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A light, inextensible string is wound around the rim of a uniform…
Question
A light, inextensible string is wound around the rim of a uniform disc of mass 4 kg that is free to rotate about a fixed horizontal axis through its centre. A block of mass 2 kg hangs from the free end of the string and is released from rest, so the string unwinds without slipping and the disc rotates. Taking g = 10 m/s^2 and moment of inertia of the disc as (1/2)MR^2, find the downward acceleration (in m/s^2) of the hanging block.
✓ Verified answer: 5checked by our engine — not a guess
Step-by-step solution
For the block: mg - T = m a. For the disc: T R = I alpha = (1/2)M R^2 (a/R), so T = (1/2)M a. Substituting: m g - (1/2)M a = m a, giving a = m g/(m + M/2) = 2*10/(2 + 2) = 20/4 = 5 m/s^2.
Final answer5
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