Rolling motion on horizontal Surface Questions in English

(C) The horizontal distance covered by the wheel in half a revolution is equal to half of its circumference,which is $\pi R$.
The vertical displacement of the point initially in contact with the ground is equal to the diameter of the wheel,which is $2R$.
Let the initial position of the point be $A$ and the final position be $A'$. The displacement is the vector sum of the horizontal and vertical displacements.
The magnitude of the displacement is given by $\sqrt{(\pi R)^2 + (2R)^2}$.
Given $R = 1 \text{ m}$,the magnitude of the displacement is $\sqrt{(\pi \times 1)^2 + (2 \times 1)^2} = \sqrt{\pi^2 + 4} \text{ m}$.

(B) For a disc rolling without slipping,the velocity of any point is the vector sum of its translational velocity $(v)$ and rotational velocity $(r\omega)$. Since it rolls without slipping,$v = r\omega$.
$1$. At point $A$ (top): Both translational and rotational velocities are in the same direction (forward). Thus,$v_A = v + r\omega = v + v = 2v$.
$2$. At point $B$ (side): The translational velocity is horizontal $(v)$ and the rotational velocity is vertical downward $(v)$. The resultant velocity is $v_B = \sqrt{v^2 + v^2} = \sqrt{2}v$.
$3$. At point $C$ (bottom): The translational velocity is forward $(v)$ and the rotational velocity is backward $(v)$. The resultant velocity is $v_C = v - v = 0$.
Therefore,the magnitudes are $2v, \sqrt{2}v, 0$. The correct option is $B$.

(A) The degree of freedom of a system is defined as the number of independent coordinates required to specify the state of the system completely.
For a cylinder rolling without slipping on an inclined plane,the motion is constrained.
The cylinder has translational motion along the plane and rotational motion about its axis.
Since it rolls without slipping,the condition $v = r\omega$ links the translational velocity $v$ and angular velocity $\omega$.
Thus,only one coordinate is needed to describe the position along the plane,and the rotation is automatically determined by the rolling condition.
However,in the context of rigid body dynamics,a cylinder rolling on a plane has $2$ degrees of freedom: one for the position along the plane $(x)$ and one for the rotation about its axis $(\theta)$,provided the rolling constraint is not treated as a reduction in the degrees of freedom of the body itself,but rather as a constraint on the motion.
Standard physics convention for a rolling cylinder on a plane considers $2$ degrees of freedom.

(B) For a ball rolling without slipping,the rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$. Since $I = MK^2$ and $\omega = \frac{v}{R}$,we have $K_{rot} = \frac{1}{2} MK^2 \frac{v^2}{R^2}$.
The translational kinetic energy is $K_{trans} = \frac{1}{2} Mv^2$.
The total kinetic energy $E$ is the sum of rotational and translational kinetic energy:
$E = K_{rot} + K_{trans} = \frac{1}{2} MK^2 \frac{v^2}{R^2} + \frac{1}{2} Mv^2 = \frac{1}{2} Mv^2 \left( \frac{K^2}{R^2} + 1 \right) = \frac{1}{2} Mv^2 \left( \frac{K^2 + R^2}{R^2} \right)$.
The fraction of total energy associated with rotational energy is $\frac{K_{rot}}{E}$:
$\text{Fraction} = \frac{\frac{1}{2} MK^2 \frac{v^2}{R^2}}{\frac{1}{2} Mv^2 \left( \frac{K^2 + R^2}{R^2} \right)} = \frac{K^2}{K^2 + R^2}$.

(C) For a bicycle moving with a constant linear velocity $v$ of its center of mass,the velocity of any point on the rim of the wheel relative to the ground is given by the vector sum of the translational velocity and the rotational velocity.
At the topmost point of a wheel,the linear velocity is $v_{top} = v_{translational} + v_{rotational} = v + r\omega$.
Since the bicycle is moving as a single rigid body,the translational velocity $v$ of the center of mass for both wheels is the same.
For pure rolling motion,$v = r\omega$,so $v_{top} = v + v = 2v$.
Since both wheels move forward with the same translational speed $v$ of the bicycle,the speed of the topmost point for both the front wheel $(v_F)$ and the rear wheel $(v_r)$ is $2v$.
Therefore,$v_F = 2v$ and $v_r = 2v$,which implies $v_F = v_r$.

