A hollow sphere and a solid sphere having same mss and same radii are rolled down a rough inclined plane.

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KollegeApply · Accepted Answer

The solid sphere reaches the bottom with greater speed. Acceleration of a sphere on the incline plane is given by a = \frac{g\sin$\theta$}{1 + \frac{I_{COM}}{m r^2}} I_{COM} for a solid sphere = $\frac{2}{5}$m r^2 \text{So, }a = \frac{g\sin$\theta$}{1 + 2m r^$\frac{2}{5}$m r^2} = $\frac{5}{7}$g\sin$\theta$ I_{COM} for a hollow sphere = $\frac{2}{3}$m r^2 \text{So, }a' = \frac{g\sin$\theta$}{1 + 2m r^$\frac{2}{3}$m r^2} = $\frac{3}{5}$g\sin$\theta$ The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

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