Rolling On Inclined Plane Questions in English

(D) By the principle of conservation of energy,the total initial kinetic energy (translational + rotational) is equal to the final potential energy at the maximum height $h$.
Initial kinetic energy $K_i = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$
Since the object is rolling without slipping,$\omega = v/R$,so $K_i = \frac{1}{2} M v^2 + \frac{1}{2} I (v/R)^2 = \frac{1}{2} M v^2 + \frac{1}{2} I \frac{v^2}{R^2}$
Final potential energy $U_f = M g h = M g \left( \frac{3 v^2}{4 g} \right) = \frac{3}{4} M v^2$
Equating $K_i = U_f$:
$\frac{1}{2} M v^2 + \frac{1}{2} I \frac{v^2}{R^2} = \frac{3}{4} M v^2$
$\frac{1}{2} I \frac{v^2}{R^2} = \frac{3}{4} M v^2 - \frac{1}{2} M v^2 = \frac{1}{4} M v^2$
$I \frac{1}{R^2} = \frac{1}{2} M \implies I = \frac{1}{2} M R^2$
The moment of inertia $I = \frac{1}{2} M R^2$ corresponds to a disc.

(A) According to the law of conservation of energy,the total initial kinetic energy (translational + rotational) must be equal to the potential energy at the maximum height $h$.
Total 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 $I = \frac{2}{5}mR^2$ and for pure rolling,$\omega = v/R$.
Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Equating this to the potential energy at height $h$ $(PE = mgh)$:
$\frac{7}{10}mv^2 = mgh$.
Solving for $v$,we get $v^2 = \frac{10gh}{7}$,so $v = \sqrt{\frac{10gh}{7}}$.
Since the sphere must reach at least height $h$,the minimum velocity is $v \ge \sqrt{\frac{10gh}{7}}$.

(B) Applying the law of conservation of energy between the top of the track $(A)$ and the horizontal section $(BC)$:
$mgh_A = mgh_B + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the sphere rolls without slipping,$I = \frac{2}{5}mR_s^2$ and $\omega = \frac{v}{R_s}$,where $R_s$ is the radius of the sphere.
$mg(h_A - h_B) = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR_s^2)(\frac{v^2}{R_s^2})$
$mg(2.4 - 1.0) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$
$mg(1.4) = \frac{7}{10}mv^2$
$v^2 = \frac{1.4 \times 10 \times g}{7} = 2g = 20 \ m^2/s^2 \implies v = \sqrt{20} \ m/s$.
Now,for the projectile motion from point $C$ to the ground:
The height $h = 1.0 \ m$. The time taken to fall is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.0}{10}} = \sqrt{0.2} \ s$.
The horizontal distance $R = v \times t = \sqrt{20} \times \sqrt{0.2} = \sqrt{4} = 2 \ m$.

(B) For an object rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$
Here,$I$ is the moment of inertia about the central axis,$M$ is the mass,and $R$ is the radius.
Since the object rolls without slipping,the torque equation $\tau = I \alpha$ and the force equation $F = Ma$ lead to the derivation where the term $\frac{I}{M R^2}$ is added to the denominator of the translational acceleration $g \sin \theta$.

(A) Acceleration of the solid sphere slipping down the incline without rolling is
${a_{slipping}} = g \sin \theta \,\,\,\,\,\,\,(i)$
Acceleration of the solid sphere rolling down the incline without slipping is
${a_{rolling}} = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}}$
(For a solid sphere,$\frac{k^2}{R^2} = \frac{2}{5}$)
$= \frac{5}{7} g \sin \theta \,\,\,\,\,(ii)$
Dividing equation $(ii)$ by equation $(i)$,we get
$\frac{a_{rolling}}{a_{slipping}} = \frac{5}{7}$

(A) The time $t$ taken by a body to roll down an inclined plane of length $l$ and inclination $\theta$ without slipping is given by:
$t = \sqrt{\frac{2l(1 + \frac{k^2}{R^2})}{g \sin \theta}}$
where $k$ is the radius of gyration and $R$ is the radius of the object.
Since $l$,$\theta$,and $g$ are the same for both,the time depends on the factor $(1 + \frac{k^2}{R^2})$.
For a disc,$I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.5 = 1.5$.
For a solid sphere,$I = \frac{2}{5}MR^2$,so $k^2 = \frac{2}{5}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.4 = 1.4$.
Since $1.4 < 1.5$,the time taken by the sphere is less than the time taken by the disc $(t_s < t_d)$.
Therefore,the sphere reaches the bottom first.

