Question

$\sf\huge\bold{Question ?}$




A thin rod of mass M and length L is bent in a semicircle as shown in figure
(a). What is its gravitational force (both magnitude and direction) on a particle with mass m at O the centre of curvature ?
(b) what would be the force on m if the rod is in the form of complete circle?​

Answer 1

Gravitational field at point O due to small element dm is :

$dE = \dfrac{G(dM)}{ {r}^{2} }$

• πr = L , so r = L/π

$\implies dE = \dfrac{G( \dfrac{M}{L} \times dl )}{ {r}^{2} }$

$\implies dE = \dfrac{G( \dfrac{M}{L} \times r d\theta )}{ {r}^{2} }$

$\implies dE = \dfrac{GM \times d\theta }{ r L}$

$\implies dE = \dfrac{GM\pi \times d\theta }{ {L}^{2} }$

Now, consider 2 opposite segments as shown and only $\sin(\theta)$ components will be added up.

$\implies dE_{net} = \dfrac{2GM\pi \times d\theta }{ {L}^{2} } \sin( \theta)$

$\implies dE_{net} = \dfrac{2GM\pi }{ {L}^{2} } \sin( \theta) d \theta$

Integration and putting limits :

$\rm \displaystyle \implies \int_{0}^{E}dE_{net} = \dfrac{2GM\pi }{ {L}^{2} } \int_{0}^{ \frac{\pi}{2} }\sin( \theta) d \theta$

$\rm \displaystyle \implies E_{net} = \dfrac{2GM\pi }{ {L}^{2} } \bigg\{ - \cos( \theta) \bigg\}_{0}^{ \frac{\pi}{2} }$

$\rm \displaystyle \implies E_{net} = \dfrac{2GM\pi }{ {L}^{2} } \bigg\{ \cos( \theta) \bigg\}_{ \frac{\pi}{2} }^{0}$

$\rm \displaystyle \implies E_{net} = \dfrac{2GM\pi }{ {L}^{2} } \bigg\{ 1 - 0 \bigg\}$

$\boxed{ \rm \displaystyle \implies E_{net} = \dfrac{2GM\pi }{ {L}^{2} } }$

Now, net force will be :

$\rm \displaystyle \implies F_{net} = \dfrac{2GM\pi }{ {L}^{2} } \times m$

$\boxed{\rm \displaystyle \implies F_{net} = \dfrac{2GMm\pi }{ {L}^{2} } }$

Answer 2

Answer:

here's your answer

Explanation:

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