COMEDK · Physics · 8. Rotational Motion
An object of mass \(1 \mathrm{~kg}\) is allowed to hang tangentially from the rim of the wheel of radius R. When released from the rest, the block falls vertically through \(4 \mathrm{~m}\) height in 2 seconds. The moment of inertia is \(1 \mathrm{~kg} \mathrm{~m}^2\). The radius of the wheel \(\mathrm{R}\) is

- A 0.025 m
- B 0.5 m
- C 1 m
- D 0.25 m
Answer & Solution
Correct Answer
(B) 0.5 m
Step-by-step Solution
Detailed explanation
Let \(m = 1 \text{ kg}\) be the mass of the block, \(I = 1 \text{ kg m}^2\) be the moment of inertia of the wheel, \(R\) be the radius of the wheel, and \(h = 4 \text{ m}\) be the distance fallen in time \(t = 2 \text{ s}\). The block starts from rest, so the distance fallen is…
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