(A) The total energy at the top is potential energy,$PE = mgh$.
At the bottom,this energy is converted into translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.
$mgh = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mR^2$ and $\omega = \frac{v}{R}$.
Substituting these,$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Thus,$v^2 = \frac{10gh}{7}$.
The rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{5}mv^2$.
Substituting $v^2 = \frac{10gh}{7}$,we get $K_r = \frac{1}{5}m(\frac{10gh}{7}) = \frac{2}{7}mgh$.
Given $m = 3 \; kg$,$g = 10 \; m/s^2$,and $h = 7 \; m$:
$K_r = \frac{2}{7} \times 3 \times 10 \times 7 = 60 \; J$.

(B) When rolling down an inclined plane,the acceleration of a body is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $a_{solid} = \frac{g \sin \theta}{1 + 0.4} = \frac{g \sin \theta}{1.4} \approx 0.71g \sin \theta$.
For a hollow sphere,the moment of inertia $I = \frac{2}{3}MR^2$,so $a_{hollow} = \frac{g \sin \theta}{1 + 0.67} = \frac{g \sin \theta}{1.67} \approx 0.60g \sin \theta$.
Since the acceleration of the solid sphere is greater,it will reach the bottom of the inclined plane faster than the hollow sphere.

(B) For a body rolling without slipping,the total kinetic energy $E_{total}$ is the sum of translational kinetic energy $E_t$ and rotational kinetic energy $E_r$.
$E_t = \frac{1}{2}mv^2$
$E_r = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$
$E_{total} = E_t + E_r = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$
The fraction of total energy that is rotational is given by:
$\frac{E_r}{E_{total}} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})}$
$= \frac{K^2/R^2}{(R^2 + K^2)/R^2} = \frac{K^2}{K^2 + R^2}$

(B) For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5}MR^2$.
Since the sphere is rolling without slipping,its linear velocity $v$ and angular velocity $\omega$ are related by $v = R\omega$.
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 $K_{total} = K_{rot} + K_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{7}{10}Mv^2$.
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 correct option is $B$.

(C) Case $(i)$: When the hollow cylinder slides without rotation,it possesses only translational kinetic energy.
$K_T = \frac{1}{2}mv^2$
Case (ii): When it rolls without slipping,it possesses both translational and rotational kinetic energy.
The total kinetic energy is given by $K_R = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$.
For a thin hollow cylinder,the radius of gyration $k = R$,so $\frac{k^2}{R^2} = 1$.
Thus,$K_R = \frac{1}{2}mv^2(1 + 1) = mv^2$.
The ratio of the kinetic energies is:
$\frac{K_T}{K_R} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$.

(A) The acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
where $K$ is the radius of gyration and $R$ is the radius of the body.
For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$.
Since $I = MK^2$,we have $K^2 = \frac{2}{5}R^2$,which implies $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting this into the acceleration formula:
$a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5}{7}g \sin \theta$.

(B) For a body rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the body is a solid sphere,its moment of inertia $I = \frac{2}{5}mR^2$ and the condition for rolling without slipping is $v = R\omega$.
Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$.
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Solving for $v$: $v^2 = \frac{10}{7}gh$,which gives $v = \sqrt{\frac{10}{7}gh}$.
However,using the general formula $v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$ where $K^2 = \frac{2}{5}R^2$ for a solid sphere:
$v = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{2gh}{7/5}} = \sqrt{\frac{10}{7}gh}$.
Note: The provided option $(b)$ $\sqrt{\frac{4}{3}gh}$ corresponds to a hollow sphere or a ring-like distribution where $I = \frac{2}{3}mR^2$ or $I = mR^2$. For a solid sphere,the correct result is $\sqrt{\frac{10}{7}gh}$.

(A) The time taken to reach the bottom of an inclined plane is given by $t = \frac{1}{\sin \theta} \sqrt{\frac{2h}{g} \left( 1 + \frac{K^2}{R^2} \right)}$.
For a solid sphere,the radius of gyration factor is $\frac{K^2}{R^2} = \frac{2}{5} = 0.4$.
For a hollow spherical shell,the radius of gyration factor is $\frac{K^2}{R^2} = \frac{2}{3} \approx 0.67$.
Since the time $t$ is directly proportional to $\sqrt{1 + \frac{K^2}{R^2}}$,the object with the smaller value of $\frac{K^2}{R^2}$ will take less time to reach the bottom.
Because $\left( \frac{K^2}{R^2} \right)_{\text{solid}} < \left( \frac{K^2}{R^2} \right)_{\text{hollow}}$,the solid sphere reaches the bottom first.

