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A solid cylinder of mass m and radius R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

1. \( \sqrt{2gh} \)
2. \( \sqrt{4gh/3} \)
3. \( \sqrt{3gh/4} \)
4. \( \sqrt{4g/h} \)

Correct answer: \( \sqrt{4gh/3} \)

Solution

For a rolling object, the total energy is conserved. The potential energy at the top, mgh, is converted into translational kinetic energy (1/2 mv²) and rotational kinetic energy (1/2 Iω²) at the bottom. For a solid cylinder, I = (1/2)mR² and ω = v/R. Solving for v gives v = √(4gh/3).

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