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 botto

Question

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 inclined plane. [2008]

Smart Achievers · Accepted Answer

(4)

In case of pure rolling on inclined plane acceleration of a body

$a = \frac{g \sin θ}{1 + \frac{I}{M R^{2}}}$

For hollow cylinder $\frac{I}{M R^{2}} = \frac{M R^{2}}{M R^{2}} = 1$

For solid cylinder $\frac{I}{M R^{2}} = \frac{\frac{1}{2} M R^{2}}{M R^{2}} = \frac{1}{2}$

Hence acceleration of solid cylinder is more than hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.

In the case of rolling there will be no work done by friction. Therefore total mechanical energy remains conserved.

$∴$ ${(K . E)}_{solid} = {(K . E)}_{hollow} = decrease in P.E = m g h$

Solution

1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

2 Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

3 Statement-1 is True, Statement-2 is False

4 Statement-1 is False, Statement-2 is True

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