Question:

One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$. The total kinetic energy of the ring is

Updated On: Jan 1, 2023
  • $M \omega_{0}^{2} R^{2}$
  • $\frac{1}{2} M \omega_{0}^{2}(R-r)^{2}$
  • $M \omega_{0}^{2}(R-r)^{2}$
  • $\frac{3}{2} M \omega_{0}^{2}(R-r)^{2}$
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The Correct Option is C

Solution and Explanation

Answer (c) $M \omega_{0}^{2}(R-r)^{2}$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].