Question

20. A thin rod of mass M and length L gravitationally interacts with a point mass m that is a perpendicular distance a away from its left end (see the figure). The rod is non-uniform, and its linear density (mass per unit length) increases with the distance from its left end according to 2(x) = 2Mx/L?, where x is the horizontal coordinate along the rod (so that x = 0 is at its left end and x=L is at its right end). What is the component of the gravitational force the rod exerts on the point mass perpendicular to the rod? m 2GmM L x2dx L2 Jo (x2+a2)3/2 2GmM L xdx L2 Jo x2+a2 2GmM L xdx L JO (x2+a2)3/2 2GmM L xdx al2 Jo (x2+a2)1/2 2GmMa L xdx L2 Jo (x2+a2)3/2 L- >

Answer 1

soln: dx • X - Let us consider an elementh of length dx x x L at a distance x from K L its left end. . So the gavitational force exerted by this small element of length dx of mass dM on the particle of mass n is given by - df = Gm. dm 1. R2 Now, R²=x²+a2 a dM = 14) dx ĐdM= 2M Xdx Ľ dF = GM 2M (xdx) = 29 mm (xdx ) (x² +0² The gavitational force exerted by the whole Hod is given by 12 - C F = 2GMm ( xax - ² x²fa2 Scanned with CamScanner