Question:

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter.

The correct statement for this situation is

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Use energy conservation with rolling: \(v^2 = \dfrac{2gh}{1 + I/(mR^2)}\); a smaller \(I/(mR^2)\) gives a larger speed.
Updated On: Nov 4, 2026
  • All of them will have same velocity.
  • The ring has greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.
  • The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
  • The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.
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The Correct Option is C

Solution and Explanation

Energy conservation for rolling without slipping: \(mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 mv^2\left(1 + \dfrac{I}{mR^2}\right)\).
So \(v = \sqrt{\dfrac{2gh}{1 + I/(mR^2)}}\).
Ring: \(I/mR^2 = 1\), \(v = \sqrt{gh}\).
Solid cylinder: \(I/mR^2 = \tfrac12\), \(v = \sqrt{4gh/3}\).
Solid sphere: \(I/mR^2 = \tfrac25\), \(v = \sqrt{10gh/7}\).
Since \(1 \lt 4/3 \lt 10/7\) the sphere is fastest and the ring slowest.
Answer: (3) The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
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