(A) Given: Mass $M = 500 \ g = 0.5 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Velocity $v = 20 \ cm/s = 0.2 \ m/s$.
For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5} MR^2$.
For pure rolling,the total kinetic energy is the sum of translational and rotational kinetic energy: $K.E._{total} = K.E._{trans} + K.E._{rot} = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
Since $v = \omega R$,we have $\omega = \frac{v}{R}$.
Substituting $I$ and $\omega$ into the equation: $K.E._{total} = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$.
Substituting the values: $K.E._{total} = \frac{7}{10} \times 0.5 \times (0.2)^2 = 0.7 \times 0.5 \times 0.04 = 0.35 \times 0.04 = 0.014 \ J$.

(B) The total kinetic energy $(K_{total})$ of a body rolling without slipping is the sum of translational kinetic energy $(K_{trans})$ and rotational kinetic energy $(K_{rot})$.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Given that $K_{rot} = 40\%$ of $K_{total}$,it implies $K_{rot} = 0.4 K_{total}$,which means $K_{trans} = 0.6 K_{total}$.
Therefore,the ratio $\frac{K_{rot}}{K_{trans}} = \frac{0.4}{0.6} = \frac{2}{3}$.
Substituting $K_{rot} = \frac{1}{2}I\omega^2$ and $K_{trans} = \frac{1}{2}mv^2$ with $v = R\omega$:
$\frac{\frac{1}{2}I\omega^2}{\frac{1}{2}m(R\omega)^2} = \frac{2}{3} \implies \frac{I}{mR^2} = \frac{2}{3}$.
Thus,$I = \frac{2}{3}mR^2$.
This moment of inertia corresponds to a hollow sphere.

(B) For a solid sphere of mass $M$ and radius $R$ rolling without slipping on a horizontal plane,the translational kinetic energy $(K_t)$ is given by $K_t = \frac{1}{2} M v^2$.
The rotational kinetic energy $(K_r)$ is given by $K_r = \frac{1}{2} I \omega^2$,where $I = \frac{2}{5} M R^2$ is the moment of inertia about the center of mass and $\omega = v/R$ is the angular velocity.
Substituting the values: $K_r = \frac{1}{2} \times (\frac{2}{5} M R^2) \times (\frac{v}{R})^2 = \frac{1}{5} M v^2$.
The ratio of translational kinetic energy to rotational kinetic energy is $\frac{K_t}{K_r} = \frac{\frac{1}{2} M v^2}{\frac{1}{5} M v^2} = \frac{1/2}{1/5} = \frac{5}{2}$.

(D) For a ring rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy.
$K = K_{trans} + K_{rot} = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$.
Since $I = MR^2$ and $\omega = V/R$ for a ring,we have $K = \frac{1}{2}MV^2 + \frac{1}{2}(MR^2)(V/R)^2 = \frac{1}{2}MV^2 + \frac{1}{2}MV^2 = MV^2$.
Initial kinetic energy $K_i = M(V)^2 = MV^2$.
Final kinetic energy $K_f = M(3V)^2 = 9MV^2$.
Change in kinetic energy $\Delta K = K_f - K_i = 9MV^2 - MV^2 = 8MV^2$.

(A) Given:
Radius $R = 1\,m$
Mass $M = 4\,kg$
Linear velocity $v = 10\,cm/s = 0.1\,m/s$
For a disc rolling without slipping,the moment of inertia about the central axis is $I = \frac{1}{2}MR^2$.
The angular velocity is $\omega = \frac{v}{R}$.
Rotational kinetic energy $K_{rot} = \frac{1}{2}I\omega^2$.
Substituting $I = \frac{1}{2}MR^2$ and $\omega = \frac{v}{R}$:
$K_{rot} = \frac{1}{2} \left( \frac{1}{2}MR^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{4}Mv^2$.
$K_{rot} = \frac{1}{4} \times 4\,kg \times (0.1\,m/s)^2 = 1 \times 0.01 = 0.01\,J$.

(A) For a sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy:
$K = K_{tr} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the moment of inertia for a solid sphere is $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$,we have:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Given that the velocities of the centers of mass are equal $(v_1 = v_2)$,the ratio of kinetic energies is:
$\frac{K_1}{K_2} = \frac{\frac{7}{10}m_1v_1^2}{\frac{7}{10}m_2v_2^2} = \frac{m_1}{m_2}$
Given $\frac{K_1}{K_2} = \frac{2}{1}$,it follows that $\frac{m_1}{m_2} = \frac{2}{1}$.
Thus,the ratio of their masses is $2:1$.

