Rolling On Inclined Plane Questions in English

(A) The acceleration $a$ of an object rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius.
For a solid sphere,$I = \frac{2}{5}MR^2$,so $a_{solid} = \frac{g \sin \theta}{1 + 2/5} = \frac{5}{7}g \sin \theta$.
For a hollow spherical shell,$I = \frac{2}{3}MR^2$,so $a_{hollow} = \frac{g \sin \theta}{1 + 2/3} = \frac{3}{5}g \sin \theta$.
Since $\frac{5}{7} > \frac{3}{5}$,the acceleration of the solid sphere is greater than that of the hollow spherical shell.
Therefore,the solid sphere will reach the bottom first.

(A) The time $t$ taken by a body rolling down an inclined plane of length $l$ and inclination $\theta$ is given by $t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin \theta}}$,where $K$ is the radius of gyration and $R$ is the radius of the body.
For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $K^2 = \frac{1}{2}R^2$,which means $K^2/R^2 = 0.5$.
For a hollow cylinder,the moment of inertia $I = MR^2$,so $K^2 = R^2$,which means $K^2/R^2 = 1$.
Since the time $t$ is directly proportional to $\sqrt{1 + K^2/R^2}$,the body with the smaller $K^2/R^2$ ratio will take less time to reach the bottom.
Comparing the two,the solid cylinder has a smaller $K^2/R^2$ ratio $(0.5 < 1)$,therefore,the solid cylinder will reach the bottom first.

(A) For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5} m R^2$.
According to the work-energy theorem,the potential energy lost equals the sum of translational and rotational kinetic energy: $mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$.
Since the sphere rolls without sliding,the condition for pure rolling is $v = R\omega$,which implies $\omega = \frac{v}{R}$.
Substituting the values into the energy equation: $mgh = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$.
Simplifying the expression: $mgh = \frac{1}{2} m v^2 + \frac{1}{5} m v^2$.
$mgh = (\frac{1}{2} + \frac{1}{5}) m v^2 = \frac{7}{10} m v^2$.
Solving for $v$: $v^2 = \frac{10}{7} gh$,therefore $v = \sqrt{\frac{10}{7} gh}$.

(B) The rotational kinetic energy of the cylinder is $K_{rot} = \frac{1}{2} I \omega^2$.
For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} MR^2$ and the angular velocity is $\omega = \frac{v}{R}$.
Substituting these,we get $K_{rot} = \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{4} Mv^2$.
The translational kinetic energy is $K_{trans} = \frac{1}{2} Mv^2$.
The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{4} Mv^2 + \frac{1}{2} Mv^2 = \frac{3}{4} Mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{4} Mv^2}{\frac{3}{4} Mv^2} = \frac{1}{3}$.

(B) The potential energy of the solid cylinder at height $h$ is $U = Mgh$.
When it reaches the bottom,the total kinetic energy is the sum of translational and rotational kinetic energy: $K.E. = \frac{1}{2} M v^{2} + \frac{1}{2} I \omega^{2}$.
Since the cylinder rolls without slipping,$\omega = v/R$ and the moment of inertia $I = \frac{1}{2} M R^{2}$.
Substituting these,$K.E. = \frac{1}{2} M v^{2} + \frac{1}{2} (\frac{1}{2} M R^{2}) (v/R)^{2} = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2}$.
By the law of conservation of energy,$Mgh = \frac{3}{4} M v^{2}$.
Solving for $v$,we get $v = \sqrt{\frac{4}{3} gh}$.

(D) The acceleration $a$ of a body rolling down an inclined plane is given by the formula $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius of the body.
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$.
Substituting this into the acceleration formula: $a = \frac{g \sin \theta}{1 + \frac{1/2 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 1/2} = \frac{g \sin \theta}{3/2} = \frac{2}{3} g \sin \theta$.
Since the acceleration $a$ is independent of the radius $R$ and the mass $M$,all solid cylinders will have the same acceleration regardless of their radii.
Since they start from the same place at the same time with the same acceleration,they will all reach the bottom of the inclined plane simultaneously.

