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Sir

Pl solve question no-14

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Asked by shobhit 23rd June 2018, 12:02 AM
Answered by Expert
Answer:
Plank is pushed by the force F from left to right. This force makes the plank to move from left to right.
 
Hence at point of contact, friction force f will act towards left in the plank as shown in figure. 
 
At plank we have, ( F - f ) = m2×a ......................(1)
where a2 is acceleration of the plank.
 
At point of contact between plank and cylinder, as per newton's thrid law, there is a reaction force f acting on the cylinder.
This reaction force is towards right as shown in figure.  ( This reaction force is opposite to friction force acting on the plank. )

This reaction force f is trying to move the point of contact by an acceleration aCM = f / m1 towards right. 
 
Also this reaction force f is trying to rotate the cylinder of radius R about its centre of mass in clockwise direction. 
 
hence we have,  begin mathsize 12px style f cross times R space equals space I cross times alpha space equals space fraction numerator m subscript 1 R squared over denominator 2 end fraction cross times alpha space space............. left parenthesis 2 right parenthesis end style
from eqn.(2) we have,  begin mathsize 12px style alpha cross times R space equals space fraction numerator 2 space f over denominator m subscript 1 end fraction end style
acceleration at point of contact P, begin mathsize 12px style a subscript P space equals space a subscript C M space end subscript minus space alpha cross times R space equals space f over m subscript 1 space minus space fraction numerator 2 f over denominator m subscript 1 end fraction space equals space minus f over m subscript 1 space................ left parenthesis 3 right parenthesis end style
as per eqn.(3), point of contact P move towards left. Hence friction force f ' acts towards right as shown in figure.
 
Let a1 be the acceleration of Centre of mass of cylinder.
 
If there is no slipping between plank and cylinder, we have begin mathsize 12px style a subscript 2 space equals space 2 cross times a subscript 1 space................. left parenthesis 4 right parenthesis end style
 
By applying Newton's law to cylinder, we can write, begin mathsize 12px style m subscript 1 a subscript 1 space equals space f space plus space f space apostrophe end style ......................(5)
Also these forces trying to rotate the cylinder and due to rotational motion, we can write,
 
 begin mathsize 12px style T o r q u e space tau space space equals space f cross times R space minus space f space apostrophe cross times R space equals space I cross times alpha space equals space 1 half m subscript 1 R squared space cross times a subscript 1 over R
h e n c e space open parentheses f minus f space apostrophe close parentheses space equals space 1 half m subscript 1 a subscript 1 space..................... left parenthesis 6 right parenthesis end style

we can solve eqns.(5) and (6) to get f and f ' ;   begin mathsize 12px style f space equals space 3 over 4 m subscript 1 space a subscript 1 space end subscript space a n d space space f space apostrophe space equals space 1 fourth m subscript 1 a subscript 1 end style ....................(7)
using eqn.(1), (4) and (7), we write

 begin mathsize 12px style F minus space 3 over 4 space m subscript 1 a subscript 1 space equals space 2 space m subscript 2 a subscript 1 space space end subscript space o r space space a subscript 1 space equals space fraction numerator 4 F over denominator open parentheses 3 space m subscript 1 space plus space 8 space m subscript 2 close parentheses end fraction

h e n c e space a subscript 2 equals space fraction numerator 8 F over denominator open parentheses 3 space m subscript 1 space plus space 8 space m subscript 2 close parentheses end fraction end style

magnitudes of friction forces f and f ' can be calculated using eqn.(7) by substituting the value of a1 

Answered by Expert 26th June 2018, 4:29 PM
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