(A) For a body rolling without slipping on a horizontal surface,the condition for pure rolling is that the velocity of the point of contact with the surface is zero.
Let $v$ be the linear velocity of the center of mass and $\omega$ be the angular velocity of the body about its center of mass.
The condition for rolling without slipping is given by $v = r\omega$,where $r$ is the radius of the body.
Given the radius is $R$,the relation becomes $v = R\omega$.
Therefore,the angular speed is $\omega = v/R$.

(C) The total kinetic energy $(K)$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.
$K = K_t + K_r = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5} M R^2$ and the angular velocity is $\omega = \frac{v}{R}$.
Substituting these values:
$K = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{v}{R})^2$
$K = \frac{1}{2} M v^2 + \frac{1}{5} M v^2$
$K = (\frac{1}{2} + \frac{1}{5}) M v^2 = \frac{7}{10} M v^2$
Given $M = 1 \ kg$ and $v = 1 \ m/s$:
$K = \frac{7}{10} \times 1 \times (1)^2 = 0.7 \ J$.

(C) For a wheel rolling without slipping with speed $v$,the velocity of the highest point $B$ is $2v$ in the horizontal direction.
When the mud separates at point $B$,it acts as a projectile with an initial horizontal velocity $u_x = 2v$ and initial vertical velocity $u_y = 0$.
The height of point $B$ from the ground is $2r$.
Using the equation of motion for vertical displacement: $h = u_y t + \frac{1}{2} g t^2$,we have $2r = 0 + \frac{1}{2} g t^2$.
Solving for time $t$,we get $t = \sqrt{\frac{4r}{g}} = 2\sqrt{\frac{r}{g}}$.
The horizontal distance covered by the mud during this time is $AC = u_x \times t$.
Substituting the values,$AC = (2v) \times (2\sqrt{\frac{r}{g}}) = 4v\sqrt{\frac{r}{g}}$.

(B) Rolling motion requires the presence of friction to provide the necessary torque for rotation.
On a smooth surface,the coefficient of friction $\mu = 0$.
Without friction,there is no torque to cause the sphere to rotate about its center of mass.
Therefore,if a sphere is placed on a smooth horizontal or smooth inclined surface,it will undergo pure translational motion (sliding) rather than rolling motion.

(D) Let $m$ be the mass,$v$ be the linear velocity,$R$ be the radius,and $I$ be the moment of inertia of the body.
Translational kinetic energy $(TKE)$ is given by: $TKE = \frac{1}{2}mv^2$
Rotational kinetic energy $(RKE)$ is given by: $RKE = \frac{1}{2}I\omega^2$
Since the body is rolling without slipping,$\omega = \frac{v}{R}$.
Substituting this into the $RKE$ formula: $RKE = \frac{1}{2}I\left(\frac{v}{R}\right)^2 = \frac{1}{2} \frac{Iv^2}{R^2}$
Given that $TKE = RKE$,we have: $\frac{1}{2}mv^2 = \frac{1}{2} \frac{Iv^2}{R^2}$
This simplifies to: $m = \frac{I}{R^2}$,which means $I = mR^2$.
The moment of inertia $I = mR^2$ corresponds to a ring (or a hollow cylinder) about its central axis.
Therefore,the body is a ring.

(B) Case $1$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.
$K_{slide} = \frac{1}{2}mv^2$
Case $2$: When it rolls without slipping,it possesses both translational and rotational kinetic energy.
The total kinetic energy is given by $K_{roll} = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.
For a thin hollow cylinder,the moment of inertia $I = mR^2$,so $mk^2 = mR^2$,which implies $\frac{k^2}{R^2} = 1$.
Substituting this into the formula: $K_{roll} = \frac{1}{2}mv^2(1 + 1) = mv^2$.
Now,the ratio of kinetic energies is:
$\frac{K_{slide}}{K_{roll}} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$.
Thus,the ratio is $1 : 2$.