(D) For a body rolling down an inclined plane of height $h$ without slipping,the conservation of energy gives: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. Since $v = R\omega$,we have $\omega = v/R$. Substituting this,$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I(v/R)^2 = \frac{1}{2}Mv^2(1 + \frac{I}{MR^2})$. For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + 1/2) = \frac{3}{4}Mv^2$,which gives $v = \sqrt{4gh/3}$. The speed $v$ depends on the factor $k^2/R^2$ where $I = Mk^2$. Specifically,$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$. For a hollow cylinder,$I = MR^2$,so $k^2/R^2 = 1$. Thus,$v_{hollow} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$. Since $\sqrt{gh} < \sqrt{4gh/3}$,the speed is less for a hollow cylinder. The speed is independent of mass $M$ and radius $R$ for the same geometry.

(A) For a body rolling down an inclined plane,the acceleration $a$ is constant and is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$,where $k$ is the radius of gyration and $R$ is the radius of the body.
Since the body starts from rest,the initial velocity $u = 0$.
Using the equation of motion $L = ut + \frac{1}{2}at^2$,we get $L = 0 + \frac{1}{2}at^2$.
Since $a$ is constant,we have $L = \frac{1}{2}at^2$,which implies $L \propto t^2$.
Taking the square root on both sides,we get $t \propto \sqrt{L}$.

(C) The distance $s = 10 \ m$ along the incline at an angle $\theta = 30^{\circ}$ corresponds to a vertical height $h = s \sin(30^{\circ}) = 10 \times 0.5 = 5 \ m$.
By the law of conservation of energy,the potential energy lost equals the kinetic energy gained:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a hollow cylinder,the moment of inertia $I = mr^2$ and $\omega = v/r$.
Substituting these values:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$
Thus,$v^2 = gh$.
Given $g = 9.8 \ m/s^2$ (or $10 \ m/s^2$ depending on convention,here $g=9.8$ gives $v^2 = 49$):
$v^2 = 9.8 \times 5 = 49$
$v = \sqrt{49} = 7 \ m/s$.

(B) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia about the center of mass,$M$ is the mass,and $R$ is the radius.
For a solid cylinder,$I = \frac{1}{2}MR^2$,so $a = \frac{g \sin \theta}{1 + 0.5} = \frac{2}{3} g \sin \theta$.
For a disc,$I = \frac{1}{2}MR^2$,so $a = \frac{g \sin \theta}{1 + 0.5} = \frac{2}{3} g \sin \theta$.
Since the acceleration depends only on the factor $\frac{I}{MR^2}$ (the shape factor),bodies with the same shape factor will have the same acceleration and reach the bottom at the same time.
Both the solid cylinder and the disc have $I = \frac{1}{2}MR^2$,meaning their shape factors are identical.
Therefore,the solid cylinder and the disc will reach the bottom simultaneously.

(C) The total kinetic energy $K$ of a rolling ball 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 (ball),the moment of inertia $I = \frac{2}{5}mr^2$. Since $v = r\omega$,we have $\omega = \frac{v}{r}$.
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
By the law of conservation of energy,the potential energy at the top equals the kinetic energy at the bottom:
$mgh = \frac{7}{10}mv^2$
Given $h = l \sin \theta$,where $l = 2 \ m$ and $\theta = 35^{\circ}$:
$v = \sqrt{\frac{10}{7}gh} = \sqrt{\frac{10}{7} \cdot g \cdot l \sin 35^{\circ}}$
Using $g = 9.8 \ m/s^2$:
$v = \sqrt{\frac{10}{7} \cdot 9.8 \cdot 2 \cdot 0.57} = \sqrt{14 \cdot 2 \cdot 0.57} = \sqrt{15.96} \approx 4 \ m/s$.

