Two uniform solid spheres having unequal mass | vedclass

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Question

(C) The acceleration $a$ of a body rolling without slipping on an inclined plane of angle $	heta$ is given by the formula:$a = \frac{g \sin 	heta}{1

Vedclass · Accepted Answer

the two spheres reach the bottom together

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(C) The acceleration $a$ of a body rolling without slipping on an inclined plane of angle $	heta$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$where $K$ is the radius of gyration and $R$ is the radius of the sphere.For a uniform 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:$a = \frac{g \sin 	heta}{1 + \frac{2}{5}} = \frac{g \sin 	heta}{\frac{7}{5}} = \frac{5}{7} g \sin 	heta$Since the acceleration $a$ is independent of both the mass $M$ and the radius $R$,both spheres will have the same acceleration. Given they start from rest from the same height,they will reach the bottom at the same time. Thus,option $C$ is correct.

Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

Similar Questions

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. When it reaches the foot of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

$A$ solid cylinder is placed on a rough surface of inclination $\theta$. The minimum value of the coefficient of static friction between the cylinder and the surface so that the cylinder rolls without slipping is:

$A$ thin hollow cylinder is open at both ends.
$(i)$ It slides without rotation.
(ii) It rolls without slipping.
If the speed is the same in both cases,the ratio of the kinetic energies is:

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