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(A) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.The moment of inertia $I$ for

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The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

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(A) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.The moment of inertia $I$ for the objects are: $I_{	ext{ring}} = mR^2$,$I_{	ext{cylinder}} = \frac{1}{2}mR^2$,and $I_{	ext{sphere}} = \frac{2}{5}mR^2$.Comparing the moments of inertia: $I_{	ext{ring}} > I_{	ext{cylinder}} > I_{	ext{sphere}}$.Since $a$ is inversely proportional to the factor $(1 + \frac{I}{mR^2})$,the acceleration follows the order: $a_{	ext{ring}} Using the kinematic equation $v^2 = u^2 + 2as$ with $u = 0$,the final velocity $v = \sqrt{2as}$ is directly proportional to the square root of acceleration.Therefore,the final velocities at the bottom of the plane follow the order: $v_{	ext{ring}} Thus,the sphere has the greatest velocity and the ring has the least velocity.

Consider a situation in which a ring,a solid cylinder,and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and have identical diameters. The correct statement for this situation is:

Similar Questions

$A$ solid cylinder of mass $m$ is wrapped with an inextensible light string and is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is: [The coefficient of static friction,$\mu_{s}$,is $0.4$]

$Assertion$ : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane,compared to when it rolls down the same plane.
$Reason$ : In rolling down,a body acquires both kinetic energy of translation and rotation.

$A$ solid sphere rolls down without slipping on an inclined plane,then the percentage of rotational kinetic energy of the total energy will be ........ $\%.$

$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

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