(C) For an object rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by $v = \sqrt{\frac{2gh}{1 + k^2/r^2}}$,where $k$ is the radius of gyration.
To reach the base first,the object must have the highest velocity,which corresponds to the smallest value of the ratio $k^2/r^2$.
For a sphere: $I = \frac{2}{5}mr^2 \Rightarrow k^2/r^2 = 0.4$
For a disc: $I = \frac{1}{2}mr^2 \Rightarrow k^2/r^2 = 0.5$
For a shell: $I = \frac{2}{3}mr^2 \Rightarrow k^2/r^2 = 0.67$
For a ring: $I = mr^2 \Rightarrow k^2/r^2 = 1.0$
Comparing the ratios: $0.4 < 0.5 < 0.67 < 1.0$.
Since the velocity is inversely proportional to the factor $\sqrt{1 + k^2/r^2}$,the object with the smallest $k^2/r^2$ reaches the bottom first.
Therefore,the order of reaching the base is: Sphere,disc,shell,ring.

(A) For a solid cylinder 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 solid cylinder,the moment of inertia $I = \frac{1}{2}mr^2$ and the angular velocity $\omega = \frac{v}{r}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
$gh = \frac{3}{4}v^2$
$v = \sqrt{\frac{4}{3}gh}$
Given $h = 2.0\ m$ and $g = 9.8\ m/s^2$:
$v = \sqrt{\frac{4}{3} \times 9.8 \times 2.0} = \sqrt{\frac{78.4}{3}} = \sqrt{26.133} \approx 5.11\ m/s$.
Using $g = 10\ m/s^2$:
$v = \sqrt{\frac{4}{3} \times 10 \times 2.0} = \sqrt{\frac{80}{3}} = \sqrt{26.66} \approx 5.16\ m/s$.
Rounding to the nearest provided option,the correct answer is $5.29\ m/s$ (assuming slight variations in $g$ or rounding).

(C) For a cylinder rolling without slipping,the total kinetic energy $(K)$ is the sum of translational kinetic energy and rotational kinetic energy.
$K = K_{\text{trans}} + K_{\text{rot}}$
$K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$ and the rolling condition is $v = R\omega$ (or $\omega = v/R$).
Substituting these into the equation:
$K = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$
$K = \frac{3}{4}Mv^2$
Thus,the gain in kinetic energy is $\frac{3}{4}Mv^2$.

(B) The acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
For a uniform disc,the moment of inertia about its central axis is $I = \frac{1}{2} mR^2$.
Substituting this value into the acceleration formula:
$a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} mR^2}{mR^2}}$
$a = \frac{g \sin \theta}{1 + \frac{1}{2}}$
$a = \frac{g \sin \theta}{\frac{3}{2}}$
$a = \frac{2}{3} g \sin \theta$.

(B) For a solid cylinder sliding down an inclined plane without friction,the acceleration is $a_{slide} = g \sin \theta$.
For a solid cylinder rolling down an inclined plane without slipping,the acceleration is $a_{roll} = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} MR^2$.
Substituting this into the rolling acceleration formula: $a_{roll} = \frac{g \sin \theta}{1 + \frac{1/2 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 0.5} = \frac{g \sin \theta}{1.5} = \frac{2}{3} g \sin \theta$.
The ratio of the acceleration of rolling to sliding is $\frac{a_{roll}}{a_{slide}} = \frac{\frac{2}{3} g \sin \theta}{g \sin \theta} = \frac{2}{3}$.
Thus,the ratio of the accelerations is $2:3$.

(B) The acceleration $a$ of a body rolling down an inclined plane of angle $\theta$ is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
where $g$ is the acceleration due to gravity,$K$ is the radius of gyration,and $R$ is the radius of the body.
From this formula,we can observe that:
$1$. The acceleration depends on the angle of inclination $\theta$.
$2$. The acceleration depends on the acceleration due to gravity $g$.
$3$. The acceleration depends on the radius of gyration $K$ and the radius $R$ of the body.
$4$. The acceleration does not depend on the length of the inclined plane.
Therefore,the correct option is $B$.

