InterviewSolution
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InterviewSolution
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A disc of circumference `s` is at rest at a point `A` on a horizontal surface when a constant horizontal force begins to act on its centre. Between `A` and `B` there is sufficient friction toprevent slipping, and the surface is smooth to the right of `B.AB = s`. The disc moves from `A` to `B` in time `T`. To the right of `B`, .A. the angular acceleration of the disc will disappear, linear acceleration will remain unchanged.B. linear acceleration of the disc will increaseC. the disc will make one rotation in time `T//2`D. the disc will cover a distance greater than `s` in further time `T`. |
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Answer» Correct Answer - B::C::D (B) To the right of `B`, friction vanishes, so linear acceleration of disc will increase. ( c) From `(A)` to `B` it complete one rotation in time `T` `theta = (1)/(2) alpha t^(2), 2 pi = (1)/(2) alpha T^(2), alpha = (4 pi)/(T^(2))` ....(1) & `omega = alpha T`...(2) from (1) & (2) `omega = (4 pi)/(T)` after `B, omega` becomes constant there fore time taken to complete one rotation after `B` `theta = omega t, 2 pi = (4 pi)/(T) t, t = (T)/(2)` (D) After `B` disc is linearly accelerated so it will cover greater distance than `s` in further time `T`. |
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