A particle of mass m moves under the central repulsive force ((mb)/r^3 ) and is initially moving at a distance ′a′ from the origin of a force with velocity ′v′ at right angle to ′a′. show that rcos pθ=a where p =(b/(a^2 v^2 ))+1.


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 191868 by Spillover last updated on 02/May/23

$A p a r t i c l e o f m a s s m m o v e s u n d e r t h e c e n t r a l$ $r e p u l s i v e f o r c e \frac{m b}{r^{3}} a n d i s i n i t i a l l y m o v i n g$ $a t a d i s t a n c e^{'} a^{'} f r o m t h e o r i g i n o f a f o r c e$ $w i t h v e l o c i t y^{'} v^{'} a t r i g h t a n g l e t o^{'} a^{'} .$ $s h o w t h a t$ $r \cos p θ = a w h e r e p = \frac{b}{a^{2} v^{2}} + 1 .$
	Answered by mr W last updated on 03/May/23

	Commented by Spillover last updated on 29/Jun/23

	$t h a n k s f o r t h e s k e t c h$
	Answered by Spillover last updated on 15/Jul/24

	$T h e p r e s e n c e o f t h e c e n t r a l f o r c e i m p l i e s t h a t$ $t h e a n g u l a r m o m e n t u m L i s c o n s e r v e d$ $F (r) = \frac{m b}{r^{3}} L = {m r}^{2} θ$ $G i v e n i n i t i a l c o n d i t i o n$ $◼ i n i t i a l d i s t a n c e f r o m t h e o r i g i n r = a$ $◼ i n i t i a l v e l o c i t y v p e r p e n d i c u l a r t o a i s g i v e n$ $L = m a v$ $T o t a l e n e r g y o f t h e p a r t i c l e i s c o n s e r v e d c o n s i s t$ $K . E a n d e f f e c t i v e P . E [V_{e f f} (r)]$ $V_{e f f} (r) = \frac{L^{2}}{2 {m r}^{2}} + \frac{m b}{2 r^{2}}$ $L = m a v$ $V_{e f f} (r) = \frac{L^{2}}{2 {m r}^{2}} + \frac{m b}{2 r^{2}}$ $V_{e f f} (r) = \frac{{(m a v)}^{2}}{2 {m r}^{2}} + \frac{m b}{2 r^{2}} = \frac{m (a^{2} v^{2} + b)}{2 r^{2}}$ $f r o m R a d i a l e q u a t i o n o f t h e m o t i o n$ $\frac{d θ}{d t} = \frac{L^{2}}{{m r}^{2}} = \frac{a v}{r^{2}}$ $T h e r a d i a l e q u a t i o n o f t h e m o t i o n f r o m$ $e f f e c t i v e p o t e n t i a l$ $m \frac{d^{2} r}{{d t}^{2}} = m \frac{d^{2} θ}{{d t}^{2}} = \frac{L^{2}}{{m r}^{3}} + \frac{m b}{r^{3}}$ $\frac{d^{2} r}{{d t}^{2}} = \frac{a^{2} v^{2} + b}{r^{3}}$ $f r o m \frac{d θ}{d t} = \frac{L^{2}}{{m r}^{2}} = \frac{a v}{r^{2}}$ $u = \frac{1}{r} \frac{d u}{d θ} = - \frac{1}{r} \frac{d r}{d θ}$ $\frac{d^{2} r}{{d t}^{2}} = \frac{d}{d θ} (\frac{d r}{d θ} . \frac{d θ}{d t})$ $\frac{d^{2} r}{{d t}^{2}} = \frac{d}{d θ} (\frac{d r}{d θ}) . {(\frac{d θ}{d t})}^{2} + \frac{d r}{d θ} . \frac{d^{2} θ}{{d t}^{2}}$ $b u t \frac{d^{2} θ}{{d t}^{2}} = 0$ $(\frac{d^{2} r}{d θ^{2}}) . {(\frac{a v}{r^{2}})}^{2}$ $s o l v e d . e f o r u$ $\frac{d^{2} θ}{{d t}^{2}} + u = \frac{a^{2} v^{2} + b}{a^{2} v^{2}} u^{3}$ $\frac{d^{2} θ}{{d t}^{2}} + u = (1 + \frac{b}{a^{2} v^{2}}) u^{3}$ $g i v e n t h e b o u n d a r y c o n d i t i o n$ $u = \frac{1}{r} a t θ = 0$ $u (θ) = \frac{1}{a} \cos (p θ) w h e r e p = 1 + \frac{b}{a^{2} v^{2}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum