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Blazing goIITian

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13 Apr 2009 16:49:27 IST

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393

integration and gravitation

Kinematics , NewtonsLawsCirMotion , Work Power Energy collisions , Rotation , Gravitation , Simple Harmonic Motion , Fluid Mechanics , Materials Wave Sound , Super Position , units dimensions , com momentum , shm , wave sound

find th gravitational force acting on an obkect of mass m placed in the centre of a thin wire of mass M and length L

pls help

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Comments (9)

VARUN RAJ

Blazing goIITian

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13 Apr 2009 22:28:03 IST

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could anyone pls solve it

VARUN RAJ

Blazing goIITian

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14 Apr 2009 17:47:41 IST

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the object is placed at the centre of the wire

VARUN RAJ

Blazing goIITian

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15 Apr 2009 18:24:28 IST

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could anyne pls solve

NugoRama

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15 Apr 2009 18:28:39 IST

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centre of wire ???makes no sence unless the wire is in form of a circular loop...is it so?

VARUN RAJ

Blazing goIITian

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16 Apr 2009 22:17:26 IST

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yes

Prakhar Banga

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16 Apr 2009 22:18:58 IST

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If the wire is a circular loop then there will be no force acting on the mass because the force due to any element will exactly cancel that due to an element of the same mass at the diametrically opposite end of it.

NugoRama

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16 Apr 2009 22:20:57 IST

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unless there is no current in the wire .....answer is ZERO..cos we have to consider gravitational force only..if nothing is specified ..

VARUN RAJ

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16 Apr 2009 22:23:36 IST

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see this is the orignal quesfind the force of attraction on a particle of mass m placed at the centre of the semicucular wire of masss M and Lthe answer is2pieGMm?L^2

BALGANESH

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16 Apr 2009 22:39:53 IST

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consider en element dl from the semicircular wire.....

mass per unti length =M/L...

and the element is at an angle @ from the horizontal...

let R be the radius....

force due to this element dF=Gm M/L dl / r^2 sin@....

but l=rd@

So df = GmMrd@ sin@ / L R^2 [ bcos all the cos@ forces cancle out each other]

...integrating with limits as 0 to pie..

.we get 2GMm/LR...

now pie R = L.....

therefore F= 2pieGMm/L^2

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