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A thin uniform circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclination plane will be

1. g / 2
2. g / 3
3. g / 4
4. 2g / 3

Correct answer: g / 3

Solution

For a rolling object, the linear acceleration is determined by both translational and rotational motion. Using the equation for rolling motion and the moment of inertia of a ring, the linear acceleration is a = g sin(θ) / (1 + I/(mR²)). For a ring, I = mR², so a = g sin(30°) / 2 = g/3.

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