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Blazing goIITian

Joined:	12 Jun 2007
Post:	1078

12 Dec 2007 23:36:32 IST

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verma-centre of mass

None

A uniform chain of mass M and length L is held vertically in such a way that its lower end just touches the floor. the chain is released from rest in this position. Any portion that strikes the ground comes to rest. Assuming that the chain does not form a heap, calculate the force exerted by it on the floor when a length x has reached the floor.

Ans- 3Mgx/l

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Comments (5)

Karthik M

Blazing goIITian

Joined: 1 May 2007

Posts: 2830

12 Dec 2007 23:52:30 IST

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See, here we will have two forces, one due to the own weight of the chain, and the other due to momentum transfer,

Now, consider a length x of the chain. The topmost part of this will acquire a velocity of v =

2gx (As u = 0)
So, momentum acquired when it reaches the bottom = mxv/L.
Now, weight of this part = mx/L.

So force due to its own weight = mgx/L.

Now, force = dP/dt where P is the momentum.

So, Force = d/dt (mxv/L) = mvLdx/dt (dont differentiate v, it is constant for a particular length of the chain).

But we have dx/dt = v.

So force due to momentum transfer = mv²/L = 2mgx/L

So, net force = force due to own weight + force due to momentum transfer =

2mgx/L + mgx/L = 3mgx/L

Karthik M

Blazing goIITian

Joined: 1 May 2007

Posts: 2830

12 Dec 2007 23:57:48 IST

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feel free to ask if you didn't follow my solution anywhere.

Anand Hegde

Blazing goIITian

Joined: 12 Jun 2007

Posts: 1078

13 Dec 2007 00:12:57 IST

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wht is ' force due to momentum transfer??
Is it the thrust force??????

abc xyz

Blazing goIITian

Joined: 3 Aug 2007

Posts: 603

13 Dec 2007 00:21:39 IST

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momentum transfer means that since chains comes to rest, it has transferred is MV to the floor.

sowjanya gudipati

Cool goIITian

Joined: 13 Oct 2007

Posts: 75

13 Dec 2007 00:23:13 IST

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hey i am using the normal reaction offered by the floor on the chain which is equal to the force exerted by the chain on the floor

N=N_wt+N_thrust

N_wt=(m/l)xg

N_thrust=V_relative(dm/dt)

V_relative=u-v = v

dm/dt=(dm/dy)(dy/dt)=m/l(v)

N_thrust=v²(m/l)

v=(2gx)^1/2

N_thrust=2gx(m/l)

N=(m/l)gx+2gx(m/l)=3(m/l)gx

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