Mass m is distributed uniformly along a wire of length 2L . A particle of mass m0 is at a point that is ….

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

Numerical

(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

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Ans: (b)

Sol: Consider a element of length dx , mass of this element is $\displaystyle dm = \frac{m}{2L}dx $

Numerical

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 .