(D) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$.
For a solid sphere,the radius of gyration $K$ is given by $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
Given $\theta = 30^\circ$,we have $\sin 30^\circ = 1/2$.
Substituting these values into the formula:
$a = \frac{g \sin 30^\circ}{1 + 2/5} = \frac{g(1/2)}{7/5} = \frac{g}{2} \times \frac{5}{7} = \frac{5g}{14}$.
Thus,the linear acceleration is $5g/14$.

(A) For a body rolling without slipping down an inclined plane of height $h$,the conservation of mechanical energy gives: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the body is a solid sphere,its moment of inertia about the center of mass is $I = \frac{2}{5}mR^2$.
For rolling without slipping,$\omega = \frac{v}{R}$.
Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$.
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Solving for $v$,we get $v^2 = \frac{10}{7}gh$,which implies $v = \sqrt{\frac{10}{7}gh}$.

(D) For a solid cylinder rolling down an inclined plane without slipping,the linear velocity $v$ at the base is given by the conservation of energy: $v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.
For a solid cylinder,the radius of gyration $K$ is related to the radius $R$ by $K^2 = \frac{R^2}{2}$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
Substituting the values: $v = \sqrt{\frac{2 \times 10 \times 2}{1 + 0.5}} = \sqrt{\frac{40}{1.5}} = \sqrt{26.66} \approx 5.16 \ m/s$.
However,using the standard approximation $v = \sqrt{\frac{4}{3}gh} = \sqrt{\frac{4}{3} \times 10 \times 2} = \sqrt{26.66} \approx 5.16 \ m/s$.
Angular velocity $\omega = \frac{v}{R}$. Given diameter $= 30 \ cm$,so radius $R = 0.15 \ m$.
$\omega = \frac{5.16}{0.15} \approx 34.4 \ rad/s$. The closest integer value is $34 \ rad/s$.

(B) The time taken to reach the bottom is given by $t = \sqrt{\frac{2h(1 + K^2/R^2)}{g \sin^2 \theta}}$.
Since $h$,$g$,and $\theta$ are constant,$t \propto \sqrt{1 + K^2/R^2}$.
The values of $K^2/R^2$ are:
For a solid sphere: $K^2/R^2 = 0.4$
For a disc: $K^2/R^2 = 0.5$
For a ring: $K^2/R^2 = 1.0$
Comparing the values: $(K^2/R^2)_{\text{sphere}} < (K^2/R^2)_{\text{disc}} < (K^2/R^2)_{\text{ring}}$.
Therefore,the order of time taken is $t_{\text{sphere}} < t_{\text{disc}} < t_{\text{ring}}$.
Thus,the sphere reaches the bottom first,followed by the disc,and finally the ring.

(C) When an object of mass $m$ slides down an inclined plane without friction,the potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv^2$,which gives $v = \sqrt{2gh}$.
When the object is a ring rolling down the inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}I\omega^2$.
For a ring,the moment of inertia $I = mR^2$ and $\omega = \frac{v_{ring}}{R}$.
Substituting these,we get: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}(mR^2)(\frac{v_{ring}}{R})^2 = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}mv_{ring}^2 = mv_{ring}^2$.
Thus,$v_{ring} = \sqrt{gh}$.
Since $v = \sqrt{2gh}$,we can write $\sqrt{gh} = \frac{v}{\sqrt{2}}$.
Therefore,the velocity of the ring is $\frac{v}{\sqrt{2}}$.

(A) The time taken by an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h}{g \sin^2 \theta} (1 + \frac{K^2}{R^2})}$,where $K$ is the radius of gyration and $R$ is the radius of the object.
For a solid sphere,$\frac{K^2}{R^2} = \frac{2}{5} = 0.4$.
For a disc,$\frac{K^2}{R^2} = \frac{1}{2} = 0.5$.
For a solid cylinder,$\frac{K^2}{R^2} = \frac{1}{2} = 0.5$.
Since the solid sphere has the minimum value of $\frac{K^2}{R^2}$,it will have the minimum time of descent.
Therefore,the solid sphere reaches the bottom first.

