A solid sphere,a disc,and a solid cylinder st | vedclass

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Question

(A) The time taken by an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h}{g \sin^2 	heta} (1 + \frac{K^2}{R^2})}$,where $K$ is

Vedclass · Accepted Answer

The solid sphere will reach the bottom first.

Vedclass · Answer

(A) The time taken by an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h}{g \sin^2 	heta} (1 + \frac{K^2}{R^2})}$,where $K$ is the radius of gyration and $R$ is the radius of the object.For a solid sphere,$\frac{K^2}{R^2} = \frac{2}{5} = 0.4$.For a disc,$\frac{K^2}{R^2} = \frac{1}{2} = 0.5$.For a solid cylinder,$\frac{K^2}{R^2} = \frac{1}{2} = 0.5$.Since the solid sphere has the minimum value of $\frac{K^2}{R^2}$,it will have the minimum time of descent.Therefore,the solid sphere reaches the bottom first.

$A$ solid sphere,a disc,and a solid cylinder start rolling down an inclined plane from rest. All three objects are made of the same material and have the same mass. Then,

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