(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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System of Particles and Rotational Motion

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Rolling On Inclined Plane

$A$ ring of radius $3a$ is fixed rigidly on a table. $A$ small ring of mass $m$ and radius $a$ rolls inside it without slipping,as shown in the figure. The small ring is released from position $A$ (the horizontal position). What will be the speed of the center of the small ring when it reaches the lowest point?

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$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

DifficultJEE MAIN 2023

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$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

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$STATEMENT-1$: Two cylinders,one hollow (metal) and the other solid (wood) with the same mass and identical dimensions,are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
$STATEMENT-2$: By the principle of conservation of energy,the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

DifficultIIT 2008

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$A$ circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane,it takes $t \ s$. When it rolls down the plane,it takes $\left(\frac{\alpha}{2}\right)^{1/2} t \ s$,where $\alpha$ is:

DifficultJEE MAIN 2024

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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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$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

$A$ circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane,it takes $t \ s$. When it rolls down the plane,it takes $\left(\frac{\alpha}{2}\right)^{1/2} t \ s$,where $\alpha$ is:

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