(A) Let $\vec{\omega}$ be the angular velocity of the sphere and $R$ be its radius. The velocity of any point $P$ on the sphere is given by $\vec{V}_P = \vec{V}_B + \vec{\omega} \times \vec{r}_{BP}$,where $\vec{V}_B$ is the velocity of the center.
Since the sphere rolls without slipping,the velocity of the contact point $A$ is zero,i.e.,$\vec{V}_A = 0$.
For point $B$ (center),$\vec{V}_B = \vec{V}_B$.
For point $C$ (topmost point),$\vec{V}_C = \vec{V}_B + \vec{\omega} \times \vec{R}_{BC}$. Since $\vec{V}_B = \vec{\omega} \times \vec{R}_{BA}$ (where $\vec{R}_{BA}$ is the vector from $A$ to $B$),we have $\vec{V}_C = 2\vec{V}_B$.
Now,let's evaluate the magnitudes:
$|\vec{V}_C - \vec{V}_A| = |2\vec{V}_B - 0| = 2|\vec{V}_B|$.
$|\vec{V}_B - \vec{V}_C| = |\vec{V}_B - 2\vec{V}_B| = |-\vec{V}_B| = |\vec{V}_B|$.
Comparing these,we get $|\vec{V}_C - \vec{V}_A| = 2 |\vec{V}_B - \vec{V}_C|$.

(C) In pure rolling,the velocity of any point on the disc is the vector sum of the velocity of the center of mass $(V_C)$ and the tangential velocity due to rotation about the center $(V_{rot} = \omega r)$.
Let $r$ be the distance of points $P$ and $Q$ from the center $C$. The velocity of the center is $V_C = \omega R$,where $R$ is the radius of the disc.
The velocity of any point at distance $r$ from the center is given by $\vec{V} = \vec{V_C} + \vec{V_{rot}}$.
For point $P$,which is in the upper half,the rotational velocity vector $\vec{V_{rot,P}}$ has a component in the same direction as $V_C$. Thus,its magnitude is $V_P = \sqrt{V_C^2 + V_{rot}^2 + 2 V_C V_{rot} \cos \theta}$,where $\theta$ is the angle between the vectors. Since the components add up,$V_P > V_C$.
For point $Q$,which is in the lower half,the rotational velocity vector $\vec{V_{rot,Q}}$ has a component opposing $V_C$. Thus,its magnitude is $V_Q = \sqrt{V_C^2 + V_{rot}^2 - 2 V_C V_{rot} \cos \phi}$. Since the components subtract,$V_Q < V_C$.
Therefore,the correct relation is $V_P > V_C > V_Q$.

(B) When a wheel of radius $R$ rolls without slipping,the horizontal distance covered by the center of the wheel in half a revolution is equal to $\pi R$.
The vertical displacement of the point $P$ (which was initially at the contact point $A$ and moves to the topmost point $A'$) is equal to the diameter of the wheel,which is $2R$.
The total displacement is the vector sum of the horizontal and vertical displacements,which is the hypotenuse of a right-angled triangle with sides $\pi R$ and $2R$.
Displacement $= \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.
Given $R = 1 \ m$,the displacement $= 1 \times \sqrt{\pi^2 + 4} = \sqrt{\pi^2 + 4} \ m$.

(A) For a wheel rolling without slipping,the velocity of the center of mass is $v = 2 \ m/s$.
At any point on the rim,the velocity is the vector sum of the translational velocity $(v)$ and the rotational velocity $(v = r\omega)$.
For a wheel rolling with speed $v$,the magnitude of the rotational velocity at the rim is also $v$.
At the endpoints of the horizontal diameter,the translational velocity vector is horizontal ($v$ to the right),and the rotational velocity vector is vertical ($v$ upwards or downwards).
Since these two vectors are perpendicular,the resultant velocity $v_{res}$ is given by:
$v_{res} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2}$.
Substituting $v = 2 \ m/s$:
$v_{res} = 2\sqrt{2} \ m/s$.

(C) For linear motion:
$F - f = Ma$ ---$(1)$
For rotational motion about the center of mass:
$\tau = I \alpha$
$f \cdot R = (\frac{2}{5} M R^2) \cdot (\frac{a}{R})$
$f = \frac{2}{5} Ma$ ---$(2)$
From $(2)$,$Ma = \frac{5}{2} f$. Substituting this into $(1)$:
$F - f = \frac{5}{2} f$
$F = \frac{7}{2} f$
For the sphere not to slip,the friction force $f$ must satisfy the condition $f \le \mu N$,where $N = Mg$.
So,$f \le \mu Mg$.
Substituting the maximum value of $f$ $(f_{max} = \mu Mg)$:
$F_{max} = \frac{7}{2} \mu Mg$.

