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Homework Help: Velocity in a Spiral Path

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img534.imageshack.us/img534/6164/questionv.jpg [Broken]

    2. Relevant equations

    ω = dθ/dt

    v = dS/dt

    3. The attempt at a solution

    I used the second expression for the tangential speed:

    [itex]v = \frac{dS}{dt} = r \frac{d \dot{\theta}}{d t} = (b-ct) \frac{d(kt)}{dt}[/itex]

    [itex]\therefore \ v(t) = (b-ct) k[/itex]

    So is this a correct expression for speed as a function of time? :confused:

    So when r=0, the velocity would also be 0?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 14, 2012 #2

    pcm

    User Avatar

    in polar coordinates v2=(dr/dt)2+(rd[itex]\theta[/itex]/dt)2
     
  4. Apr 14, 2012 #3

    pcm

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    is it dimensionally correct?
     
  5. Apr 16, 2012 #4
    I just substituted the rate of change of radial distance into the equation v = r.dθ/dt, and then I differentiated it. What's wrong with that? :confused:
     
  6. Apr 25, 2012 #5
    I think it's dimensionally correct. Why? Did I use the wrong equations?
     
  7. Apr 25, 2012 #6
    In this case, I think this formula isn't correct.
    So, [itex]
    \vec{r}=r\cos(\theta)\hat{x}+r\sin(\theta)\hat{y}
    \\
    \vec{v}=\frac{d\vec{r}}{dt}
    \\
    v_x=\frac{d(r\cos(\theta))}{dt}=\cos(\theta)\frac{dr}{dt}+r\frac{d(\cos\theta)}{dt}=\cos(\theta) \frac{dr}{dt}-r\sin\theta \frac{d\theta}{dt}
    \\
    v_y=\frac{d(r\sin(\theta))}{dt}=\sin(\theta)\frac{dr}{dt}+r\cos\theta\frac{d\theta}{dt}
    \\
    v^2=(v_x)^2+(v_y)^2=(\frac{dr}{dt})^2+(r\frac{d \theta}{dt})^2

    [/itex]
     
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