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