Question-215608

Question Number 215608 by ajfour last updated on 11/Jan/25

Commented by ajfour last updated on 11/Jan/25

The charged masses released as shown. Find their speeds when they get in line. I wonder if this gets periodic for certain appropriate ratios!

T h e c h a r g e d m a s s e s r e l e a s e d a s s h o w n .

F i n d t h e i r s p e e d s w h e n t h e y g e t i n l i n e .

I w o n d e r i f t h i s g e t s p e r i o d i c f o r c e r t a i n

a p p r o p r i a t e r a t i o s!

Commented by mahdipoor last updated on 11/Jan/25

w h y n e g a t i v e m a s s r i s i n g u p ?

Commented by ajfour last updated on 11/Jan/25

come?on + one pulls it upwards while the negative pair repel..

c o m e ? o n + o n e p u l l s i t u p w a r d s w h i l e

t h e n e g a t i v e p a i r r e p e l . .

Answered by mr W last updated on 12/Jan/25

Commented by mr W last updated on 12/Jan/25

say at time t the postion of M is (0, y_1 ) and the position of m is (±x_2 , y_2 ). d_1 ^2 =x_2 ^2 +(y_1 −y_2 )^2 d_2 ^2 =4x_2 ^2 F_1 =((k_e Qq)/d_1 ^2 )=((k_e Qq)/(x_2 ^2 +(y_1 −y_2 )^2 )) F_2 =((k_e q^2 )/d_2 ^2 )=((k_e q^2 )/(4x_2 ^2 )) V=−(dy_1 /dt) (↓) A=(dV/dt)=−(d^2 y_1 /dt^2 ) u=(dx_2 /dt) (→) v=(dy_2 /dt) (↿) a_x =(du/dt)=(d^2 x_2 /dt^2 ) a_y =(dv/dt)=(d^2 y_2 /dt^2 ) MA=Mg+2F_1 sin θ ⇒(d^2 y_1 /dt^2 )=−g−((2k_e Qq(y_1 −y_2 ))/(M[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) )) ma_y =F_1 sin θ−mg ⇒(d^2 y_2 /dt^2 )=((k_e Qq(y_1 −y_2 ))/(m[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) ))−g ma_x =F_2 −F_1 cos θ ⇒(d^2 x_2 /dt^2 )=((k_e q^2 )/(4mx_2 ^2 ))−((k_e Qqx_2 )/(m[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) )) at t=0: y_1 =b x_2 =a, y_2 =0 V=−(dy_1 /dt)=0 u=(dx_2 /dt)=0,v= (dy_2 /dt)=0 ..... ============= i don′t think the problem is solvable.

s a y a t t i m e t t h e p o s t i o n o f M i s

(0, y_{1}) a n d t h e p o s i t i o n o f m i s

(\pm x_{2}, y_{2}) .

d_{1}^{2} = x_{2}^{2} + {(y_{1} - y_{2})}^{2}

d_{2}^{2} = 4 x_{2}^{2}

F_{1} = \frac{k_{e} Q q}{d_{1}^{2}} = \frac{k_{e} Q q}{x_{2}^{2} + {(y_{1} - y_{2})}^{2}}

F_{2} = \frac{k_{e} q^{2}}{d_{2}^{2}} = \frac{k_{e} q^{2}}{4 x_{2}^{2}}

V = - \frac{{d y}_{1}}{d t} (↓)

A = \frac{d V}{d t} = - \frac{d^{2} y_{1}}{{d t}^{2}}

u = \frac{{d x}_{2}}{d t} (\to)

v = \frac{{d y}_{2}}{d t} (↿)

a_{x} = \frac{d u}{d t} = \frac{d^{2} x_{2}}{{d t}^{2}}

a_{y} = \frac{d v}{d t} = \frac{d^{2} y_{2}}{{d t}^{2}}

M A = M g + 2 F_{1} \sin θ

\Rightarrow \frac{d^{2} y_{1}}{{d t}^{2}} = - g - \frac{2 k_{e} Q q (y_{1} - y_{2})}{M {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}}

{m a}_{y} = F_{1} \sin θ - m g

\Rightarrow \frac{d^{2} y_{2}}{{d t}^{2}} = \frac{k_{e} Q q (y_{1} - y_{2})}{m {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}} - g

{m a}_{x} = F_{2} - F_{1} \cos θ

\Rightarrow \frac{d^{2} x_{2}}{{d t}^{2}} = \frac{k_{e} q^{2}}{4 {m x}_{2}^{2}} - \frac{k_{e} {Q q x}_{2}}{m {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}}

a t t = 0 :

y_{1} = b

x_{2} = a, y_{2} = 0

V = - \frac{{d y}_{1}}{d t} = 0

u = \frac{{d x}_{2}}{d t} = 0, v = \frac{{d y}_{2}}{d t} = 0

\dots . .

