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Dynamics: Particle confined in slot and a pole moving

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img692.imageshack.us/img692/91/1391.gif [Broken]


    2. Relevant equations
    [tex] \Sigma F_{r} = ma_{r} = m(\ddot{r}-r\dot{\theta}^2) [/tex]
    [tex] \Sigma F_{\theta} = ma_{\theta} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})[/tex]


    3. The attempt at a solution
    http://img21.imageshack.us/img21/6157/freebody.png [Broken]
    (The green angle is theta...couldn't write that on Paint)
    I used the picture above as my free body diagram instead of the one provided by the solution (in picture in problem). I did this:
    For [tex]\theta[/tex] direction:
    [tex]F_{\theta} + mg\sin{\phi} - N\sin{\phi} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})[/tex]

    And for r direction:
    [tex] mg\cos{\phi} - N\cos{\phi} = m(\ddot{r}-r\dot{\theta}^2)[/tex]

    Where [tex]\phi = \frac {\pi}{2} - \theta [/tex] and [tex] r = \frac{0.5}{\cos{\theta}}[/tex]

    Plugging the numbers in, I solve that F = 1.833N and N = 6.248N
    The solution book gives it as F = 1.78N and N = 5.79N

    Where I'm confused
    One thing I'm confused is that in the diagram the solution manual gives, they didnt draw the normal force perpendicular to the surface. I thought the basic definition of a normal force was it's supposed to be perpendicular to the surface of the object?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 12, 2009 #2

    ehild

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    The normal force is perpendicular to the surface "the constraint" on which the object moves.

    As for your solution, it is so complicated, that I can not follow it. How do you take into account that the particle can move only inside the slot?

    I found a very simple derivation, considering the motion of the particle one-dimensional, along the slot, which I called x direction. I expressed x, the position of the particle, with h and theta, calculated the second derivative of x with respect to time, from that I got the resultant force, which is equal to the vector sum of G and the normal forces of both from the slot and the rod.

    ehild
     
  4. Nov 13, 2009 #3
    I'm using the so-called "standard" coordinates in the book, so it might be a little confusing without a description.

    So basically they start out with this
    http://img269.imageshack.us/img269/9218/axis.gif [Broken]
    And they set
    [tex] \textbf{r} = r \textbf{u}_{r}[/tex]
    Then they start taking derivatives with respect to time to get velocity:
    [tex] \textbf{v} = \dot{\textbf{r}} = \dot{r}\textbf{u}_{r} + r \dot{\textbf{u}}_{r}[/tex]
    http://img43.imageshack.us/img43/619/urutheta.gif [Broken]
    Referring to the image above, and that the derivative of
    [tex]\dot{\textbf{u}} = \dot{\theta} \textbf{u}_{\theta}[/tex]
    So the equation of velocity becomes
    [tex] \textbf{v} = v_{r}\textbf{u}_{r} + v_{\theta}\textbf{u}_{\theta} [/tex]

    And now for acceleration. Basically doing the same thing and using the picture below, you get
    http://img8.imageshack.us/img8/5599/urutheta2.gif [Broken]
    [tex] a = a_{r}\textbf{u}_{r} + a_{\theta}\textbf{u}_{\theta} [/tex]
    Where
    [tex]a_{r} = \ddot{r} - r\dot{\theta}^2[/tex]
    And
    [tex]a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta} [/tex]

    That's what my accelerations are.

    Btw, in the book, it says that the normal force is always perpendicular to the tangent of the path at the given instant. Even so, the solution manual's normal force isnt...which I find puzzling.

    I was using Normal force, but I think I failed to set it up properly. And I think the image I provided was confusing because I didnt put on my axis. There should be a "r" axis which is parallel to the black line, and a theta axis perpendicular to the r axis.
     
    Last edited by a moderator: May 4, 2017
  5. Nov 13, 2009 #4

    ehild

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    The accelerations contain the first and second time-derivatives of r. I do not see how did you calculate their values at the given angle.

    ehild
     
  6. Nov 13, 2009 #5
    So I mentioned in the first post that
    [tex] r = \frac{0.5}{\cos{\theta}} [/tex]
    And so
    [tex] \frac{dr}{dt} = \frac{dr}{d\theta} \frac{d\theta}{dt}
    \Rightarrow .5\sec{\theta}\tan{\theta}\frac{d\theta}{dt} [/tex]
    And
    [tex] \frac{d^2r}{dt^2} = \frac{.5\sin{\theta}^2+.5}{\cos{\theta}^3}(\frac{d\theta}{dt})^2 + 0.5\sec{\theta}\tan{\theta}\frac{d^2\theta}{dt^2}[/tex]

    One more equation I forgot to give is that, the angle between the tangent line and the r axis is defined as
    [tex]\tan{\psi} = \frac{r}{\frac{dr}{d\theta}}[/tex]
    And the phi I used in my equation is 90 degrees minus psi, so
    [tex]\phi = 90^\circ - \psi [/tex]

    Tell me if you need any more information

    [add]
    At first I didnt add in the extra [tex]\dot{\theta}[/tex] and [tex]\ddot{\theta}[/tex]
    but after re-doing it, I still don't get the correct answer.
    Btw, could you explain why the solution manual setup its free body diagram like that?
     
    Last edited: Nov 13, 2009
  7. Nov 13, 2009 #6

    ehild

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    Now I understand what you did, but it is rather complicated, can not find the mistake in it for the time being. I do not understand the free-body diagram of the solution manual, but I have got the correct solution with an easy method. Maybe it is of some use for you.

    So, instead of polar coordinates, I use the normal x,y axes, x being horizontal, parallel to the slot and y is vertical.

    There are three forces acting on the body: the push from the rod (F), gravity (G) and the the normal force of the slot (N). I assumed this normal force upward. F is perpendicular to the rod, and it makes the angle theta with the positive x axis.

    [tex]x=h\tan(\theta)[/tex]

    [tex]\dot x=h\frac{\omega}{\cos(\theta)^2}[/tex]

    [tex]\ddot x=2h\omega^2\frac{\sin(\theta)}{\cos^3(\theta)}[/tex]

    The x component of the push from the rod is equal to mass times second derivative of x:

    [tex]M\ddot x=F\cos(\theta) \rightarrow F=2hM\omega^2\frac{\sin(\theta)}{\cos^4(\theta)}= 1.78 N[/tex]

    The y component of F cancels with the normal force and gravity:

    [tex]Mg-N+F\sin(\theta)=0 \rightarrow N=5.79 N [/tex]

    ehild
     
    Last edited: Jun 29, 2010
  8. Nov 13, 2009 #7
    Oh cool. I didn't expect to solve it this easily. I guess sometimes I shouldn't be mislead to use the coordinates that the book uses in that particular section. But I think the purpose of this is simply to practice the polar coordinates.
     
  9. Nov 13, 2009 #8

    ehild

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    Well, it is good to practice polar coordinates, but not on one-dimensional motion.
    Meanwhile I have found a mistake in your derivation:
    ----------
    For theta direction:
    [tex]
    F_{\theta} + mg\sin{\phi} - N\sin{\phi} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})
    [/tex]

    And for r direction:
    [tex]
    F_{\theta} + mg\sin{\phi} - N\sin{\phi} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta})
    [/tex]

    --------------------

    It should be theta everywhere instead of phi. But it was a good job you did.

    ehild
     
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