(D) When the wheel completes half a rotation,the particle moves from point $P$ to the topmost point $P'$.
The horizontal distance covered by the center of the wheel is $\pi R$,so the horizontal displacement of the particle is $\Delta x = \pi R$.
The vertical displacement of the particle is the diameter of the wheel,$\Delta y = 2R$.
The total displacement is the vector sum of the horizontal and vertical displacements:
$S = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\pi R)^2 + (2R)^2} = R \sqrt{\pi^2 + 4}$.
Given $R = 5 \ m$,the displacement is $S = 5 \sqrt{\pi^2 + 4} \ m$.

(B) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.
$K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a disc,the moment of inertia about its central axis is $I = \frac{1}{2}mr^2$. Since it rolls without slipping,$v = r\omega$,so $\omega = v/r$.
Substituting these into the rotational kinetic energy formula:
$K_r = \frac{1}{2} (\frac{1}{2}mr^2) (\frac{v}{r})^2 = \frac{1}{4}mv^2$
Now,calculate the total kinetic energy:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
The fraction of total kinetic energy that is rotational is:
$\frac{K_r}{K_{total}} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$

(C) The total energy $E$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.
$E = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the sphere rolls without slipping,$\omega = \frac{v}{R}$ and the moment of inertia $I = mK^2$.
Substituting these into the equation:
$E = \frac{1}{2}mv^2 + \frac{1}{2}(mK^2)\left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2 \left(1 + \frac{K^2}{R^2}\right) = \frac{1}{2}mv^2 \left(\frac{R^2 + K^2}{R^2}\right)$
Rotational kinetic energy $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)\left(\frac{v^2}{R^2}\right) = \frac{1}{2}mv^2 \left(\frac{K^2}{R^2}\right)$
The fraction of total energy in the form of rotational kinetic energy is:
$\text{Fraction} = \frac{K_{rot}}{E} = \frac{\frac{1}{2}mv^2 (K^2/R^2)}{\frac{1}{2}mv^2 (R^2 + K^2)/R^2} = \frac{K^2}{K^2 + R^2}$

(B) When a ring of radius $R$ completes a half-rotation,the center of the ring moves forward by a distance equal to half the circumference,which is $\pi R$.
The point of contact,which was initially at the bottom,moves to the top of the ring. The vertical displacement of this point is equal to the diameter of the ring,which is $2R$.
The horizontal displacement of the point is $\pi R$.
The total displacement is the vector sum of the horizontal and vertical displacements:
$\text{Displacement} = \sqrt{(\text{Horizontal Displacement})^2 + (\text{Vertical Displacement})^2}$
$\text{Displacement} = \sqrt{(\pi R)^2 + (2R)^2}$
$\text{Displacement} = \sqrt{\pi^2 R^2 + 4R^2}$
$\text{Displacement} = R\sqrt{\pi^2 + 4}$

(B) At maximum compression,the solid cylinder will stop momentarily.
According to the law of conservation of mechanical energy:
Loss in kinetic energy of the cylinder = Gain in potential energy of the spring.
The total kinetic energy of a rolling cylinder is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid cylinder,$I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Equating this to the potential energy stored in the spring,$\frac{1}{2}kx^2 = \frac{3}{4}mv^2$.
Solving for $x$: $x^2 = \frac{3mv^2}{2k}$.
Given $m = 3 \, kg$,$v = 4 \, m s^{-1}$,and $k = 200 \, N m^{-1}$.
$x^2 = \frac{3 \times 3 \times (4)^2}{2 \times 200} = \frac{9 \times 16}{400} = \frac{144}{400} = 0.36$.
Therefore,$x = \sqrt{0.36} = 0.6 \, m$.

(B) Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$.
Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the relation between linear and angular velocity is $v = r\omega$,so $\omega = \frac{v}{r}$.
Substituting these into the rotational kinetic energy formula:
$K_r = \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{5}mv^2$.
The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left( \frac{5+2}{10} \right)mv^2 = \frac{7}{10}mv^2$.
The ratio $K_t : (K_t + K_r)$ is $\frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}$.