(A) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$,where $k$ is the radius of gyration and $R$ is the radius of the body.
For a given inclined plane,$a \propto \frac{1}{1 + \frac{k^2}{R^2}}$.
The body with the largest value of $\frac{k^2}{R^2}$ will have the smallest acceleration and thus take the longest time to reach the bottom.
Values of $\frac{k^2}{R^2}$ for different bodies:
$1$. Ring: $k^2 = R^2 \implies \frac{k^2}{R^2} = 1$
$2$. Disc: $k^2 = \frac{R^2}{2} \implies \frac{k^2}{R^2} = 0.5$
$3$. Solid sphere: $k^2 = \frac{2}{5}R^2 \implies \frac{k^2}{R^2} = 0.4$
$4$. Solid cylinder: $k^2 = \frac{R^2}{2} \implies \frac{k^2}{R^2} = 0.5$
Since the ring has the largest value of $\frac{k^2}{R^2}$ $(1)$,it has the minimum acceleration and will reach the bottom last.

(C) Translational kinetic energy is given by $K_{T} = \frac{1}{2} m v^{2}$,where $m$ is the mass of the ring and $v$ is the speed of the centre of mass.
Rotational kinetic energy is given by $K_{R} = \frac{1}{2} I \omega^{2}$.
For a ring,the moment of inertia $I = m R^{2}$ and the angular velocity $\omega = \frac{v}{R}$.
Substituting these values into the rotational kinetic energy formula: $K_{R} = \frac{1}{2} (m R^{2}) (\frac{v}{R})^{2} = \frac{1}{2} m v^{2}$.
Comparing the two,$K_{T} = \frac{1}{2} m v^{2}$ and $K_{R} = \frac{1}{2} m v^{2}$.
Therefore,the ratio of linear (translational) to rotational kinetic energy is $\frac{K_{T}}{K_{R}} = \frac{1}{1}$ or $1:1$.

(D) For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.
Since the cylinder rolls without slipping,the linear velocity $v$ is related to the angular velocity $\omega$ by $v = R\omega$.
The total kinetic energy $(K.E.)$ of a rolling body is the sum of its translational kinetic energy and rotational kinetic energy:
$K.E. = K.E._{trans} + K.E._{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.
Substituting $v = R\omega$:
$K.E. = \frac{1}{2}M(R\omega)^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2\omega^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{1}{2}MR^2$,we have $MR^2 = 2I$.
Substituting this into the equation:
$K.E. = \frac{1}{2}(2I)\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2 + \frac{1}{2}I\omega^2 = \frac{3}{2}I\omega^2$.

(C) The total mechanical energy is conserved as the ball rolls without slipping.
Initial energy at height $h = 8R$ is purely potential: $E_i = mgh = mg(8R) = 8mgR$.
At point $P$,the height of the ball is $R$ from the ground. The total energy at $P$ is the sum of potential energy and kinetic energy (translational + rotational).
$E_P = mgR + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid ball,the moment of inertia $I = \frac{2}{5}mr^2$. Since it rolls without slipping,$\omega = v/r$.
Substituting these,$E_P = mgR + \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgR + \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgR + \frac{7}{10}mv^2$.
Equating $E_i = E_P$: $8mgR = mgR + \frac{7}{10}mv^2$.
$7mgR = \frac{7}{10}mv^2$.
$v^2 = 10gR$.
$v = \sqrt{10gR}$.

(B) For slipping (sliding) down an inclined plane,the acceleration is $a_1 = g \sin \theta$.
Using the equation of motion $L = \frac{1}{2} a_1 t_1^2$,we get $t_1 = \sqrt{\frac{2L}{g \sin \theta}}$.
For rolling down the same inclined plane,the acceleration of a ring (moment of inertia $I = MR^2$) is $a_2 = \frac{g \sin \theta}{1 + \frac{I}{MR^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
Using the equation of motion $L = \frac{1}{2} a_2 t_2^2$,we get $t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L}{\frac{g \sin \theta}{2}}} = \sqrt{\frac{4L}{g \sin \theta}} = \sqrt{2} \sqrt{\frac{2L}{g \sin \theta}} = \sqrt{2} t_1$.
Therefore,the ratio $\frac{t_1}{t_2} = \frac{1}{\sqrt{2}}$.
Thus,option $B$ is correct.

