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Mass moving back and forth at the bottom of a circle (Polar)

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass ##m## at the bottom of a circle of radius R moves back and forth with no friction and the follows the equation (where ##\alpha(t)## is small) ##\theta(t)=\frac{3\pi}{2}+\alpha(t)##. Find a differential equation using polar coordinates for ##\alpha(t)## which is linear.

    untiled99.png

    2. Relevant equations
    ##r=r\hat{r}##
    ##v=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}##
    ##a=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}##

    3. The attempt at a solution

    Since ##r## is the constant ##R## we have ##r=R\hat{r}##, we also know that the tangential acceleration ##a_{\theta}=(2\dot{r}\dot{\theta}+r\ddot{\theta})## which from the diagram we can also see that ##a_{\theta}=mg\sin\alpha(t)##, the radial acceleration is given by ##a_{r}=(\ddot{r}-r\dot{\theta}^2)## which I believe is equal to (where ##N## is the normal force) ##a_{r}=N-mg\cos\alpha(t)##. I cant quite see where to go from here.
     
  2. jcsd
  3. Feb 22, 2016 #2

    ehild

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    You know that θ = 3π/2+ α, so the derivatives of θ are the same as those of α. What is the tangential acceleration in terms of α, taking into account that r=R = constant?
    We measure the angles anti-clockwise. The tangential acceleration is of opposite direction as mgsinα.
     
  4. Feb 25, 2016 #3
    I managed to get to the solution. Since the radial acceleration in this case will always be equal to zero there is no point in integrating it. The angle ##\alpha## will be determined by ##a_{\phi}## which is given by ##a_{\phi}=-g\sin\alpha## and from our relationships we have ##R\ddot{\alpha}=-g\sin\alpha\iff\ddot{\alpha}=-\frac{g}{R}\sin\alpha##, now the small angle approximation can be made and we obtain ##\ddot{\alpha}=-\frac{g}{R}\alpha##
     
  5. Feb 25, 2016 #4

    ehild

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    The radius is constant, so its derivatives are zero. The radial acceleration is not, it is ##a_r = -r \dot \theta^2 ##, the centripetal acceleration.
    This is correct.
     
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