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Circular acceleration and Integration

  1. Mar 17, 2005 #1
    Consider a bead of mass m that is free to move on a thin, circular wire of
    radius r. The bead is given an initial speed v0, and there is a coefficient of
    kinetic friction [tex]F_k[/tex]. The experiment is performed in a spacecraft drifting inspace. Find the speed of the bead at any subsequent time t.

    Ok so you need to find the velocity of the object.

    Well the acceleration would be: [tex] a = -F_k v^2 / r [/tex]

    To find velocity, you would have to integrate that...

    [tex]dv/dt = \int -F_k v^2 / r [/tex]

    and that is....

    [tex]v^-2 F_k r^-1[/tex]

    now what do i do?
     
    Last edited: Mar 17, 2005
  2. jcsd
  3. Mar 17, 2005 #2
    I'm not sure what you mean by [tex]v^-2 F_k r^-1[/tex]. Separate the equation, integrate both sides and put in the boundary condition you've been given. Also, the RHS of the line before makes no sense, cos you're not integrating wrt any variable.
     
    Last edited: Mar 17, 2005
  4. Mar 17, 2005 #3
    You forgot that [tex]a[/tex] and [tex]\frac{dv}{dt}[/tex] are the same thing.

    Let's denote by [tex]\mu[/tex] the coefficient of friction (because F reminds us of force) :smile:

    Then you have a differential equation that you can solve for v in terms of t:

    [tex]\frac{dv}{dt} = -\mu \frac{v^2}{r}[/tex]

    The 'v' side is to be integrated from the initial velocity [tex]v_0[/tex] to [tex]v[/tex] at any instance of time.
     
  5. Mar 17, 2005 #4
    What is the v side? Do you mean dV? If so, how is that calculated?
     
  6. Mar 17, 2005 #5
    dv is a differential if you remember it from calculus :wink: .

    So take all 'v's to the left hand side and all 't's to the right hand side, and then integrate over the same limits:

    [tex]\frac{dv}{v^2} = -\frac{\mu}{r}dt[/tex]

    [tex]\int_{v_0}^{v} \frac{1}{v^2} dv = -\frac{\mu}{r} \int_{t_0}^{t} dt = -\frac{\mu}{r}t[/tex]

    [tex]-\frac{1}{v} - (-\frac{1}{v_0}) = -\frac{\mu}{r}t[/tex]

    Here you solve for v in terms of [tex]v_0[/tex] and t.
     
  7. Mar 17, 2005 #6
    I assume you find t and then find v0?

    So, for t,

    [tex]t = \mu r / v+ v_0 ?[/tex]
     
    Last edited: Mar 17, 2005
  8. Mar 17, 2005 #7
    From the question it is clear that it is asked v in terms of t, right? So time is given and we don't need to derive it! It is asked v(t), where constant [tex]v_0[/tex] comes in.

    Take it conceptually. You give the bead an initial velocity and it moves with friction around the circle until it stops due to the deceleration. We got:

    [tex]\frac{1}{v} = \frac{1}{v_0} + \frac{\mu}{r}t[/tex]

    You can see that with the passage of time [tex]\frac{1}{v}[/tex] increases, so v decreases. Also you can notice that the units agree (m/s).
    Hope that this makes it quite clear.
     
  9. Mar 17, 2005 #8
    So [tex]v(t) = v_0 + r / \mu t[/tex]? Thanks for the help :smile:
     
    Last edited: Mar 17, 2005
  10. Mar 17, 2005 #9
    No, it is wrong because there's addition in the middle.

    Remember this:

    [tex]\frac{1}{\frac{1}{a} + \frac{1}{b}} [/tex] is not the same as [tex]a + b[/tex]
     
  11. Mar 17, 2005 #10
    This is frustrating. More so for you probably! Can you please walk through it?
     
  12. Mar 17, 2005 #11
    The identity you claim simply doesn't hold! You are not dealing with multiplication

    [tex]\frac{1}{\frac{1}{a} \frac{1}{b}}[/tex],

    but with addition

    [tex]\frac{1}{\frac{1}{a} + \frac{1}{b}}[/tex]

    If it were otherwise we would for example have written the formula for lenses as f = v + u, and not as,

    [tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex].
     
    Last edited: Mar 17, 2005
  13. Mar 17, 2005 #12
    ahhh. Got it :)

    Cheers
     
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