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Find the force on a particle of mass m placed at the centre of a semicircular wire of length L and mass M.
Hi Sanchit,
The answer is :
let R be raidus of wire , then , ∏R = L --- (1)
element of length RdΦ at angle Φ from centre will exert a force dF = Gm(M/L)(RdΦ)/R2
Similar force will be exerted by a element at angle -Φ from the centre.
Thus net force on m will be : 2dFcosΦ
Thus Force F = ∫2dFcosΦ where Φ varies from 0 to 90
Thus F = 2∫(Gm(M/L)dΦ/R)cosΦ
F = 2GmM/LR ∫cosΦ dΦ where Φ varies from 0 to 90
F = 2GmM/LR ( 1 - 0 )
Thus force on particle due to wire will be F = 2GMm/LR = 2GMm∏/L2 ( from (1) )
Please feel free to ask as many questions you have.
Puneet
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