(D) Given that the linear kinetic energy $(K_T)$ is equal to the rotational kinetic energy $(K_R)$.
$K_T = K_R$
$\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2$
Since $I = mk^2$ and $v = R\omega$,we have $\omega = v/R$.
$\frac{1}{2}mv^2 = \frac{1}{2}(mk^2)(v/R)^2$
$1 = \frac{k^2}{R^2}$
This implies the radius of gyration $k = R$.
For a hollow cylinder,the moment of inertia $I = mR^2$,so $mk^2 = mR^2$,which gives $k = R$.
Therefore,the object is a hollow cylinder.

(A) The time taken by a body to roll down an inclined plane is given by the formula $t = \sqrt{\frac{2L(1 + \frac{k^2}{R^2})}{g \sin \theta}}$,where $k$ is the radius of gyration and $R$ is the radius of the object.
For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $\frac{k^2}{R^2} = 0.5$.
For a hollow cylinder,the moment of inertia $I = MR^2$,so $k^2 = R^2$,which gives $\frac{k^2}{R^2} = 1$.
Since the hollow cylinder has a larger value of $\frac{k^2}{R^2}$,it will have a larger value of $t$.
Therefore,the hollow cylinder takes more time to reach the bottom.

(A) For a body rolling down an inclined plane without slipping,the linear acceleration $a$ is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
For a solid cylinder,the radius of gyration $K$ is related to the radius $R$ by $K^2 = \frac{R^2}{2}$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
Given the angle of inclination $\theta = 30^\circ$,we have $\sin 30^\circ = 1/2$.
Substituting these values into the formula:
$a = \frac{g \sin 30^\circ}{1 + 1/2} = \frac{g(1/2)}{3/2} = \frac{g}{3}$.

(B) Since the surface is frictionless,there is no torque to change the angular velocity of the sphere. The rotational kinetic energy remains constant throughout the motion.
Applying the law of conservation of mechanical energy:
Total initial energy = Total final energy at height $h$
$\frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} = \frac{1}{2}I{\omega ^2} + mgh$
Since the sphere must reach at least height $h$,the initial translational kinetic energy must be sufficient to overcome the potential energy gain:
$\frac{1}{2}m{v^2} \ge mgh$
$v^2 \ge 2gh$
$v \ge \sqrt {2gh}$

(B) When the cylinder rolls up the inclined plane,its linear velocity $v$ is directed up the incline and its angular velocity $\omega$ is clockwise. Since it is moving up,its linear velocity decreases,meaning it experiences a linear deceleration. To maintain rolling without slipping,it requires an angular deceleration. The torque provided by the frictional force $f$ acting up the incline creates a counter-clockwise torque,which provides the necessary angular deceleration.
When the cylinder rolls down the inclined plane,its linear velocity $v$ is directed down the incline and its angular velocity $\omega$ is counter-clockwise. As it moves down,its linear velocity increases,meaning it experiences linear acceleration. To maintain rolling without slipping,it requires an angular acceleration. The frictional force $f$ acting up the incline creates a torque that results in the necessary angular acceleration to increase the counter-clockwise angular velocity.

(C) The acceleration of an object rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a given height and angle,the time taken to reach the bottom is $t = \sqrt{\frac{2s}{a}}$,where $s$ is the distance along the incline.
Thus,$t \propto \sqrt{1 + \frac{I}{MR^2}}$.
The object with the smallest moment of inertia $(I)$ coefficient will have the largest acceleration and reach the bottom first.
The coefficients of $I$ for the objects are:
$1$. Sphere: $I = \frac{2}{5}MR^2 = 0.4 MR^2$
$2$. Disc: $I = \frac{1}{2}MR^2 = 0.5 MR^2$
$3$. Hollow sphere: $I = \frac{2}{3}MR^2 \approx 0.67 MR^2$
$4$. Ring: $I = MR^2 = 1.0 MR^2$
Since the order of coefficients is $0.4 < 0.5 < 0.67 < 1.0$,the order of reaching the bottom is Sphere,Disc,Hollow sphere,Ring.

(C) The time taken for an object to roll down an inclined plane of height $h$ and angle $\theta$ is given by the formula: $t = \frac{1}{\sin \theta} \sqrt{\frac{2h}{g} \left( 1 + \frac{K^2}{R^2} \right)}$.
For a solid sphere,the radius of gyration squared is $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
For a disc,the radius of gyration squared is $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
The ratio of the time taken by the sphere $(t_S)$ to the time taken by the disc $(t_D)$ is: $\frac{t_S}{t_D} = \sqrt{\frac{1 + (K^2/R^2)_S}{1 + (K^2/R^2)_D}}$.
Substituting the values: $\frac{t_S}{t_D} = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$.
Thus,the ratio is $\sqrt{14}:\sqrt{15}$.

