A ring,a solid sphere,and a disc are rolling | vedclass

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

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Sphere,disc,ring

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(B) The time taken for an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h(1 + \frac{k^2}{R^2})}{g \sin^2 	heta}}$.Since $h$,$g$,and $	heta$ are constant,$t \propto \sqrt{1 + \frac{k^2}{R^2}}$.For a solid sphere,$\frac{k^2}{R^2} = 0.4$.For a disc,$\frac{k^2}{R^2} = 0.5$.For a ring,$\frac{k^2}{R^2} = 1$.Comparing the values of $\frac{k^2}{R^2}$,we have $0.4 Therefore,the time taken follows the order $t_{sphere} Thus,the sphere reaches the surface first,followed by the disc,and finally the ring.

$A$ ring,a solid sphere,and a disc are rolling down from the top of an inclined plane of the same height. What is the sequence in which they reach the surface?

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