Question

Q. What is the potential energy of interaction between the arc of ring of radius R and mass m and the particle of mass [tex] m_{0} [\tex] placed at centre of curvature?

#ANSWER THE QUESTION WITH PROPER EXPLANATION

#No Spamming Plz​

Answer 1

Solution :

To Find :-

Potential energy of interaction b/w the arc of ring of radius R (mass M) and the particle of mass m$_o$ placed at the centre of curvature.

Angle with arc = 60° = π/3 rad

Concept :-

Formula of potential energy of interaction between two point masses is given by

$\underline{\boxed{\bf{\pink{U=-\dfrac{GM_1M_2}{r}}}}}$

; r = distance b/w point masses

Calculation :-

If we consider the arc of ring to have continuous distribution of matter, the summation in the formula of potential energy should be replaced by integration.

First we have to find out mass of each small continent in terms of linear mass density($\lambda$).

Let mass of each component be dM and length of each component be dx.

$\displaystyle\implies\bf\:\lambda=\dfrac{Mass\:of\:arc}{Length\:of\:arc}\\ \\ \implies\sf\:\lambda=\dfrac{M}{angle\times radius}\\ \\ \implies\sf\:\lambda=\dfrac{M}{\frac{\pi}{3}\times R}\\ \\ \implies\sf\:\red{\lambda=\dfrac{3M}{\pi R}}\\ \\ \mapsto\bf\:dU=-\dfrac{G(dM)m_o}{R}\\ \\ \mapsto\sf\:dU=-\dfrac{G(\lambda\times dx)m_o}{R}\\ \\ \mapsto\sf\:dU=-\dfrac{G(\frac{3M}{\pi R})m_o}{R}\times (dx)\\ \\ \mapsto\sf\:dU=-\dfrac{3GMm_o}{\pi R^2}\times (dx)\\ \\ \mapsto\sf\:U=\int{dU}={\tiny\lim(0\to\frac{\pi R}{3})}\int{-\dfrac{3GMm_o}{\pi R^2}\times (dx)}\\ \\ \mapsto\sf\:U=-\dfrac{3GMm_o}{\pi R^2}\int{(dx)}\\ \\ \mapsto\sf\:U=-\dfrac{3GMm_o}{\pi R^2}[x]{\tiny\lim({0\to\frac{\pi R}{3}})}\\ \\ \mapsto\sf\:U=-\dfrac{3GMm_o}{\pi R^2}\times\dfrac{\pi R}{3}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{U=-\dfrac{GMm_o}{R}}}}}$

Answer 2

${\huge{\boxed{\overline{\mid{Answer:-}}}}}}$

$\implies$$\frac{ - {2gm}^{2} }{r}$

$\implies$$\frac{ {gm}^{2} }{\pi r}$

$\implies$$v = - \frac{gmm}{r} = - \frac{ {gm}^{2} }{r}$