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

[Potatochip911]

Homework Statement

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.

Homework Equations

$r=r\hat{r}$
$v=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$
$a=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}$

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 can't quite see where to go from here.

 

[ehild]

Homework Statement

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.

Homework Equations

$r=r\hat{r}$
$v=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$
$a=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}$

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 can't quite see where to go from here.

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.

 

[Potatochip911]

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.

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

 

[ehild]

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

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

This is correct.