=============

i {d o n}^{'} t t h i n k t h e p r o b l e m i s s o l v a b l e .

Commented by ajfour last updated on 12/Jan/25

Gravitation free space or on frictionless horizontal flat plane i meant..sir.

G r a v i t a t i o n f r e e s p a c e o r o n f r i c t i o n l e s s

h o r i z o n t a l f l a t p l a n e i m e a n t . . s i r .

Commented by mr W last updated on 12/Jan/25

ok. then g=0 in the equstions. it′s still too hard to solve, i think.

o k . t h e n g = 0 i n t h e e q u s t i o n s .

{i t}^{'} s s t i l l t o o h a r d t o s o l v e, i t h i n k .

Commented by mr W last updated on 12/Jan/25

is it similar to the three−body problem, that means there is no general closed−form solution?

i s i t s i m i l a r t o t h e t h r e e - b o d y

p r o b l e m, t h a t m e a n s t h e r e i s n o

g e n e r a l c l o s e d - f o r m s o l u t i o n ?

Commented by ajfour last updated on 12/Jan/25

cant say, i have imagined and constructed all of it on my own. Think first we'll determine incline side as a function of angle theta..i have tried but could solve i ll try again.

Commented by ajfour last updated on 13/Jan/25

Say y+z+ssin θ=b (dz/dt)=(((2m)/M))v (dy/dt)=v (dx/dt)=u ⇒ dy(1+((2m)/M))+d(ssin θ)=0 scos θ=x m(((udu)/dx))=((kq^2 )/x^2 )−((kqQ)/s^2 )cos θ ⇒ mu^2 =((kq^2 )/s_0 ^2 )∫((s_0 /(scos θ)))^2 {1−((Q/q))cos^3 θ}d(scos θ) m(((vdv)/dy))=((kqQ)/s^2 )sin θ ((M(2v)d(2v))/dz)=2(((kqQ)/s^2 ))sin θ mv^2 =−((kq^2 )/s_0 ^2 )((M/(M+2m)))∫((Q/q))((s_0 /(ssin θ)))^2 sin^3 θd(ssin θ) ....& even from energy conservation U_0 =((kq^2 )/(4a^2 ))−((2kQq)/((a^2 +b^2 )))=((kq^2 )/s_0 ^2 )(((a^2 +b^2 )/4)−((2Q)/q)) U=((kq^2 )/s^2 )((1/(2cos θ)))^2 −((2kQq)/s^2 )+mu^2 +(m+2M)v^2 or differentiating this ((kq^2 )/(4s^2 ))(2sec^2 θtan θ)(dθ/ds)−((2kq^2 )/s^3 )(((sec^2 θ)/4)−((2Q)/q)) +2mudu+2(m+2M)vdv=0 ....

S a y y + z + s \sin θ = b

\frac{d z}{d t} = (\frac{2 m}{M}) v \frac{d y}{d t} = v \frac{d x}{d t} = u

\Rightarrow d y (1 + \frac{2 m}{M}) + d (s \sin θ) = 0

s \cos θ = x

m (\frac{u d u}{d x}) = \frac{{k q}^{2}}{x^{2}} - \frac{k q Q}{s^{2}} \cos θ

\Rightarrow {m u}^{2} = \frac{{k q}^{2}}{s_{0}^{2}} \int {(\frac{s_{0}}{s \cos θ})}^{2} {1 - (\frac{Q}{q}) \cos^{3} θ} d (s \cos θ)

m (\frac{v d v}{d y}) = \frac{k q Q}{s^{2}} \sin θ

\frac{M (2 v) d (2 v)}{d z} = 2 (\frac{k q Q}{s^{2}}) \sin θ

{m v}^{2} = - \frac{{k q}^{2}}{s_{0}^{2}} (\frac{M}{M + 2 m}) \int (\frac{Q}{q}) {(\frac{s_{0}}{s \sin θ})}^{2} \sin^{3} θ d (s \sin θ)

\dots . & e v e n f r o m e n e r g y c o n s e r v a t i o n

U_{0} = \frac{{k q}^{2}}{4 a^{2}} - \frac{2 k Q q}{(a^{2} + b^{2})} = \frac{{k q}^{2}}{s_{0}^{2}} (\frac{a^{2} + b^{2}}{4} - \frac{2 Q}{q})

U = \frac{{k q}^{2}}{s^{2}} {(\frac{1}{2 \cos θ})}^{2} - \frac{2 k Q q}{s^{2}} + {m u}^{2} + (m + 2 M) v^{2}

o r d i f f e r e n t i a t i n g t h i s

\frac{{k q}^{2}}{4 s^{2}} (2 \sec^{2} θ \tan θ) \frac{d θ}{d s} - \frac{2 {k q}^{2}}{s^{3}} (\frac{\sec^{2} θ}{4} - \frac{2 Q}{q})

+ 2 m u d u + 2 (m + 2 M) v d v = 0

\dots .

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