IIT JEE , AIEEE Forum - Physics , Mechanics

anandghegde

http://mypage.skrbl.com/rotation.html

hsbhatt

Let the tension in the rope be T and the friction be f.

Since the rope is inextensible, the block and the cylinder are displaced by the same amounts in opp directions. Hence the modulus of acceleration is the same for both, call it a

For the block, the force eqn is F-T-f = Ma

For the cylinder it is T-f = ma

Also f = I_cyl

Since the cylinder doesn't slip, we have a = r

Hence f = 1/2 mr²

= ma/2

Hence T = 3ma/2

Hence F = 3ma/2+ma/2+Ma = a(M+3m)

Thus, a = F/(3m+M)

anchitsaini

F-T-f=Ma ----1
T-f=ma ----2
fR=I?
?=2f/mR [I=mRR/2)
also for rolling R?-a=a
therefore ?=2a/R=2f/mR
therefore f=ma

1+2----
F-2f=(M+m)a
F-2ma=Ma+ma
F=(M+3m)a
therefore a=F/M+3m

Cheers!!

anchitsaini

in the above soln
?=alpha

anchitsaini

hsbhatt's soln is incorrect

anandghegde

whats wrong with HSBhat sir's solution?

anchitsaini

first of all look at the last line
F = 3ma/2+ma/2+Ma = a(M+3m) ----this is incorrect itself
as it =2ma +Ma=(M+2m)a
the soln came wrong because the concept used is wrong-
here a is not equal to r*alpha
but r*alpha-a=a ie a=r*alpha/2
hope you understand!!

i don't know why you gave hsbhatt 5 pts!!!

hsbhatt

anchit is right. The condition for not slipping is that the point of contact with the surface is instantaneously at rest relative to the surface.

Hence, if

is the angular velocity of the cylinder and v is the velocity of the centre of mass of the cylinder, then the velocity of point at the bottom of the cylinder is v-r

.

Since the surface itself is moving with velocity v, we must have v-r

= -v or w = 2v/r and hence

= 2a/r. Hence f = ma and T = 2ma and hence a = F/(3m+M).

Thanx anchit.

anandghegde

thanx anchit

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