(A) For an object rolling down an inclined plane without slipping,the conservation of mechanical energy states that the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a uniform spherical shell,the moment of inertia about its center is $I = \frac{2}{3}mr^2$.
Since the object is rolling without slipping,$\omega = \frac{v}{r}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mr^2)(\frac{v}{r})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{3}mv^2$
$mgh = (\frac{1}{2} + \frac{1}{3})mv^2 = \frac{5}{6}mv^2$
$gh = \frac{5}{6}v^2$
$v^2 = \frac{6gh}{5}$
$v = \sqrt{\frac{6gh}{5}}$

(D) For a solid disk rolling down an inclined plane,the forces acting along the plane are the component of gravity $mg \sin \theta$ and the frictional force $f$.
The equation of motion for linear acceleration is: $mg \sin \theta - f = ma$ $(1)$
The torque equation about the center of mass is: $\tau = I \alpha = fR$
Since the disk rolls without slipping,$\alpha = a/R$. The moment of inertia for a solid disk is $I = \frac{1}{2} mR^2$.
Substituting these into the torque equation: $fR = (\frac{1}{2} mR^2) (\frac{a}{R})$
$fR = \frac{1}{2} maR$
$f = \frac{1}{2} ma$

(C) When the cylinder rolls up the incline,it tends to slide down due to gravity,so the static friction acts up the incline to provide the necessary torque for rolling and to oppose the sliding tendency.
At the highest point,the cylinder stops momentarily. As it starts rolling back down,the component of gravity $mg \sin \theta$ acts down the plane,and the cylinder tends to slide down. To maintain pure rolling,static friction must act up the incline to provide the torque required for rotation.
Therefore,in both the upward and downward journeys,the force of friction is directed up the plane.

(D) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$. Since the angle of inclination $\theta$ and the radius of gyration $K$ are constant for the same ball,the acceleration $a$ is constant throughout the motion. Thus,statement $(ii)$ is incorrect.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,with $u = 0$,we get $S = \frac{1}{2}at^2$,which implies $t = \sqrt{\frac{2S}{a}}$.
Let $S_P$ be the distance $PO$ and $S_Q$ be the distance $QO$. Since $S_P = \frac{h}{\sin \theta}$ and $S_Q = \frac{2h}{\sin \theta}$,we have $S_Q = 2S_P$.
Therefore,the time taken $t_Q = \sqrt{\frac{2(2S_P)}{a}} = \sqrt{2} t_P$. Thus,statement $(i)$ is incorrect.
By the law of conservation of energy,the potential energy at the starting point is converted into kinetic energy at the bottom $O$. $K.E._P = mgh$ and $K.E._Q = mg(2h) = 2mgh$. Therefore,$K.E._Q = 2 K.E._P$. Thus,statement $(iii)$ is correct.

(C) Using the conservation of mechanical energy,the total potential energy at the top is converted into translational and rotational kinetic energy at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the objects roll without slipping,$\omega = v/R$ and $I = mk^2$,where $k$ is the radius of gyration.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(v/R)^2 = \frac{1}{2}mv^2(1 + k^2/R^2)$
Translational kinetic energy $K_t = \frac{1}{2}mv^2 = \frac{mgh}{1 + k^2/R^2}$.
For a ring,$k^2 = R^2 \implies K_{ring} = \frac{mgh}{1+1} = \frac{mgh}{2}$.
For a coin (disc),$k^2 = R^2/2 \implies K_{coin} = \frac{mgh}{1+1/2} = \frac{2mgh}{3}$.
For a solid ball (sphere),$k^2 = 2R^2/5 \implies K_{ball} = \frac{mgh}{1+2/5} = \frac{5mgh}{7}$.
Taking the ratio $K_{ring} : K_{coin} : K_{ball} = \frac{1}{2} : \frac{2}{3} : \frac{5}{7}$.
Multiplying by the $LCM$ of denominators $(42)$: $21 : 28 : 30$.

(D) When a cylinder rolls down an incline with sliding, the kinetic friction acts on the body to change its translational and rotational motion.
If the initial conditions (such as initial velocity or angular velocity) are such that the friction can bring the body to the condition of pure rolling $(v = r\omega)$, then it will start pure rolling after some time.
However, if the incline is not long enough or the conditions are such that the body reaches the bottom before the pure rolling condition is satisfied, it may never start pure rolling.
Therefore, both possibilities exist depending on the initial parameters and the length of the incline.

