1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can't figure this problem out for the life of me

  1. Oct 14, 2006 #1
    Ok, let me see if I can explain this well enough. There's a horizontal circular rod governing the motion of a collar having mass .75kg. The coefficient of kinetic friction is .3 and the collar is given an initial speed of 4 m/s starting at theta=0. I need to determine the arclength the collar travels before coming to a rest.
    This problem doesn't seem all that hard until I try to solve it, mainly because the magnitude of the frictional force depends on instantaneous velocity. Can anyone help me?
     
  2. jcsd
  3. Oct 14, 2006 #2

    Fermat

    User Avatar
    Homework Helper

    The friction force is mu * ( NR + Fc)
    where
    NR = normal reaction from the rod on the collar
    Fc = centripetal force = mv²/R
    Now use Newton's 2nd law.
     
  4. Oct 14, 2006 #3
    But how do I account for the varying velocity?
     
  5. Oct 14, 2006 #4

    Fermat

    User Avatar
    Homework Helper

    You incorporate that into your eqn of motion.
    Using Newton's 2nd law, F = ma, or

    [tex]F_r = m\ddot s[/tex]
    where
    [tex]F_r = \mu(NR + F_c)[/tex]
    and
    [tex]\ddot s = \frac{d^2s}{dt^2}= \frac{dv}{dt}[/tex]

    [tex]v = \dot s[/tex] the peripheral, or circumferential velocity.
     
  6. Oct 14, 2006 #5
    Have you tried to solve the problem? It's impossible to solve for s with what you just told me. If it were that easy I'd have already done it. You come up with a=(u/m)(NR+mv^2/R), but you can't go further than that. The differential equation is unsolvable due to the v squared.
     
  7. Oct 14, 2006 #6

    Fermat

    User Avatar
    Homework Helper

    You can certainly go further than that.
    If you were to continue, you would end up with a term including v² on the denominator. This is where you could use a hyperbolic trig substutution:- v = √(gR)sinh(u). You would end up with integrating 1/cosh(u) - which is integrable. It's just not very nice.

    The better way to go is to rewrite [tex]\ddot s[/tex] as [tex]\ddot s = \frac{d^2s}{dt^2} = \frac{dv}{dt} = \frac{dv}{ds}\cdot\frac{ds}{dt} = v\cdot\frac{dv}{ds}[/tex]

    Your expression is now easily integrable and you should end up with a logarithmic function.
     
  8. Oct 14, 2006 #7
    I did think of that approach, but don't you need the acceleration in terms of the position to do it that way?
     
  9. Oct 14, 2006 #8

    Fermat

    User Avatar
    Homework Helper

    You shouldn't have any problem itegrating your expression. If you do, could you post the expression that you are working with ?
     
  10. Oct 14, 2006 #9
    I don't know what I'm doing in all honesty. That's why I'm here. What I was asking is in order to carry out ads=vdv wouldn't you need a in terms of s. What I have is a in terms of v, so it doesn't matter what expression im trying to integrate, it won't work.
     
  11. Oct 14, 2006 #10

    Fermat

    User Avatar
    Homework Helper

    You have a=(u/m)(NR+mv^2/R)
    and
    a = v.(dv/ds)
    Equate the two,
    v.(dv/ds) = (u/m)(NR+mv^2/R)
    Now integrate.

    Edit: I missed a bit out. You will have to put in a minus sign in the above eqn, since the friction force is acting in the opposite direction to direction of travel.
     
  12. Oct 15, 2006 #11
    I'm sorry, but I'm really not understanding. Maybe you're referring to something in calculus I haven't come across yet, but I don't know how to integrate an expression with two differentials in it. I only know how to integrate expressions where there is an expression with the only variable being the one you're integrating over, followed by the differential of that variable- ie f(v)dv.
     
  13. Oct 15, 2006 #12

    Fermat

    User Avatar
    Homework Helper

    Have you come across this method before ? It is called separation of variables.

    Suppose you have a differential equation [tex]\frac{dy}{dx} = \frac{y}{x}[/tex]

    then separate the variables like this,

    [tex]\frac{dy}{y} = \frac{dx}{x}[/tex]

    You keep all the y-variables (and differential operator) on one side and the x-variables on the other.

    Now integrate both sides by putting an integral sign in front of both of them, like so

    [tex]\int\frac{dy}{y} = \int\frac{dx}{x}[/tex]
    [tex]ln(y) = ln(x) + ln(K)[/tex]
    [tex]ln(y) = ln(Kx)[/tex]
    [tex]y = Kx[/tex]
    ======

    Can you see how to apply this method to your problem, using the equation in my previous post, #10 ?
     
    Last edited: Oct 15, 2006
  14. Oct 15, 2006 #13
    I am familiar with basic differential equations- but the equation is not seperable. It appears I'm having a math problem. If you really don't see why this problem could be confusing, could you just go ahead and show me how to solve it? All you've been doing is showing me basic equations and concepts that are second nature to me by now, but that I'm having trouble applying to this problem. I understand all of the principles perfectly, it's just a hard problem.
     
  15. Oct 15, 2006 #14

    Fermat

    User Avatar
    Homework Helper

    The equation, from post #10, is

    v.(dv/ds) = (u/m)(NR+mv^2/R)

    and the equation is separable.

    write it as,

    v/(NR + mv²/R) dv = (u/m) ds

    all of the v-terms are on the lhs, and all of the s-terms (if there were any) would be on the rhs, but in this case, the rhs is just a constant value.

    Put integral signs in front of both sides in the expression above and integrate.

    As I said in an earlier post, you will get a logarithmic function.
     
  16. Oct 15, 2006 #15
    Ok, I think I got it. Thanks a lot.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Can't figure this problem out for the life of me
Loading...