Pith balls problem

1. Feb 10, 2012

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.

2. Feb 11, 2012

bruce550

3. Feb 11, 2012

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!

4. Feb 11, 2012

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... plz hlp man.. gt exams..

5. Feb 11, 2012

bruce550

nd hi.... :)

6. Feb 11, 2012

tiny-tim

hi bruce550!

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

7. Feb 11, 2012

bruce550

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

8. Feb 11, 2012

tiny-tim

you first!

9. Feb 11, 2012

bruce550

i rly dont know how to attempt this part. serious!

10. Feb 11, 2012

tiny-tim

hint: first find the initial angular acceleration

11. Feb 11, 2012

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.

12. Feb 11, 2012

bruce550

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

13. Feb 11, 2012

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)

14. Feb 11, 2012

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.

15. Feb 11, 2012

tiny-tim

16. Feb 11, 2012

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.. :(

17. Feb 11, 2012

bruce550

**rate of chng of tangential velocity

18. Feb 11, 2012

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...

19. Feb 11, 2012

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

20. Feb 11, 2012

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