Question

A force f acts tangentially at the highest point of a disc of mass m kept on a rough horizontal plane. If the disc rolls without slipping, the acceleration of centre of the disc is

Answer 1

The acceleration of centre of the disc is a =    4F / 3m

Explanation:

• The force F acting on the sphere moves it towards right.
• The contact point between force and sphere will slip towards right
• Thus the static friction( force f shown ) on the sphere will act towards left.

Let

Radius of the sphere  = r

Linear acceleration  = a

Since there is slipping alpha "α"=  r a

​For linear motion F - f = ma   ----- (1)

Where "f" is the frictional force

For rotational motion, torque about center "τ" = Fr+fr = Iα=(  1/2  mr^2  )   a/r

F+f=   1/2 ma ---- (2)

From (1) and (2) adding, we get 2 F = 3 / 2 ma  

a=    4F / 3m

Thus the acceleration of centre of the disc is a =  4F / 3m

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