Question

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is \(\frac{\sqrt{x}}{2}\). Then, the value of x is _____________.

Hint:

For rolling from rest through height h, \(v^2 = \frac{2gh}{1 + I/(mR^2)}\). Use \(I/(mR^2) = 1\) for a ring and \(\frac12\) for a solid cylinder.

Answer 1

Correct Answer: 3

Energy conservation for rolling without slipping: \(mgh = \frac12 mv^2 + \frac12 I\omega^2 = \frac12 mv^2\left(1+\frac{I}{mR^2}\right)\), so \(v = \sqrt{\frac{2gh}{1+I/mR^2}}\).
Ring: \(I = mR^2\), so \(v_r = \sqrt{\frac{2gh}{2}} = \sqrt{gh}\).
Solid cylinder: \(I = \frac12 mR^2\), so \(v_c = \sqrt{\frac{2gh}{3/2}} = \sqrt{\frac{4gh}{3}}\).
Ratio: \(\frac{v_r}{v_c} = \sqrt{\frac{gh}{4gh/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).
Comparing with \(\frac{\sqrt{x}}{2}\) gives \(x = 3\).
Answer: 3