STATEMENT-1: Two cylinders,one hollow (metal) | vedclass

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(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.For a hollow cylinder,$I

Vedclass · Accepted Answer

$STATEMENT-1$ is False,$STATEMENT-2$ is True

Vedclass · Answer

(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.For a hollow cylinder,$I = mR^2$,so $a_{hollow} = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.For a solid cylinder,$I = \frac{1}{2}mR^2$,so $a_{solid} = \frac{g \sin 	heta}{1 + 0.5} = \frac{g \sin 	heta}{1.5} = \frac{2g \sin 	heta}{3}$.Since $a_{solid} > a_{hollow}$,the solid cylinder reaches the bottom first. Thus,$STATEMENT-1$ is False.By the principle of conservation of energy,the potential energy lost $(mgh)$ is converted into total kinetic energy $(K_{total} = K_{trans} + K_{rot})$. Since both have the same mass and height,their total kinetic energies at the bottom are identical. Thus,$STATEMENT-2$ is True.

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