Q: Mass m is distributed uniformly along a wire of length 2L . A particle of mass m0 is at a point that is at a distance r above the centre on its perpendicular bisector (point P in the figure). The gravitational force on the particle is
(a) $\displaystyle -\frac{G m m_0}{2r \sqrt{r^2 + L^2}} \hat{j} $
(b) $\displaystyle -\frac{G m m_0}{r \sqrt{r^2 + L^2}} \hat{j} $
(c) $\displaystyle \frac{G m m_0}{\sqrt{r^2 + L^2}} (-r\hat{j} + L \hat{i}) $
(d) Zero
Click to See Answer :
Sol: Consider a element of length dx , mass of this element is $\displaystyle dm = \frac{m}{2L}dx $
Force on m0 due to this element is , $\displaystyle dF = \frac{G m_0 (dm)}{r^2 + x^2}$
$\displaystyle dF = \frac{G m_0 (\frac{m}{2L}dx)}{r^2 + x^2}$
The component dFcosθ will be cancel out
Net Force acting on m0 will be ,
$\displaystyle F = \int_{-L}^{L} dF sin\theta $
$\displaystyle F = \int_{-L}^{L} \frac{G m_0 (dm)}{r^2 + x^2} . \frac{r}{\sqrt{r^2 + x^2}} $
$\displaystyle F = \int_{-L}^{L} \frac{G m_0 (\frac{m}{2L}dx )}{r^2 + x^2} . \frac{r}{\sqrt{r^2 + x^2}} $
$\displaystyle F = \frac{G m m_0 r }{2L} \int_{-L}^{L} \frac{dx}{(r^2 + x^2)^{3/2}} $
Now Integrating this on substituting , r = x tanα we get the result .