(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.

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(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.

Class 11 Physics System of Particles and Rotational Motion Rolling On Inclined Plane

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$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. When these bodies roll down to the foot of the inclined plane,then

A
the mechanical energy of each body is conserved
B
the velocity of centre of mass of the cylinder is $2\sqrt{\frac{gh}{3}}$
C
the velocity of centre of mass of the ring is $\sqrt{gh}$
D
All of the above

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Rolling On Inclined Plane

$A$ ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \,s$, then the height of the top of the inclined plane, in metres, is. . . . . . Take $g=10 \,m \,s^{-2}$.

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The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest is:

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$A$ solid cylinder $30\ cm$ in diameter at the top of an inclined plane $2.0\ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is.......... $m/s$.

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$A$ solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^{\circ}$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant,the new speed of the sphere at the bottom of the plane is $v_2$. The ratio of $v_1^2 : v_2^2$ is

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How is the motion of a rolling sphere down a slope? Write the general formula for the kinetic energy of a rolling body down a slope.

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$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. When these bodies roll down to the foot of the inclined plane,then

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The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest is:

$A$ solid cylinder $30\ cm$ in diameter at the top of an inclined plane $2.0\ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is.......... $m/s$.

How is the motion of a rolling sphere down a slope? Write the general formula for the kinetic energy of a rolling body down a slope.

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