Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Pith balls problem

  1. Feb 10, 2012 #1
    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. jcsd
  3. Feb 11, 2012 #2
    Help please...
     
  4. Feb 11, 2012 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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:
     
  5. Feb 11, 2012 #4
    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..
     
  6. Feb 11, 2012 #5
    nd hi.... :)
     
  7. Feb 11, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi bruce550! :smile:

    (have a theta: θ and try using the X2 button just above the Reply box :wink:)
    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))
     
  8. Feb 11, 2012 #7
    yea gt dat... tnx. but what about the next part????????????????
     
  9. Feb 11, 2012 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    you first! :wink:
     
  10. Feb 11, 2012 #9
    i rly dont know how to attempt this part. serious!
     
  11. Feb 11, 2012 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hint: first find the initial angular acceleration :wink:
     
  12. Feb 11, 2012 #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.
     
  13. Feb 11, 2012 #12
    i mean i get domega/dt =(dq/dt)^2 *(k/mgx^2) ..
     
  14. Feb 11, 2012 #13

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (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:
     
  15. Feb 11, 2012 #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.
     
  16. Feb 11, 2012 #15

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

  17. Feb 11, 2012 #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.. :(
     
  18. Feb 11, 2012 #17
    **rate of chng of tangential velocity
     
  19. Feb 11, 2012 #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...
     
  20. Feb 11, 2012 #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
     
  21. Feb 11, 2012 #20

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook