Askiitians Expert Soumyajit IIT-Kharagpur
Last Activity: 14 Years ago
Dear Bharath Devasani,
Ans:- Now draw the FBD of two bodies.
The cylinder in the bottom will have the following set of forces
f1=force of friction between the two bodies and is towards right for the cylinder and towards left for the plank.
f2=force of friction between the ground and the cylinder and is towards left.
F=applied horizontal force on the plank
a1= Acceleration of the plank towards the right
a= Acc of the cylinder towards right
ß= Angular Acc of the cylinder
Then the eqns of motion are
F-f1=M2 a1.........................(1)
f1 - f2=M1 a........................(2)
(f1+f2)r=Iβ...........................(3) (where I= momont of inertia= 1/2 M2 r²)
And the conditions of non slleeping are
a1=a+ βr..................(4)(At the top cotact of two bodies)
a2=βr..............(5)(At the contact of the cylinder and the ground)
Solving these Eqsns we get
a=4F/(8M2+3M1)
a1=8F/(8M2+3M1)
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Regards,
Askiitians Experts
Soumyajit Das IIT Kharagpur