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Question

# A constant force F acts tangentially at the highest point of a uniform disc of mass m kept on a rough horizontal surface as shown in figure. If the disc rolls without slipping, calculate the acceleration of the centre C and points A and B on the disc.

A
aC=4F3m,aA=0,aB=8F3m
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B
aC=2F3m,aA=0,aB=4F3m
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C
aC=Fm,aA=0,aB=2F3m
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D
aC=F2m,aA=0,aB=F2m
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Solution

## The correct option is A aC=4F3m,aA=0,aB=8F3m As shown in figure, the force F rotates the disc, so the point of contact has a tendency to slip towards left and the static friction on the disc will act towards right. As there is no slipping, aceeleration of point A,aA=0, and angular acceleration of disc α=ar (Here, a=Linear acceleration, α=angular acceleration) For the linear motion of the centre,∑Fx=ma F+f=ma ... (i) For the rotational motion about the centre, τcom=Iα −Fr+fr=Iα Fr−fr=(12mr2)(ar) F−f=12ma ... (ii) From (i) and (ii), 2F=32ma or a=4F3m ⇒aC=a=4F3m Acceleration of point B, aB=aC+α×r ⇒aB=a+ar×r ⇒aB=2a=2(4F3m)=(8F3m)

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