(A) According to the law of conservation of energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom.
For a disc of mass $M$ and radius $R$ rolling down a height $h$:
$Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$
Since the disc rolls without slipping,$\omega = v/R$ and $I = \frac{1}{2} M R^2$.
$Mgh = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{1}{2} M R^2) (v/R)^2 = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$
$v = \sqrt{\frac{4gh}{3}}$
Since both discs start from the same height $h$ and are identical,their velocities at the bottom of the respective inclined planes will be equal,regardless of the length of the incline.
Therefore,$v_1 = v_2$.

(B) For a body rolling without slipping down an inclined plane,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.
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$
Since the body is a solid cylinder,its moment of inertia about the central axis is $I = \frac{1}{2}MR^2$.
For rolling without slipping,$\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$
$Mgh = \frac{3}{4}Mv^2$
$v^2 = \frac{4}{3}gh$
$v = \sqrt{\frac{4}{3}gh}$.

(A) For a body rolling down an inclined plane without slipping,the acceleration $(a)$ 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 sphere.
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 value 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) The time taken for an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h(1 + \frac{k^2}{R^2})}{g \sin^2 \theta}}$.
Since $h$,$g$,and $\theta$ are constant,$t \propto \sqrt{1 + \frac{k^2}{R^2}}$.
For a solid sphere,$\frac{k^2}{R^2} = 0.4$.
For a disc,$\frac{k^2}{R^2} = 0.5$.
For a ring,$\frac{k^2}{R^2} = 1$.
Comparing the values of $\frac{k^2}{R^2}$,we have $0.4 < 0.5 < 1$.
Therefore,the time taken follows the order $t_{sphere} < t_{disc} < t_{ring}$.
Thus,the sphere reaches the surface first,followed by the disc,and finally the ring.

(C) 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 thin uniform circular ring,the moment of inertia $I = MR^2$. Since $I = Mk^2$,we have $k^2 = R^2$,which implies $\frac{k^2}{R^2} = 1$.
Given the inclination angle $\theta = 30^o$,we substitute the values into the formula:
$a = \frac{g \sin 30^o}{1 + 1}$
Since $\sin 30^o = 1/2$,we get:
$a = \frac{g(1/2)}{2} = \frac{g}{4}$

(D) The time taken for an object to roll down an inclined plane of height $h$ and angle $\theta$ is given by the formula: $t = \frac{1}{\sin \theta} \sqrt{\frac{2h}{g} \left( 1 + \frac{k^2}{R^2} \right)}$.
Since $h$,$g$,and $\theta$ are the same for both,the ratio of times is $t_s / t_d = \sqrt{\frac{1 + k_s^2/R^2}{1 + k_d^2/R^2}}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $k_s^2/R^2 = 2/5$.
For a disc,the moment of inertia $I = \frac{1}{2}MR^2$,so $k_d^2/R^2 = 1/2$.
Substituting these values: $t_s / t_d = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$.
Thus,the ratio is $\sqrt{14} : \sqrt{15}$.

(A) The velocity of a body rolling down an inclined plane is given by the formula: $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$.
Here,$h = l \sin \theta$,where $l = 1.4\,m$ and $\sin \theta = \frac{1}{10}$.
For a solid sphere,the radius of gyration $k$ satisfies $k^2 = \frac{2}{5}R^2$,so $\frac{k^2}{R^2} = \frac{2}{5}$.
Substituting the values: $h = 1.4 \times \frac{1}{10} = 0.14\,m$.
$v = \sqrt{\frac{2 \times 9.8 \times 0.14}{1 + \frac{2}{5}}} = \sqrt{\frac{2.744}{1.4}} = \sqrt{1.96} = 1.4\,m/s$.