(D) $1$. The friction force $f$ acts up the incline,while the component of gravity $Mg \sin \theta$ acts down the incline. Thus,friction opposes the translational motion of the centre of mass.
$2$. The torque $\tau$ about the centre of mass is provided by the friction force $f$,where $\tau = f \cdot R$. This torque causes the angular acceleration $\alpha$ required for rolling motion.
$3$. For pure rolling,the condition $v = \omega R$ must be satisfied. Without friction,there is no torque to produce angular acceleration $\alpha$,so the bodies would simply slide down the plane without rotating. Thus,friction is essential for rolling.
$4$. Since statements $A$,$B$,and $C$ are all correct,the correct option is $D$.

(D) For an object of mass $M$ and radius $R$ rolling down an inclined plane of angle $\theta$,the acceleration is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a ring,$I = MR^2$,so $a_{ring} = \frac{g \sin \theta}{2}$. The friction force $f_{ring} = Mg \sin \theta - Ma_{ring} = \frac{1}{2} Mg \sin \theta$.
For a solid cylinder,$I = \frac{1}{2} MR^2$,so $a_{cyl} = \frac{g \sin \theta}{1 + 0.5} = \frac{2}{3} g \sin \theta$. The friction force $f_{cyl} = Mg \sin \theta - Ma_{cyl} = \frac{1}{3} Mg \sin \theta$.
Comparing the friction forces,$f_{ring} = 0.5 Mg \sin \theta$ and $f_{cyl} = 0.33 Mg \sin \theta$. Thus,$f_{ring} > f_{cyl}$.
Also,the condition for pure rolling is $f \le \mu N = \mu Mg \cos \theta$. Since $f_{ring} > f_{cyl}$,if the friction is sufficient for the ring to roll,it is definitely sufficient for the cylinder to roll. Therefore,both $(B)$ and $(C)$ are correct.

(D) For an object rolling down an inclined plane without slipping,the mechanical energy is conserved because the work done by static friction is zero.
Using the conservation of energy: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.
Since $v = R\omega$,we have $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}Mv^2(1 + \frac{I}{MR^2})$.
For a ring,$I = MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + 1) = Mv^2$,which gives $v_{ring} = \sqrt{gh}$.
For a solid cylinder,$I = \frac{1}{2}MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + \frac{1}{2}) = \frac{3}{4}Mv^2$,which gives $v_{cylinder} = \sqrt{\frac{4gh}{3}} = 2\sqrt{\frac{gh}{3}}$.
Thus,all statements are correct.

(B) For a body rolling down an inclined plane without slipping,the forces acting on it are the component of gravity $Mg \sin \theta$ down the plane and the static friction $f$ acting up the plane.
Applying Newton's second law for linear motion: $Mg \sin \theta - f = Ma$ (where $a$ is the linear acceleration).
Applying Newton's second law for rotational motion about the center of mass: $\tau = I \alpha = fR$,where $\alpha = a/R$ is the angular acceleration.
Substituting $f = I \alpha / R = I a / R^2$ into the linear equation:
$Mg \sin \theta - \frac{I a}{R^2} = Ma$
$Mg \sin \theta = Ma + \frac{I a}{R^2} = Ma(1 + \frac{I}{M R^2})$
Solving for $a$,we get: $a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$.

(D) The radius of the ring is $R = 4a$. The radius of the disc is $r = a$.
When the disc is at the top of its path (at the same level as the center of the ring),the center of mass of the disc is at a height $h = R - r = 4a - a = 3a$ above the lowest point of the ring.
By the principle of conservation of mechanical energy,the potential energy lost by the disc is equal to the kinetic energy gained by the disc.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the disc rolls without slipping,$v = r\omega$,so $\omega = v/r$.
The moment of inertia of the disc about its center of mass is $I = \frac{1}{2}mr^2$.
Substituting these into the energy equation:
$mg(3a) = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$
$3mga = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$
$3mga = \frac{3}{4}mv^2$
$3ga = \frac{3}{4}v^2$
$v^2 = 4ga$
$v = \sqrt{4ga}$

