R}$.
The angular velocity at time $t$ is $\omega(t) = \omega_0 + \alpha t = \frac{v_0}{4R} + \frac{2\mu_K g}{R}t$.
Rolling starts when the condition $v(t) = \omega(t)R$ is satisfied.
Substituting the expressions: $v_0 - \mu_K gt = (\frac{v_0}{4R} + \frac{2\mu_K g}{R}t)R$.
This simplifies to: $v_0 - \mu_K gt = \frac{v_0}{4} + 2\mu_K gt$.
Rearranging terms: $\frac{3v_0}{4} = 3\mu_K gt$.
Solving for $t$: $t = \frac{v_0}{4\mu_K g}$.
Plugging in the values: $t = \frac{49}{4 \times 0.25 \times 9.8} = \frac{49}{9.8} = 5 \text{ s}$.