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

1. Feb 21, 2016

Potatochip911

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.

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. Feb 22, 2016

ehild

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.

3. Feb 25, 2016

Potatochip911

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 $\ddot{\alpha}=-\frac{g}{R}\alpha$

4. Feb 25, 2016

ehild

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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