(B) Using the conservation of energy,the potential energy at the top equals the total kinetic energy at the bottom:
$mgl\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = \frac{1}{2}mR^2$ and $\omega = \frac{v}{R}$ for rolling without slipping:
$mgl\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
Solving for $v$,we get $v = \sqrt{\frac{4gl\sin\theta}{3}}$.
The angular momentum $L$ about the point of contact is given by $L = I_{cm}\omega + mvR$:
$L = (\frac{1}{2}mR^2)(\frac{v}{R}) + mvR = \frac{1}{2}mvR + mvR = \frac{3}{2}mvR$
Substituting the value of $v$:
$L = \frac{3}{2}mR\sqrt{\frac{4gl\sin\theta}{3}} = \sqrt{\frac{9}{4}m^2R^2 \cdot \frac{4gl\sin\theta}{3}} = \sqrt{3m^2R^2gl\sin\theta}$.

(D) $1$. On a horizontal surface, for pure rolling without any external force, the acceleration of the center of mass is zero, and the angular acceleration is zero. Thus, the friction force is zero.
$2$. When moving upward on an inclined plane, the sphere experiences a torque due to gravity that tends to decrease its angular velocity. Friction acts in the upward direction along the incline to provide the necessary torque to maintain pure rolling.
$3$. When moving downward on an inclined plane, the sphere experiences a torque due to gravity that tends to increase its angular velocity. Friction acts in the upward direction along the incline to oppose this increase and maintain pure rolling.
$4$. Friction does not always oppose the motion of the center of mass; it acts to maintain the condition of pure rolling $(v = r\omega)$. Therefore, the statement that friction force is always opposite to the motion of the sphere is incorrect.

(A) The total mass of the system is $M = m + m = 2m$. Since the liquid is non-viscous and the inner surface is frictionless,the liquid does not rotate with the sphere. Thus,the moment of inertia of the system about the center of mass is only due to the hollow sphere,$I = \frac{2}{3}mR^2$.
For pure rolling,$v = \omega R$,so $\omega = v/R$.
The translational kinetic energy is $K_{trans} = \frac{1}{2} M v^2 = \frac{1}{2} (2m) v^2 = mv^2$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} mR^2) (v/R)^2 = \frac{1}{3} mv^2$.
The ratio is $\frac{K_{trans}}{K_{rot}} = \frac{mv^2}{\frac{1}{3} mv^2} = \frac{3}{1}$.
Thus,$x = 3$ and $y = 1$. The value of $(x+y) = 3+1 = 4$.

(D) The forces acting on the disc along the incline are the component of gravity $mg \sin \theta$ acting downwards and the kinetic friction $f_k$ acting upwards.
The equation of motion for the centre of mass is: $mg \sin \theta - f_k = ma$.
The torque equation about the centre of mass is: $f_k R = I \alpha = \left(\frac{mR^2}{2}\right) \alpha$.
From the torque equation,we get $f_k = \frac{mR \alpha}{2}$.
Substituting $f_k = \mu N = \mu mg \cos \theta$ into the torque equation:
$\frac{mR \alpha}{2} = \mu mg \cos \theta$.
Since $\mu = \tan \theta = \frac{\sin \theta}{\cos \theta}$,we have:
$\frac{mR \alpha}{2} = \left(\frac{\sin \theta}{\cos \theta}\right) mg \cos \theta = mg \sin \theta$.
Solving for $R \alpha$:
$R \alpha = 2 g \sin \theta$.

(A) When a sphere rolls up an inclined plane,its center of mass experiences a downward acceleration due to the component of gravity $(mg \sin \theta)$ and the frictional force $(f)$.
To maintain rolling motion,the sphere must have an angular deceleration (clockwise torque if the sphere is moving up the plane).
The torque about the center of mass is provided by the frictional force. For the sphere to slow down its rotation while moving up,the frictional force must act in a direction that creates a clockwise torque.
Since the point of contact is below the center of mass,a frictional force directed up the plane creates a clockwise torque,which opposes the initial angular velocity. Therefore,the frictional force is directed up the plane.

(A) For a hollow cylinder,the moment of inertia about its central axis is $I = mR^2$.
When placed on the incline,the friction force $f$ acts up the plane to oppose the slipping. The friction force is $f = \mu N = \mu mg \cos \theta = (\tan \theta) mg \cos \theta = mg \sin \theta$.
The torque $\tau$ about the centre of mass is $\tau = fR = (mg \sin \theta) R$.
Using $\tau = I\alpha$,we get $(mg \sin \theta) R = (mR^2) \alpha$,which gives the angular acceleration $\alpha = (g \sin \theta) / R$.
Since the cylinder is spinning clockwise,the friction force acts up the plane,causing an angular deceleration. The final angular velocity $\omega = \omega_0 - \alpha t$.
Setting $\omega = 0$ to find the time when the cylinder stops spinning (and thus the centre of mass remains stationary until pure rolling begins),we have $0 = \omega_0 - (g \sin \theta / R) t$.
Solving for $t$,we get $t = \omega_0 R / (g \sin \theta)$.

(D) For a solid sphere rolling down an inclined plane,the acceleration of the center of mass is $a_{cm} = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{5}{7} g \sin \theta$. The angular acceleration is $\alpha = \frac{a_{cm}}{R} = \frac{5g \sin \theta}{7R}$.
Any point on the sphere has two components of acceleration: centripetal acceleration $\omega^2 R$ directed towards the center and tangential acceleration $R\alpha$ perpendicular to the radius.
At the point of contact $C$,the acceleration is purely centripetal,directed towards the center of the sphere,which is along the line $CA$.

(B) For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$.
Since the sphere rolls without slipping,the condition $V = \omega R$ holds,which implies $\omega = \frac{V}{R}$.
The rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} MR^2) (\frac{V}{R})^2 = \frac{1}{5} MV^2$.
The translational kinetic energy is $E_t = \frac{1}{2} MV^2$.
The total kinetic energy is $E_{total} = E_r + E_t = \frac{1}{5} MV^2 + \frac{1}{2} MV^2 = \frac{7}{10} MV^2$.
The ratio of rotational kinetic energy to total energy is $\frac{E_r}{E_{total}} = \frac{\frac{1}{5} MV^2}{\frac{7}{10} MV^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.
To find the percentage,we calculate $\frac{2}{7} \times 100 \approx 28.57\%$,which rounds to $28.6\%$.

(D) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5} MR^2$.
Substituting this into the formula: $a = \frac{g \sin \theta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 2/5} = \frac{g \sin \theta}{7/5} = \frac{5}{7} g \sin \theta$.
Given $\theta = 30^o$,we have $\sin 30^o = 0.5$.
Therefore,$a = \frac{5}{7} g \times 0.5 = \frac{5}{14} g$.

(A) For a body rolling down an inclined plane without slipping,the acceleration is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.
For a solid cylinder,the radius of gyration $k$ is such that $k^2 = \frac{1}{2} R^2$,so $\frac{k^2}{R^2} = \frac{1}{2}$.
Thus,$a = \frac{g \sin \theta}{1 + 1/2} = \frac{2}{3} g \sin \theta$.
The force equation along the incline is $mg \sin \theta - f = ma$,where $f$ is the friction force.
Substituting $a$,we get $f = mg \sin \theta - m(\frac{2}{3} g \sin \theta) = \frac{1}{3} mg \sin \theta$.
Also,the torque equation about the center is $f R = I \alpha = (\frac{1}{2} m R^2)(\frac{a}{R}) = \frac{1}{2} m R a$.
Substituting $a$,$f = \frac{1}{2} m (\frac{2}{3} g \sin \theta) = \frac{1}{3} mg \sin \theta$.
Since $f \le \mu N$ and $N = mg \cos \theta$,we have $\frac{1}{3} mg \sin \theta \le \mu mg \cos \theta$.
Therefore,$\mu \ge \frac{1}{3} \tan \theta$.

(D) When a body rolls down an inclined plane without slipping,the force of static friction acts up the incline.
This frictional force opposes the translational motion of the center of mass.
Simultaneously,this force provides a torque about the center of mass that causes the body to rotate.
Thus,the frictional force acts to convert some of the translational kinetic energy into rotational kinetic energy.

(B) The velocity of a body rolling down an inclined plane is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$,where $k$ is the radius of gyration.
For a solid sphere,$k^2 = \frac{2}{5}R^2$,so $v_{solid} = \sqrt{\frac{2gh}{1 + 0.4}} = \sqrt{\frac{2gh}{1.4}}$.
For a hollow sphere,$k^2 = \frac{2}{3}R^2$,so $v_{hollow} = \sqrt{\frac{2gh}{1 + 0.67}} = \sqrt{\frac{2gh}{1.67}}$.
Since the denominator for the solid sphere is smaller,the solid sphere reaches the bottom with a greater speed.
Both spheres start from the same height,so their potential energy $mgh$ is the same. Since the total energy is conserved,the total kinetic energy (translational + rotational) at the bottom is the same for both.