One solid sphere and one disc of the same rad | vedclass

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(A) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + K^2 / R^2}$,where $K$ is the radius of gyr

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different radius of gyration

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(A) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + K^2 / R^2}$,where $K$ is the radius of gyration and $R$ is the radius.$(i)$ For a solid sphere,$I = \frac{2}{5} MR^2 = MK^2$,so $K^2 = \frac{2}{5} R^2$. The acceleration is $a_s = \frac{g \sin 	heta}{1 + 2/5} = \frac{5}{7} g \sin 	heta$.$(ii)$ For a disc,$I = \frac{1}{2} MR^2 = MK^2$,so $K^2 = \frac{1}{2} R^2$. The acceleration is $a_d = \frac{g \sin 	heta}{1 + 1/2} = \frac{2}{3} g \sin 	heta$.$(iii)$ Since $a_s > a_d$,the solid sphere reaches the bottom earlier. The difference in acceleration is due to the different values of the radius of gyration $K$ for the two bodies.

One solid sphere and one disc of the same radius are rolling down an inclined plane without slipping. One reaches the bottom earlier than the other due to:

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