JEE Challenger
More from System of Particles and Rotational Motion
PhyHardMarks to AllAdv 2017 • Paper 2

Calculate Total Kinetic Energy of Ring Rolling on Rotating Finger

Comprehension Passage

Consider a circular ring of mass MM and radius RR twirled near the tip of a finger as illustrated. During this motion, the finger constantly maintains contact with the inner rim of the ring. The finger traces out the surface of a cone. The contact point between the ring and the finger traces a circular path of radius rr. The finger rotates with an angular speed ω0\omega_0. The ring performs rolling without slipping on the outer boundary of the smaller circle described by the contact point. The coefficient of friction between the ring and finger is μ\mu, and the acceleration due to gravity is gg.

Determine the total kinetic energy of the rotating ring.

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Question Diagram 1

Options

A

Mω02R2M\omega_0^2 R^2

B

12Mω02(R−r)2\frac{1}{2} M\omega_0^2 (R - r)^2

C

Mω02(R−r)2M\omega_0^2 (R - r)^2

D

32Mω02(R−r)2\frac{3}{2} M\omega_0^2 (R - r)^2

Topics & Concepts

System of Particles and Rotational MotionRolling MotionFriction

Step-by-Step Solution

To determine the total kinetic energy of the rotating ring, we analyze its motion by separating it into translation of the center of mass and rotation about the center of mass.

1. Motion of the Center of Mass

The ring has mass MM and radius RR. The point of contact between the inner rim of the ring and the finger moves in a circular path of radius rr.

The distance from the central axis (origin) to the center of mass (CMCM) of the ring is: RCM=R−rR_{CM} = R - r

Since the line joining the origin to the center of the ring revolves with an angular velocity ω0\omega_0, the speed of the center of mass vCMv_{CM} is: vCM=ω0RCM=ω0(R−r)v_{CM} = \omega_0 R_{CM} = \omega_0 (R - r)

2. Translational Kinetic Energy

The translational kinetic energy KtransK_{\text{trans}} of the ring is given by: Ktrans=12MvCM2=12Mω02(R−r)2K_{\text{trans}} = \frac{1}{2} M v_{CM}^2 = \frac{1}{2} M \omega_0^2 (R - r)^2

3. Rotational Motion and Condition for No Slipping

Let ω\omega be the spin angular velocity of the ring about its own center of mass.

For rolling without slipping on the surface, the instantaneous velocity of the point of contact PP on the ring relative to the center of rotation must be zero: v⃗P=v⃗CM+ω⃗×r⃗P/CM=0\vec{v}_P = \vec{v}_{CM} + \vec{\omega} \times \vec{r}_{P/CM} = 0

Taking magnitudes along the tangential direction: vCM−ωR=0v_{CM} - \omega R = 0

Substitute vCM=ω0(R−r)v_{CM} = \omega_0 (R - r): ω0(R−r)−ωR=0  ⟹  ω=(R−rR)ω0\omega_0 (R - r) - \omega R = 0 \implies \omega = \left( \frac{R - r}{R} \right) \omega_0

4. Rotational Kinetic Energy

, The moment of inertia of a circular ring about its central axis passing through its center of mass is: ICM=MR2I_{CM} = M R^2

The rotational kinetic energy KrotK_{\text{rot}} about the center of mass is: Krot=12ICMω2=12(MR2)[(R−r)Rω0]2=12Mω02(R−r)2K_{\text{rot}} = \frac{1}{2} I_{CM} \omega^2 = \frac{1}{2} (M R^2) \left[ \frac{(R - r)}{R} \omega_0 \right]^2 = \frac{1}{2} M \omega_0^2 (R - r)^2

5. Total Kinetic Energy

The total kinetic energy KtotalK_{\text{total}} of the ring is the sum of its translational and rotational kinetic energies: Ktotal=Ktrans+KrotK_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}

Ktotal=12Mω02(R−r)2+12Mω02(R−r)2=Mω02(R−r)2K_{\text{total}} = \frac{1}{2} M \omega_0^2 (R - r)^2 + \frac{1}{2} M \omega_0^2 (R - r)^2 = M \omega_0^2 (R - r)^2

Thus, the total kinetic energy of the rotating ring is Mω02(R−r)2M\omega_0^2 (R - r)^2, which corresponds to Option [C].

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