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Question:

A straight rod of length L extends from x=a to x=L+a. The gravitational force is exerted on a point mass 'm' at x=0, if the mass per unit length of the rod is A+Bx² is given by.

Gm[A(1/a+L-1/a)-BL]

Gm[A(1/a+L-1/a)+BL]

Gm[A(1/a-1/(a+L))+BL]

Gm[A(1/a-1/(a+L))-BL]

Solution:

dm=(A+Bx²)dx
df=GMdm/x²=F=∫a+La GM/x²(A+Bx²)dx=GM[−A/x+Bx]a+L
Gm[A(1/a-1/(a+L))+BL]