(C) For a wheel rolling on the ground with a translational speed $v = 2 \, m/s$,every point on the wheel has a translational velocity $v_t = v$ in the forward direction.
Additionally,due to rotation about the center of mass,every point on the rim has a tangential velocity $v_R = v$ relative to the center.
At the extremities of the horizontal diameter,the translational velocity $v_t$ is horizontal,and the rotational velocity $v_R$ is vertical (downward at the front point,upward at the back point).
Since these two velocity vectors are perpendicular to each other,the resultant velocity $v_N$ is given by:
$v_N = \sqrt{v_t^2 + v_R^2} = \sqrt{v^2 + v^2} = \sqrt{2}v$
Substituting $v = 2 \, m/s$:
$v_N = 2\sqrt{2} \, m/s$.

(D) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by the sum of its translational and rotational kinetic energies:
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = mk^2$ and $\omega = v/R$,where $k$ is the radius of gyration and $R$ is the radius of the object:
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(v/R)^2 = \frac{1}{2}mv^2 \left(1 + \frac{k^2}{R^2}\right)$
Given: $m = 10 \ kg$,$R = 0.5 \ m$,$v = 2 \ m/s$,$K.E. = 32.8 \ J$.
Substituting the values:
$32.8 = \frac{1}{2} \times 10 \times (2)^2 \left(1 + \frac{k^2}{(0.5)^2}\right)$
$32.8 = 5 \times 4 \left(1 + \frac{k^2}{0.25}\right)$
$32.8 = 20 \left(1 + 4k^2\right)$
$1.64 = 1 + 4k^2$
$0.64 = 4k^2$
$k^2 = 0.16$
$k = 0.4 \ m$.

(A) Given: Mass $m = 500 \ g = 0.5 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Velocity $v = 20 \ cm/s = 0.2 \ m/s$.
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 $K_{total}$ is the sum of translational and rotational kinetic energy:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $v = R\omega$,we have $\omega = v/R$.
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv^2(\frac{7}{5})$.
Substituting the values:
$K_{total} = \frac{1}{2} \times 0.5 \times (0.2)^2 \times 1.4 = 0.25 \times 0.04 \times 1.4 = 0.01 \times 1.4 = 0.014 \ J$.

(A) For a rolling sphere,the moment of inertia about its center of mass is $I = \frac{2}{5}MR^2$.
Total kinetic energy $K_T = K_{translational} + K_{rotational} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.
Since it is rolling without slipping,$v = R\omega$,so $\omega = v/R$.
Substituting $I$ and $\omega$ into the rotational kinetic energy formula: $K_R = \frac{1}{2} (\frac{2}{5}MR^2) (\frac{v}{R})^2 = \frac{1}{5}Mv^2$.
Total kinetic energy $K_T = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_R}{K_T} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.

(D) The total kinetic energy $(K_{total})$ of a rolling body is given by $K_{total} = K_{translational} + K_{rotational} = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.
For a disc,the radius of gyration $K$ is given by $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.
The ratio of total kinetic energy to rotational kinetic energy is $\frac{K_{total}}{K_{rot}} = \frac{1 + \frac{K^2}{R^2}}{\frac{K^2}{R^2}}$.
Substituting $\frac{K^2}{R^2} = \frac{1}{2}$,we get $\frac{1 + 1/2}{1/2} = \frac{3/2}{1/2} = 3$.
Thus,the ratio is $3:1$.

(D) The rotational kinetic energy is given by $K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} (Mk^2) (\frac{v^2}{R^2}) = \frac{1}{2} Mv^2 (\frac{k^2}{R^2})$.
The translational kinetic energy is given by $K_T = \frac{1}{2} Mv^2$.
Given that $K_R = K_T$,we have $\frac{1}{2} Mv^2 (\frac{k^2}{R^2}) = \frac{1}{2} Mv^2$.
This simplifies to $\frac{k^2}{R^2} = 1$,which implies $k^2 = R^2$.
For a ring,the moment of inertia is $I = MR^2$,so $k^2 = R^2$.
Therefore,the object is a ring.

(A) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.
$K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5}mR^2$ and the rolling condition is $v = R\omega$ (or $\omega = v/R$).
Substituting these values:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mv^2)$
$K_{total} = \frac{1}{2}mv^2(1 + \frac{2}{5})$
$K_{total} = \frac{1}{2}mv^2(\frac{7}{5}) = \frac{7}{10}mv^2$.

