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Question

# A disc of circumference 's' is 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 to prevent slipping and the surface is smooth to the right of B. Given that 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 aceleration will remain unchanged
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B
linear acceleration of the disc will increase
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C
the disc will make one rotation in time T2
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D
the disc will cover a distance greater than s in a further time T
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Solution

## The correct options are B linear acceleration of the disc will increase C the disc will make one rotation in time T2 D the disc will cover a distance greater than s in a further time TBetween A and B, friction force will act backwards because the contact point tends to slide in forward direction due to force F and also after B there is no friction and disc will do only translational motion. ma1=Fâˆ’f and ma2=F â‡’a1=Fâˆ’fm and a2=Fm Thus a2>a1 i.e linear acceleration will increase. Now for one rotation between A and B, 2Ï€=0+12Î±T2 â‡’2Ï€=12Î±T2 ........(1) Also w=0+Î±Tâ‡’w=Î±T For one rotation after B 2Ï€=wt 2Ï€=(Î±T)tâ‡’t=2Ï€Î±T For one rotation after B 2Ï€=wt 2Ï€=(Î±T)tâ‡’t=2Ï€Î±T t=2Ï€4Ï€T=T2 (from 1) Now, s=0+12Î±1T2â‡’s=12Î±1T2 and v=a1T Also, where sâ€² is the distance covered after the point B sâ€²=vT+12Î±2T2=Î±1T2+12Î±2T2 sâ€²=2s+12Î±2T2â‡’sâ€²>s

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