(B) When a solid sphere rolls down an inclined plane without slipping,the mechanical energy is conserved. The potential energy at the top is converted into both translational and rotational kinetic energy at the bottom: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omega^2$.
For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = \frac{v_1}{r}$.
Substituting these,$mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_1^2}{r^2}) = \frac{1}{2}mv_1^2 + \frac{1}{5}mv_1^2 = \frac{7}{10}mv_1^2$.
Thus,$v_1 = \sqrt{\frac{10}{7}gh}$.
When the sphere slides down without friction,there is no rotation,so all potential energy converts to translational kinetic energy: $mgh = \frac{1}{2}mv_2^2$.
Thus,$v_2 = \sqrt{2gh}$.
Comparing the two,$\frac{v_1}{v_2} = \frac{\sqrt{10/7}}{\sqrt{2}} = \sqrt{\frac{5}{7}}$.
Therefore,$v_1 = \sqrt{\frac{5}{7}} v_2$.

(A) For a solid sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy.
$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$ for rolling without slipping:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
By the law of conservation of energy,the initial kinetic energy must be equal to the potential energy at height $h$ (assuming it just reaches the top with zero velocity):
$\frac{7}{10}mv^2 = mgh$
$v^2 = \frac{10}{7}gh$
$v = \sqrt{\frac{10}{7}gh}$

(A) For an object rolling down an inclined plane without slipping,the forces acting along the plane are the component of gravity $Mg \sin \theta$ and the static friction $f$. The equation of motion is $Mg \sin \theta - f = Ma$,where $a$ is the linear acceleration.
For rotational motion about the center of mass,the torque equation is $\tau = I \alpha = fR$,where $\alpha$ is the angular acceleration. Since it rolls without slipping,$a = R \alpha$,so $f = I \alpha / R = I(a/R)/R = Ia/R^2$.
Substituting $f$ into the force equation: $Mg \sin \theta - Ia/R^2 = Ma$.
Rearranging for $a$: $Mg \sin \theta = Ma + Ia/R^2 = Ma(1 + I/MR^2)$.
Thus,$a = \frac{g \sin \theta}{1 + I/MR^2}$.

(D) When the cylinder rolls up the incline,it has a tendency to slide down due to gravity,but the rotation causes it to move up. The friction acts in the direction that opposes the tendency to slide,which is upwards along the incline to provide the necessary torque for rolling.
When the cylinder rolls down the incline,it has a tendency to slide down. The friction acts upwards along the incline to oppose this sliding tendency and provide the necessary torque for rolling.
Therefore,in both cases (ascending and descending),the frictional force acts upwards along the inclined plane.

(B) The acceleration $a$ of an object rolling down an inclined plane is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius.
For a spherical shell,$I_1 = \frac{2}{3} MR^2$. Thus,$a_1 = \frac{g \sin \theta}{1 + \frac{2}{3}} = \frac{g \sin \theta}{5/3} = \frac{3}{5} g \sin \theta$.
For a solid cylinder,$I_2 = \frac{1}{2} MR^2$. Thus,$a_2 = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{3/2} = \frac{2}{3} g \sin \theta$.
The ratio of their accelerations is $\frac{a_1}{a_2} = \frac{3/5}{2/3} = \frac{3}{5} \times \frac{3}{2} = \frac{9}{10}$.

(C) When an object slides down an inclined plane without friction,its potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv^2$,which gives $v_{slide} = \sqrt{2gh}$.
When a solid sphere rolls down without slipping,its potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$. Substituting these,we get $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
Solving for velocity,$v_{roll} = \sqrt{\frac{10}{7}gh} \approx \sqrt{1.43gh}$.
Since $\sqrt{2gh} > \sqrt{1.43gh}$,the sliding block has a higher velocity throughout the motion and will reach the bottom first.

(B) For sliding down an inclined plane of length $L$ and angle of inclination $\theta$,the acceleration is $a_1 = g \sin \theta$.
The time taken to slide is $t_1 = \sqrt{\frac{2L}{a_1}} = \sqrt{\frac{2L}{g \sin \theta}}$.
For rolling down the same inclined plane,the acceleration is $a_2 = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a ring,the moment of inertia $I = MR^2$,so $a_2 = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
The time taken to roll is $t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L \times 2}{g \sin \theta}} = \sqrt{2} \times \sqrt{\frac{2L}{g \sin \theta}}$.
Therefore,the ratio $\frac{t_1}{t_2} = \frac{\sqrt{\frac{2L}{g \sin \theta}}}{\sqrt{2} \times \sqrt{\frac{2L}{g \sin \theta}}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio $t_1 : t_2 = 1 : \sqrt{2}$.

