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

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Homework Help Overview

The discussion revolves around a physics problem involving two pith balls of mass m, which are charged and suspended from a common point. The original poster seeks to find an expression for the distance x between the balls and their instantaneous relative speed as they lose charge over time.

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  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the angles, forces, and distances involved in the problem. Some attempt to derive expressions for x using trigonometric relationships and force equations, while others express uncertainty about their methods and seek clarification on the next steps.

Discussion Status

Several participants have provided hints and guidance on how to approach the problem, particularly regarding the relationships between the angles and the forces acting on the pith balls. There is an ongoing exploration of the physics concepts involved, including centripetal acceleration and angular motion, with no clear consensus on the final approach yet.

Contextual Notes

Participants have noted the assumption of small angles and the rate of charge loss, which may influence the dynamics of the system. There is also a mention of imposed homework rules that guide the nature of the assistance provided.

bruce550
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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.
 
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Help please...
 
welcome to pf!

hi bruce550! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
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..
 
nd hi... :)
 
hi bruce550! :smile:

(have a theta: θ and try using the X2 button just above the Reply box :wink:)
bruce550 said:
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 :smile:

(btw, "θ is very small" means tanθ = x/L, instead of tanθ = x/√(L2 - x2))
 
yea gt dat... tnx. but what about the next part??
 
you first! :wink:
 
i rly don't know how to attempt this part. serious!
 
  • #10
hint: first find the initial angular acceleration :wink:
 
  • #11
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
i mean i get domega/dt =(dq/dt)^2 *(k/mgx^2) ..
 
  • #13
(have an omega: ω :smile:)

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) :wink:
 
  • #14
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
  • #16
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
**rate of chng of tangential velocity
 
  • #18
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
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
(forget θ = tanθ, all you need is sinθ = tanθ = x/L)

you don't seem to know what centripetal acceleration is :redface:

have you done any questions on it?

you need to apply F = ma along the string
 
  • #21
got the answer.. tnks..
 

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