Pith Balls Problem: Find Expression for x & Instantaneous Relative Speed

[bruce550]

If 2 similar pith balls of mass m are hung from a common point with the help of 2 long silk threads. The balls carry similar charges q. Assuming theta is very small find an expression for x. Also assuming that each ball loses charge at a rate of 1*10^-9 c/s at what instantaneous relative speed (dx/dt) do the balls approach each other initially?? Please help.

 

[bruce550]

Help please...

 

[tiny-tim]

welcome to pf!

hi bruce550! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[bruce550]

tcos(theta)=mg and tsintheta=F
dividing we get tantheta=f/mg =(kq^2/x^2 )*1/mg ... - (1)
now since theta is very small so theta=tantheta.
equating 1... we get
x^2=kq^2/mgtheta and hence we get the value for x .. m not sure if m correct. but if i m then how to find the second part of the question.. m stuckd... please hlp man.. gt exams..

 

[bruce550]

nd hi... :)

 

[tiny-tim]

hi bruce550!

(have a theta: θ and try using the X² button just above the Reply box )

tcos(theta)=mg and tsintheta=F
dividing we get tantheta=f/mg =(kq^2/x^2 )*1/mg ... - (1)
now since theta is very small so theta=tantheta.
equating 1... we get
x^2=kq^2/mgtheta

your method is fine, but you haven't gone far enough …

θ depends on x, so you still need to get rid of it!

use tanθ = x/L ​

(btw, "θ is very small" means tanθ = x/L, instead of tanθ = x/√(L² - x²))

 

[bruce550]

yea gt dat... tnx. but what about the next part??

 

[tiny-tim]

you first!

 

[bruce550]

i rly don't know how to attempt this part. serious!

 

[tiny-tim]

hint: first find the initial angular acceleration

 

[bruce550]

ohkay.. i did t. α=d(omega)/dt=d^2θ/dt^2 right?
then ... using previous values from the first aprt of the question i get..
(dq/dt)^2 *(k/mgx^2) .. right? now to find the instantaneous speed .. how do i find the mass and x? thanks again.

 

[bruce550]

i mean i get domega/dt =(dq/dt)^2 *(k/mgx^2) ..

 

[tiny-tim]

(have an omega: ω )

no, you need to do F = ma all over again

this time, the vertical acceleration isn't 0, so you'll need to do F = ma in the radial direction (because you do know the centripetal acceleration)

 

[bruce550]

hhey... i m not getting your point.. can u jst solv it.. i mean if u knw d right way 2 solv it. tnks.

 

[tiny-tim]

no, that's not the way it works on this forum

do you know about centripetal acceleration?​

 

[bruce550]

of course... its the rate of tangential velocity. ohk .. can you tell me to what parts i m corrct?? i mean i did dat domega/dt =(dq/dt)^2 *(k/mgx^2) .. and you are saying about centripetal acceleration.. :(

 

[bruce550]

**rate of chng of tangential velocity

 

[bruce550]

mmm.. do you mean a=dv/dt and w^2 *r=dv/dt ... if it is den.. how do i find ω here?? i know dat ω=v/r...

 

[bruce550]

hey i think i have got it... chek dis..
tao=pEsintheta.. so sintheta=theta right??
alfa=qxetheta/mx^2=qetheta/mx now since theta=x/√l^2-x^2 ... so putting we get...
alfa=qE/m√l^2-x^2

 

[tiny-tim]

(forget θ = tanθ, all you need is sinθ = tanθ = x/L)

you don't seem to know what centripetal acceleration is

have you done any questions on it?

you need to apply F = ma along the string

 

[bruce550]

got the answer.. tnks..