Question 211: Gravitation

Question

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

Accepted Answer

dm = (A + Bx
2
)dx
dF = $${{GMdm} \over {{x^2}}}$$
F = $$\int_a^{a + L} {{{GM} \over {{x^2}}}} $$ (A + Bx
2
)dx
= GM$$\left[ { - {A \over x} + Bx} \right]_a^{a + L}$$
= GM$$\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$

Answer

$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) - BL} \right]$$

Answer

$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$

Answer

$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) + BL} \right]$$

Answer

$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]$$

Gravitation

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

Choose the correct answer: