Rotation Ques 7
7 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 $A+B x^2$, is given by
(2019 Main, 12 Jan I)
(a) $G m\left[A\left(\frac{1}{a+L}-\frac{1}{a}\right)-B L\right]$
(b) $G m\left[A\left(\frac{1}{a+L}-\frac{1}{a}\right)+B L\right]$
(c) $G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]$
(d) $G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)-B L\right]$
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Answer:
Correct Answer: 7.( c )
Solution:
Force of attraction between mass ’ $m$ ’ and an elemental mass ’ $d m$ ’ of rod is
$ d F=\frac{G m d m}{x^2}=\frac{G m\left(A+B x^2\right) d x}{x^2} $
Total attraction force is sum of all such differential forces produced by elemental parts of rod from $x=a$ to $x=a+L$.
$\therefore F =\int d F=\int_{x=a}^{x=a+L} \frac{G m\left(A+B x^2\right)}{x^2} d x $
$ =G m \int_{x=a}^{x=a+L}\left(\frac{A}{x^2}+B\right) d x $
$ =G m\left[-\frac{A}{x}+B x\right]_{x=a}^{x=a+L} $
$ =G m\left(\frac{-A}{a+L}+B(a+L)+\frac{A}{a}-B a\right) $
$ =G m (\frac{A}{a}-\frac{A}{a+L}+B L )=G m {A (\frac{1}{a}-\frac{1}{a+L} )+B L }$
BETA