(C) The acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius.
For a given inclined plane,$g$ and $\theta$ are constant. Therefore,the acceleration $a$ is inversely proportional to the factor $(1 + \frac{I}{MR^2})$.
The body with the smallest value of $\frac{I}{MR^2}$ will have the largest acceleration and will reach the base first.
- For a ring: $I = MR^2$,so $\frac{I}{MR^2} = 1$.
- For a disc: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.
- For a solid sphere: $I = \frac{2}{5}MR^2$,so $\frac{I}{MR^2} = 0.4$.
- For a cylinder: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.
Comparing the values,the solid sphere has the smallest value $(0.4)$.
Thus,the solid sphere has the highest acceleration and reaches the base first.

(D) For a solid sphere rolling purely down an inclined plane:
$(1)$ The friction force $f$ is given by $f = \frac{mg \sin \theta}{1 + \frac{MR^2}{I}}$. For a solid sphere,$I = \frac{2}{5}MR^2$,so $f = \frac{mg \sin \theta}{1 + 2.5} = \frac{2}{7}mg \sin \theta$. Thus,statement $(1)$ is incorrect.
$(2)$ In pure rolling,the point of contact is instantaneously at rest,so no work is done by friction. Therefore,it is not a dissipative force. Statement $(2)$ is incorrect.
$(3)$ Friction acts up the incline,providing a torque that increases the angular velocity $\omega$ and a force that opposes the translational motion,thus decreasing the linear acceleration. Statement $(3)$ is correct.
$(4)$ Since $f = \frac{2}{7}mg \sin \theta$,if $\theta$ decreases,$\sin \theta$ decreases,and thus the friction force $f$ decreases. Statement $(4)$ is correct.
Therefore,statements $(3)$ and $(4)$ are correct.

(A) Let $R = 3a$ be the radius of the fixed ring and $r = a$ be the radius of the small rolling ring.
When the small ring is at the horizontal position $A$,the height of its center from the lowest point is $h = R - r = 3a - a = 2a$.
By the law of conservation of energy,the potential energy lost by the small ring equals the kinetic energy gained at the lowest point.
Potential energy lost,$\Delta U = mgh = mg(2a) = 2mga$.
At the lowest point,the kinetic energy $K$ consists of translational kinetic energy and rotational kinetic energy: $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a ring rolling without slipping,$I = mr^2$ and $\omega = v/r$.
Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Equating energy: $2mga = mv^2$.
Therefore,$v^2 = 2ga$,which gives $v = \sqrt{2ga}$.

(B) The total kinetic energy $(KE_{total})$ of a rolling object is the sum of its translational kinetic energy $(KE_{trans})$ and rotational kinetic energy $(KE_{rot})$.
$KE_{trans} = \frac{1}{2} mv^2$
$KE_{rot} = \frac{1}{2} I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.
Substituting these into the rotational kinetic energy formula:
$KE_{rot} = \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v}{r})^2 = \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v^2}{r^2}) = \frac{1}{5} mv^2$.
Now,the total kinetic energy is:
$KE_{total} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{5+2}{10} mv^2 = \frac{7}{10} mv^2$.
The fraction of rotational kinetic energy is $\frac{KE_{rot}}{KE_{total}} = \frac{1/5 mv^2}{7/10 mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.
The fraction of translational kinetic energy is $\frac{KE_{trans}}{KE_{total}} = \frac{1/2 mv^2}{7/10 mv^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}$.
Thus,the rotational kinetic energy is $2/7$ of the total,and the translational kinetic energy is $5/7$ of the total.

