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We have $m\ddot{\bar{r}} = F(r)\hat{r}$

Now in my lecture notes they arrive at the following equation

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

However I'm not entirely sure as to how they arrive at that, from the usual relations $x = r\cos(\theta)$ and $y = r\sin(\theta)$

I would have to take the double time derivative of the position vector $\bar{r}$, however I'm not used to this sort of notation.

How could one derive the above equation?

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  • $\begingroup$ ocw.mit.edu/courses/mechanical-engineering/… $\endgroup$ – user126422 May 15 '17 at 8:03
  • $\begingroup$ $\hat{r}$ is time dependent, not constant. Use the product rule when taking the derivatives. $\endgroup$ – Bill N May 15 '17 at 16:38
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Consider a small variation $(\textrm{d}r, \textrm{d}\theta)$ of vector $\bar{r}$. Along $\hat{r}$, this will leed to a variation of $\textrm{d}r$. However, along $\hat{\theta}$, the value of $r$ will affect the variation, and you can see that it will be $r\textrm{d}\theta$, so finally

$$\textrm{d}\bar{r} = \textrm{d}r \hat{r} + r\textrm{d}\theta\hat{\theta}$$

which gives $$\frac{\textrm{d}}{\textrm{d}t}\bar{r} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$$

Now, since $\hat{r}$ is unit vector, the first equation gives

$$ \frac{\textrm{d}}{\textrm{d}t}\hat{r} = 0\hat{r} + 1\dot{\theta}\hat{\theta} $$

and similarly, since $\hat{\theta}$ is a unit vector, but normal to $\hat{r}$:

$$\frac{\textrm{d}}{\textrm{d}t}\hat{\theta} = 0\hat{\theta} - \dot{\theta}\hat{r}$$

Finally, this gives

$$\frac{\textrm{d}^2}{\textrm{d}t^2} = \ddot{r}\hat{r} + \dot{r}\dot{\theta}\hat{\theta} + \dot{r}\dot{\theta}\hat{\theta} + r\ddot{\theta}\hat{\theta} - r\dot{\theta}\dot{\theta}\hat{r} $$

which is the result you need.

This solution is easy to remember, and can be well understood with a little figure, but it is not really rigourous...

Since $$\begin{cases} \hat{r} = \cos(\theta)\hat{x} + \sin(\theta)\hat{y}\\ \hat{\theta} = -\sin(\theta)\hat{x} + \cos(\theta)\hat{y} \end{cases}$$

then $$\begin{cases} \frac{\textrm{d}}{\textrm{d}t}\hat{r} = \dot{\theta}\left(-\sin(\theta)\hat{x} + \cos(\theta)\hat{y}\right) = \dot{\theta}\hat{\theta}\\ \frac{\textrm{d}}{\textrm{d}t}\hat{\theta} = \dot{\theta}(-\cos(\theta)\hat{x} - \sin(\theta)\hat{y}) = -\dot{\theta}\hat{r} \end{cases}$$

Now, knowing that $\bar{r} = r\hat{r}$, you can differentiate it, and you will find the same result as before.

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Now in my lecture notes they arrive at the following equation

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

This is not true in general, even in the special case of central forces. The three body problem (or higher) is not necessarily constrained to planar motion. As long as the angle $\theta$ is measured with respect to some reference line on the plane in which the body is constrained to move, it is true in the case of a single body problem where the body is subject to a central force emanating from the origin. It is also in a two body problem where the origin is at the center of mass and where the two bodies are subject to a central force that obeys Newton's third law.

What is true in all cases except straight line motion or motion through the origin is that $$\begin{align} \vec r &\equiv r \hat r \\ \vec v &= \frac d{dt} \vec r = \dot r \hat r + r\dot {\hat r} \equiv \dot r \hat r + r\dot\theta {\hat \theta} \\ \vec a &= \frac d{dt}\vec v = \ddot r \hat r + 2 \dot r \dot {\hat \theta} + r \ddot {\hat r} = \ddot r \hat r + (2 \dot r \dot\theta + r \ddot \theta){\hat \theta} + r \dot \theta \dot {\hat \theta} \end{align}$$ where $$\begin{align} r &= ||\vec r|| \\ \hat r &= \vec r \,/\, r \\ \dot \theta &= ||\dot {\hat r}|| \\ \hat \theta &= \dot {\hat r} \,/\, \dot \theta \end{align}$$

Note that $\hat r$ is undefined when $\vec r = 0$, and that $\hat \theta$ is undefined when $\vec v$ is directed along or against $\hat r$. Note also that because $||\hat r|| = 1$, $\hat \theta$ is necessarily orthogonal to $\hat r$.

What about $\dot {\hat \theta}$? Because $\hat \theta$ is a unit vector, its time derivative $\dot {\hat \theta}$ must necessarily either be zero or be orthogonal to $\hat \theta$. This means it can have a component parallel/antiparallel to $\hat r$, but it can also have a component orthogonal to both $\hat r$ and $\hat \theta$. That second component vanishes in the special case of planar motion, in which case $\dot{\hat \theta} = -\dot \theta \hat r$, and thus $$ \vec a = \ddot r \hat r + 2 \dot r \dot {\hat \theta} + r \ddot {\hat r} = (\ddot r \hat r - r \dot \theta^2) \hat r + (2 \dot r \dot\theta + r \ddot \theta){\hat \theta} $$ Multiplying both sides of the above by mass yields the desired result.

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So you have

$$\mathbf{r}=\begin{pmatrix} r\cos\theta \\ r\sin\theta \end{pmatrix}$$

You take the derivative for each component separately which gives you:

$$\dot{\mathbf{r}}=\begin{pmatrix} \dot{r}\cos\theta-r\dot{\theta}\sin\theta \\ \dot{r}\sin\theta+r\dot{\theta}\cos\theta \end{pmatrix}$$ and for the second derivative: $$ \ddot{\mathbf{r}}=\begin{pmatrix} \ddot{r}\cos\theta-2\dot{r}\dot{\theta}\sin\theta-r\ddot{\theta}\sin\theta-r\dot{\theta}^2\cos\theta \\ \ddot{r}\sin\theta+2\dot{r}\dot{\theta}\cos\theta+r\ddot{\theta}\cos\theta -r\dot{\theta}^2\sin\theta \end{pmatrix}$$

With the (perpendicular) unit vectors: $$\hat{r}\equiv \begin{pmatrix} \cos\theta\\ \sin\theta \end{pmatrix}\\ \hat{\theta}\equiv \begin{pmatrix} -\sin\theta\\ \cos\theta \end{pmatrix}$$

You can write this as: $$\ddot{r}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}$$

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  • $\begingroup$ You forgot to make the $\ddot r$ in the LHS of the last equation bold. This could be confusing. $\endgroup$ – Sha Vuklia May 15 '17 at 19:23

protected by Qmechanic May 15 '17 at 14:56

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