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please answer this

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Asked by Prashant DIGHE 18th April 2020, 9:37 PM
Answered by Expert
Answer:
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
 
begin mathsize 14px style a subscript C space equals space F over M space open curly brackets fraction numerator 1 space plus space begin display style r over R end style over denominator begin display style fraction numerator I subscript C over denominator M R squared end fraction end style space plus space 1 end fraction close curly brackets end style............................(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
 
begin mathsize 14px style f space equals F space open curly brackets fraction numerator I subscript C minus space M space R space r over denominator I subscript C plus space M space R squared end fraction close curly brackets end style .................... (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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