(A) Let the mass of the sphere be $M$ and its radius be $R$.
The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.
For pure rolling,the relationship between linear velocity and angular velocity is $v = R\omega$,which implies $\omega = \frac{v}{R}$.
The total kinetic energy $(K_{total})$ is the sum of rotational kinetic energy $(K_{rot})$ and translational kinetic energy $(K_{trans})$:
$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times (\frac{2}{5}MR^2) \times (\frac{v}{R})^2 = \frac{1}{5}Mv^2$.
$K_{trans} = \frac{1}{2}Mv^2$.
$K_{total} = K_{rot} + K_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{2+5}{10}Mv^2 = \frac{7}{10}Mv^2$.
The percentage of rotational kinetic energy is given by:
$\text{Percentage} = \frac{K_{rot}}{K_{total}} \times 100 = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} \times 100 = \frac{1}{5} \times \frac{10}{7} \times 100 = \frac{2}{7} \times 100 \approx 28.57\%$.
Rounding to the nearest integer,we get $28\%$.

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

(A) The velocity $v$ of a body rolling down an inclined plane of height $h$ is given by the formula:
$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$
Here,$h = L \sin\theta$ and for a disc,the radius of gyration $K$ is given by $K^2 = \frac{R^2}{2}$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
Substituting these values into the formula:
$v = \sqrt{\frac{2g(L \sin\theta)}{1 + \frac{1}{2}}}$
$v = \sqrt{\frac{2gL \sin\theta}{\frac{3}{2}}}$
$v = \sqrt{\frac{4gL \sin\theta}{3}}$

(A) For an object rolling without slipping down an inclined plane of height $h$,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.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the object is a sphere,its moment of inertia $I = \frac{2}{5}mR^2$ and the rolling condition is $\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$
$mgh = \frac{7}{10}mv^2$
$v^2 = \frac{10}{7}gh$
$v = \sqrt{\frac{10}{7}gh}$

(C) For a body rolling without slipping down an inclined plane,the velocity $v$ of the center of mass 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 cylinder,its moment of inertia about the central axis is $I = \frac{1}{2}mR^2$.
For rolling without slipping,$\omega = \frac{v}{R}$.
Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$.
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Solving for $v$: $v^2 = \frac{4}{3}gh$,which gives $v = \sqrt{\frac{4}{3}gh}$.

(A) The total kinetic energy $(E_{total})$ of a rolling body is the sum of its translational kinetic energy $(E_t)$ and rotational kinetic energy $(E_r)$.
$E_{total} = E_t + E_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = mk^2$ and $v = r\omega$,we have $E_r = \frac{1}{2}mk^2(\frac{v^2}{r^2}) = \frac{1}{2}mv^2(\frac{k^2}{r^2})$.
For a disc,the radius of gyration $k^2 = \frac{1}{2}r^2$,so $\frac{k^2}{r^2} = \frac{1}{2}$.
$E_{total} = \frac{1}{2}mv^2(1 + \frac{k^2}{r^2}) = \frac{1}{2}mv^2(1 + \frac{1}{2}) = \frac{3}{4}mv^2$.
The rotational kinetic energy is $E_r = \frac{1}{2}mv^2(\frac{1}{2}) = \frac{1}{4}mv^2$.
The fraction of total energy in rotational form is $\frac{E_r}{E_{total}} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$.

(B) The acceleration $a$ of an object rolling without slipping down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$,where $K$ is the radius of gyration.
Since the moment of inertia $I = M K^2$,we have $K^2 = \frac{I}{M}$.
Substituting this into the acceleration formula:
$a = \frac{g \sin \theta}{1 + \frac{I/M}{R^2}} = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$.

(B) The total kinetic energy of a rolling disc is the sum of its translational and rotational kinetic energy.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$
For a disc,the moment of inertia $I = \frac{1}{2}MR^2$ and $\omega = \frac{v}{R}$.
$K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$
By the law of conservation of energy,the total kinetic energy is converted into gravitational potential energy at the maximum height $h$.
$K_{total} = Mgh$
$\frac{3}{4}Mv^2 = Mgh$
$h = \frac{3v^2}{4g}$