IIT JEE , AIEEE Forum - Physics , Mechanics

atavistic

There is a system with two rollers each of mass M and radius r,linked with a chain of mass M and length l,crawls with a constant speed V as shown.The KE of the system is equal to (r<<<l)

a)3MV^2
b)2MV^2
c)MV^2
d)1.5MV^2

BALGANESH

the total ke of the system is given by
1/2 Mv^2 +1/2 Mv^2 + 1/2Mv^2 = 1.5Mv^2
the first two 1/2 Mv^2 is for the two spheres and the third one for the chain

bhartibhanushali

use
ke =1/2mv^2 +1/2IW^2 for rollers
for chain just take 1/2mv^2

atavistic

WRONG. Answer is 3MV^2

karthik2007

For one of the rollers:

E_total = 1/2mv².

Hence, KE for both the rollers = mv²

Now, considering the chain, it will move with a velocity of 2v and not v, by constraints.

Hence, we get the total KE of the system as mv² + 1/2m(2v)² = 3mv²

atavistic

But if it moves with 2v,then wont the lower part slip?

karthik2007

No buddy. That's because the rim of the roller also moves with a speed of 2v. In case you want the derivation:

see, for the roller, w = v/r - true for all particles on the roller, as all rotate with the same angular speed, but not with the same linear velocity. Each particle has its own tangential velocity.

Now, For the particle at the topmost point, ie, at the rim, we have

v = rw or v = r x v/r = v. This is wrt the center of mass of the roller, as it is the tangential velocity. As the COM itself moves with a speed of v, the net speed = v+v = 2v. Since the rim is attached to the chain, even the chain moves with the same speed

atavistic

True.But thats for the topmost point.IF u see the bottom-most point,the NET velocity of the particle is just V,but if the chain moves with 2V,relative velocity is not zero.SO slipping will occur na?

karthik2007

No dude, the speed at the bottommost point is NOT v. Just extend the same concept here.

anchitsaini

linear density of belt=p
m of belt=(2l+2pier)p
velocity of part DC=0 as belt is in contact with bottom ofroller which is contact with ground having v=0 for no slip
KE of BC=0.5lpvsquare=k1
where v=v+rw=2v
therefore k1=2lpvsquare
KE of APD + BQC=2*0.5Iw square
=2pieRpv square
therefor total ke of belt=(2l+2pieR)pvsquare=mvsquare

now ke of rollers=2*(0.5mvsquare +0.5Iwsquare)
=mvsquare+Iwsqaure
taking roller as hollow cylinder
ke=mvsquare +mvsquare=2mvsquare

therefore total ke=3mvsquare

anchitsaini

here is the picture to the previous answer

atavistic

Amazing,way to go man...thats the correct way to do this question. :)

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