AIIMS2019Physics-Rotational Motion

AIIMS 2019 Physics Moment of Inertia MCQ Question

Type: MCQ-numerical-Medium-Class 11

Given VCM=2 m/sV_{\text{CM}} = 2 \text{ m/s}, m=2 kgm = 2 \text{ kg}, R=4 mR = 4 \text{ m}.

(A ring is shown in the x-y plane, rolling on the positive x-axis. Its center of mass is moving with velocity vv in the positive x-direction.)

Find angular momentum of ring about origin if it is in pure rolling.

Question diagram
A

32 kgm2/s32 \text{ kgm}^2/\text{s}

B

24 kgm2/s24 \text{ kgm}^2/\text{s}

C

16 kgm2/s16 \text{ kgm}^2/\text{s}

D

8 kgm2/s8 \text{ kgm}^2/\text{s}

Correct Answer

Option A

Detailed Explanation

Option A is correct because the calculation of angular momentum L\mathbf{L} about the origin incorporates both the moment of inertia ICMI_{CM} and the translational motion of the center of mass, leading to the expression L=ICMω+mVCMR\mathbf{L} = I_{CM}\omega + mV_{CM}R. In this case, substituting the given values yields L=2×16×0.5+2×2×4=32kg\cdotpm2/s\mathbf{L} = 2 \times 16 \times 0.5 + 2 \times 2 \times 4 = 32 \, \text{kg·m}^2/\text{s}. Other options are not applicable as they do not provide relevant calculations or results related to the problem, focusing instead on alternative interpretations or incorrect values. Understanding the components of angular momentum, including both rotational and translational aspects, is crucial in solving problems involving rigid body dynamics.

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