Question no 9 plz explain with explained integration
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Dear Student
Let
u = mass/length =M/L
dm = dx * u
dEg = Gravitational field due to mass dm at d distance on perpendicular bisector of rod.
From symmetry of the rod around its bisector you can easily see that horizontal components of Gravitational field will cancel out as there will be equal field due to mass dm at distance x from centre on both sides of centre, but in opposite directions.
However the vertical component due to mass dm on either side of centre will add up to give
dEgy =2 * dEg sinϴ = 2 * G(dm) sinϴ/(d2 + x2) = G(dm) d/(d2 + x2)3/2
Now to find field due to complete rod we must integrate dEgy over complete rod
to do that put limit of x from 0 to L/2 and integrate it
Egy = 2GM/d(L2 + 4d2)1/2
so this is the Expession for gravitational field at a poind d on perpendicular bisector.
Regards
Dear Student
Let
u = mass/length =M/L
dm = dx * u
dEg = Gravitational field due to mass dm at d distance on perpendicular bisector of rod.
From symmetry of the rod around its bisector you can easily see that horizontal components of Gravitational field will cancel out as there will be equal field due to mass dm at distance x from centre on both sides of centre, but in opposite directions.
However the vertical component due to mass dm on either side of centre will add up to give
dEgy =2 * dEg sinϴ = 2 * G(dm) sinϴ/(d2 + x2) = G(dm) d/(d2 + x2)3/2
Now to find field due to complete rod we must integrate dEgy over complete rod
to do that put limit of x from 0 to L/2 and integrate it
Egy = 2GM/d(L2 + 4d2)1/2
so this is the Expession for gravitational field at a poind d on perpendicular bisector.
Regards