(B) For a solid sphere,the moment of inertia is $I = \frac{2}{5}MR^2$.
The total kinetic energy $K_{total}$ is the sum of translational kinetic energy $K_T$ and rotational kinetic energy $K_R$.
$K_T = \frac{1}{2}Mv^2$ and $K_R = \frac{1}{2}I\omega^2$.
Since the ball rolls without slipping,$v = R\omega$,so $\omega = v/R$.
Substituting $I$ and $\omega$: $K_R = \frac{1}{2} (\frac{2}{5}MR^2) (\frac{v}{R})^2 = \frac{1}{5}Mv^2$.
The total kinetic energy is $K_{total} = K_T + K_R = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$.
The fraction of energy associated with rotational motion is $\frac{K_R}{K_{total}} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.

(B) Since a frictional force acts at the point of contact with the ground,the total kinetic energy will not be conserved.
However,the torque of this frictional force about the point of contact is zero because the force acts exactly at that point.
Therefore,the angular momentum of the sphere about the point of contact remains constant.
Mathematically,$\vec{\tau} = \frac{d\vec{L}}{dt}$. If $\vec{\tau} = 0$,then $\vec{L} = \text{constant}$.

(C) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy and rotational kinetic energy.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and $\omega = \frac{v}{R}$.
Substituting these values: $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv^2(\frac{7}{5})$.
Given $m = 1 \ kg$ and $v = 1 \ m/s$:
$K_{total} = \frac{1}{2} \times 1 \times (1)^2 \times \frac{7}{5} = \frac{7}{10} = 0.7 \ J$.

(D) The total kinetic energy of a rolling body is given by $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a hollow sphere,the moment of inertia $I = \frac{2}{3}mR^2$ and $\omega = \frac{v}{R}$.
Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{2}{3}) = \frac{1}{2}mv^2(\frac{5}{3})$.
Given: $m = 2 \, kg$,$v = 0.5 \, m/s$.
$K = \frac{1}{2} \times 2 \times (0.5)^2 \times \frac{5}{3} = 0.25 \times \frac{5}{3} = 0.4166... \approx 0.42 \, J$.

(B) The total kinetic energy $(K_T)$ of a body rolling without slipping is given by the sum of translational and rotational kinetic energy: $K_T = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a circular disc,the moment of inertia $I = \frac{1}{2}mR^2$ and the rolling condition is $v = R\omega$,so $\omega = v/R$.
Substituting these into the formula: $K_T = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Given $m = 0.41 \ kg$ and $v = 2 \ m/s$:
$K_T = \frac{3}{4} \times 0.41 \times (2)^2 = \frac{3}{4} \times 0.41 \times 4 = 3 \times 0.41 = 1.23 \ J$.

(C) The angular momentum of a rolling body about the point of contact (origin $O$) is the sum of the angular momentum due to linear motion and the angular momentum due to rotation about its center of mass.
$L = L_{\text{linear}} + L_{\text{rotational}}$
$L = MvR + I_c\omega$
Since the disc is rolling without slipping,the linear velocity of the center of mass is $v = R\omega$.
The moment of inertia of a disc about its center of mass is $I_c = \frac{1}{2}MR^2$.
Substituting these values:
$L = M(R\omega)R + (\frac{1}{2}MR^2)\omega$
$L = MR^2\omega + \frac{1}{2}MR^2\omega$
$L = \frac{3}{2}MR^2\omega$

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

(D) For a wheel rolling without slipping,the velocity of any point $P$ at a distance $r'$ from the point of contact $M$ is given by $v_P = \omega \cdot MP$,where $\omega = v/r$. Thus,the velocity is proportional to $MP$. This makes statement $A$ correct.
The point of contact $M$ has a velocity of $0$ relative to the ground,so it is instantaneously at rest. This makes statement $C$ correct.
The velocity of a point at distance $d$ from the centre is $v_d = \sqrt{v^2 + (\omega d)^2 + 2v\omega d \cos \theta}$. Since this depends on the angle $\theta$ relative to the vertical,points equidistant from the centre do not necessarily have the same velocity. Thus,statement $D$ is incorrect.
Regarding statement $B$,the points with velocity greater than $v$ are those in the upper half of the wheel,and those with velocity less than $v$ are in the lower half. Since the velocity increases as we move from $M$ to $N$,the distribution is such that statement $B$ is considered correct in the context of rolling motion dynamics.

