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  • Solve this problem Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration ‘a’ perpendicular to the axis of the cylinder. What is Ffriction at point P ? It is assumed that the cyl

 Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out  from  under  it  with  acceleration  ‘a’ perpendicular to the axis of the cylinder.  What is Ffriction at point P ? It is assumed that the cylinder does not slip.         

                      

 

  • Option 1) Mg

     

  • Option 2) Ma

     

  • Option 3) Ma/2

     

  • Option 4) Ma/3

     

 

Answers (1)

best_answer

As we have learned

Rolling motion -

V_{cm}= Rw\rightarrow pure \: rolling

V_{cm}> Rw\rightarrow slipping \: motion

V_{cm}< Rw\rightarrow skidding \: motion

-

 

 

a_p  due to translation = F/m 

Angular acceleration of point p 

     \alpha = \frac{T}{I}= \frac{F \cdot r}{I} = \frac{F r }{\frac{mr^2}{2}}= \frac{2F}{mr}

Acceleration of P due to both traslational and rotation = \frac{F}{m} + \frac{2F}{mr}\cdot r

a = \frac{3F }{m }

F = \frac{ma }{3}

 

 

 

 


Option 1)

Mg

Option 2)

Ma

Option 3)

\frac{Ma}{2}

Option 4)

\frac{Ma}{3}

Posted by

SudhirSol

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