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Asked by Prashant DIGHE 18th April 2020, 9:37 PM
Let r be the distance from centre of mass to point A , which is point of application of force F .

Friction force f is acting at the point of contact.

Acceleration of centre of mass , aC = ( F - f ) / M .....................(1)

where M is mass of rolling object

Angular acceleration is due to torque of applied force F and friction force f.

By taking point of contact as reference point,  F(r+R)  = I α = ( IC + MR2 ) α .........................(2)

where I is moment of inertia about point of contact O and it is given by,  I = IC + MR2
where IC is moment of inertia about centre of mass.

Angular acceleration α = ( aC / R )  ..............................(3)

By substituting angular acceleration using eqn.(3) in eqn.(2) and after simplification of eqn.(2),
we get the following equation for aC

............................(4)

From eqn.(1), we write friction force as , f  = F - ( M aC )  ....................(5)

Substituting for aC using eqn.(4), we get friction force f from eqn.(5) as

.................... (6)
Equation (4) and (6) helps us to know about movement of centre of mass and direction of friction force

(1) when force F is applied above centre, r is +ve

From eqn.(4), we know that the Rolling object moves forward

If the distance r satisfies the condition r < [ IC / ( MR ) ], then  from eqn.(6) we know that friction is backward

If the distance r is such that, r > [ IC / ( MR ) ], then  friction is forward

(2) when force F is applied at centre of mass, r is zero .

From eqn.(4), we know that the Rolling object moves forward

From eqn.(6) we know that friction is backward

(3) When force F is applied below centre of mass

From eqn.(4), we know that the rolling object moves forward

From eqn.(6) we know that friction is always backward
Answered by Expert 20th April 2020, 8:09 AM
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