A circular ring and a solid sphere having the same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities (ring to sphere) when reached at the bottom of the plane is √x/5, where x= .

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Accepted Answer

3.5. Rolling without slipping from rest through a height h: v²=(2gh)/(1+k) with k=I/(MR²). Ring, k=1: vᵣ²=(2gh)/2=gh. Solid sphere, k=2/5: vₛ²=(2gh)/(7/5)=(10gh)/7. (vᵣ²)/(vₛ²)=gh·7/(10gh)=7/(10). (vᵣ)/(vₛ)=√7/(10), and 7/(10)=(3.5)/5. Comparing with √x/5 gives x=3.5. The answer is not a whole number, which is unusual for this format but is what the printed key gives. The ring is slower, so the ratio is less than 1: √0.7=0.837.

A circular ring and a solid sphere having the same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities (ring to sphere) when reached at the bottom of the plane is √x/5, where x= ______.

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