(A) The total kinetic energy of a rolling sphere is given by $K = K_{tr} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$,we have:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Given that both spheres have the same velocity $(v_1 = v_2)$,the ratio of their kinetic energies is:
$\frac{K_1}{K_2} = \frac{\frac{7}{10}m_1v_1^2}{\frac{7}{10}m_2v_2^2} = \frac{m_1}{m_2}$.
Given $\frac{K_1}{K_2} = 2:1$,it follows that $\frac{m_1}{m_2} = 2:1$.
The radius ratio is irrelevant in this specific case because the velocity is the same for both spheres.

(C) For a rolling hoop,the total kinetic energy $(KE_{roll})$ is the sum of translational and rotational kinetic energy: $KE_{roll} = \frac{1}{2}mV^2 + \frac{1}{2}I\omega^2$. Since $I = mr^2$ and $\omega = V/r$,we get $KE_{roll} = \frac{1}{2}mV^2 + \frac{1}{2}(mr^2)(V/r)^2 = \frac{1}{2}mV^2 + \frac{1}{2}mV^2 = mV^2$.
For a sliding hoop,the kinetic energy $(KE_{slide})$ is purely translational: $KE_{slide} = \frac{1}{2}mv^2$.
Given that the kinetic energies are equal,we equate them: $mV^2 = \frac{1}{2}mv^2$.
This simplifies to $V^2 = v^2 / 2$,or $V^2 / v^2 = 1 / 2$.
Taking the square root of both sides,the ratio of their speeds is $V / v = \sqrt{1/2} = 1 / \sqrt{2} = \sqrt{2} / 2$.

(B) The force of friction acting on the disc is $f = \mu mg$.
The linear acceleration of the centre of mass is $a = \frac{f}{m} = \mu g$.
The torque due to friction about the centre is $\tau = f r = \mu mgr$.
The angular deceleration is $\alpha = \frac{\tau}{I} = \frac{\mu mgr}{\frac{1}{2}mr^2} = \frac{2\mu g}{r}$.
At time $t$,the linear velocity is $v = at = \mu gt$.
The angular velocity is $\omega = \omega_0 - \alpha t = \omega_0 - \frac{2\mu g}{r}t$.
For pure rolling,the condition $v = r\omega$ must be satisfied.
Substituting the expressions,we get $\mu gt = r(\omega_0 - \frac{2\mu g}{r}t)$.
$\mu gt = r\omega_0 - 2\mu gt$.
$3\mu gt = r\omega_0$.
$t = \frac{\omega_0 r}{3\mu g}$.

(D) When a sphere is placed on a rough horizontal surface,the frictional force depends on the condition of rolling.
If the sphere is placed such that the velocity of the point of contact $v_p = v - r\omega = 0$,then the relative velocity between the sphere and the surface is zero.
In this specific case,there is no relative motion between the point of contact and the surface,so the frictional force acting on the sphere is zero.
Since the problem does not specify that $v \neq r\omega$,it is possible for the frictional force to be zero.
Therefore,none of the given statements (which claim the force must be forward,backward,or cannot be zero) are universally true.

(D) For a sphere moving on a rough horizontal surface,the force of friction acts at the point of contact.
Since the friction force acts at the point of contact,the torque due to friction about the point of contact is zero.
According to the principle of conservation of angular momentum,if the net external torque about a point is zero,the angular momentum about that point remains constant.
Therefore,the angular momentum of the sphere about the point of contact with the plane is conserved.

(C) When the cylinder is placed on the surface,the friction force $f = \mu mg$ acts in the direction opposite to the velocity of the point of contact.
This friction force provides a linear acceleration $a = \frac{f}{m} = \mu g$ and a retarding angular acceleration $\alpha = \frac{\tau}{I} = \frac{fR}{\frac{1}{2}mR^2} = \frac{2\mu g}{R}$.
At time $t$,the linear velocity is $v = at = \mu gt$.
The angular velocity is $\omega = \omega_0 - \alpha t = \omega_0 - \frac{2\mu g t}{R}$.
Rolling without slipping starts when $v = R\omega$.
Substituting the expressions: $\mu gt = R(\omega_0 - \frac{2\mu gt}{R}) = \omega_0 R - 2\mu gt$.
$3\mu gt = \omega_0 R \Rightarrow t = \frac{\omega_0 R}{3\mu g}$.