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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,

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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 T

Between 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

Between 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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