Let me preface my question by informing you this is for an assignment, so I would rather not have explicit answers but rather be given guidance in arriving at the correct solution.

The question is thus; I have a mass on a frictionless rotating turntable. I need to adopt a co-rotating frame to find the equation of motion for this mass. I am expected to solve this using only Newtonian mechanics.

As we are restricted to a plane, there is no component of motion in the $\vec{k}$ direction. The only forces acting on our mass (in the rotating frame) are the Centrifugal force, and the Coriolis force.

My attempt at a solution;

$\vec{F} = m\vec{a}$

$\vec{F_{cent}} = -m\vec{\Omega}$ x $\vec{\Omega}$ x $\vec{r}$

$\vec{F_{cor}} = -2m\vec{\Omega}$ x $\vec{v}$

$m\vec{a} = -m\vec{\Omega}$ x $\vec{\Omega}$ x $\vec{r} - 2m\vec{\Omega}$ x $\vec{v}$

As I'm working in the rotating frame, I don't see any reason to not use cartaesian coordinates (please note this is my first time using latex, so in this case $\vec{i},\vec{j},\vec{k}$ are unit vectors, and x,y,z are magnitudes in the respective directions);

$\vec{\Omega} = \Omega\vec{k}$

$\vec{r} = x\vec{i} + y\vec{j}$

$\vec{v} = \vec{\dot{r}} = \dot{x}\vec{i} + \dot{y}\vec{j}$

Substituting this in, evaluating the cross-products and simplifying yields;

$\ddot{x}\vec{i} + \ddot{y}\vec{j} = \Omega^2x\vec{i} + \Omega^2y\vec{j} - 2\Omega\dot{x}\vec{j} + 2\Omega\dot{y}\vec{i}$

And so we have two coupled second order differential equations;

$\ddot{x} = \Omega^2x + 2\Omega\dot{y}$

$\ddot{y} = \Omega^2y - 2\Omega\dot{x}$

A method we had used previously in class to solve coupled equations was to set $\Omega = 0$ and solve, then substitute this solution back in for $\dot{x} and \dot{y}$. I attempted this, however it yielded two cubic equations. The solution I am told this system has, for the initial conditions $(x(0),y(0)) = (x_0,0)$, is a spiral when mapped parametrically, namely;

$x(t) = (x_0 + v_{x0}t)\cos\Omega t + (v_{y0} + \Omega x_0)t\cos\Omega t$

$y(t) = -(x_0 + v_{x0}t)\sin\Omega t + (v_{y0} + \Omega x_0)t\sin\Omega t$

To me these appeared to be solutions gained from solving the non-homogenous linear second-order differential equation, however this did not work either.

Is my derivation of the original vector equation of motion correct? If not, where did I go wrong? If so, what method should I use to solve these equations to find the appropriate solutions?

  • $\begingroup$ Instead of $\vec{i}$, you should probably write $\widehat{i}$ (similarly for $\vec{j}$ and $\vec{k}$) to indicate that these vectors are unit vectors. Furthermore, it is my personal preference to use $\widehat{x}$, $\widehat{y}$, and $\widehat{z}$ over $\widehat{i}$, $\widehat{j}$, and $\widehat{k}$ because of confusion with quaternions. $\endgroup$ – Jonathan Gleason Sep 8 '11 at 17:19
  • $\begingroup$ Also, the centrifugal force term is $-m\mathbf{\Omega}\times \left( \mathbf{\Omega}\times \mathbf{r}\right)$. Without the parenthesis, your equation implies that the centrifugal force is $-m\left( \mathbf{\Omega}\times \mathbf{\Omega}\right) \times \mathbf{r}$, which can't possibly be the case because this is always $\mathbf{0}$. Remember, the cross-product is nonassociative. Furthermore, this is just another personal preference, but you might want to use boldface to denote vectors in \LaTeX because it makes them stand out more. $\endgroup$ – Jonathan Gleason Sep 8 '11 at 17:23
  • $\begingroup$ I couldn't find the command in LaTeX for the hats, hence my reliance on the i,j,k notation. So for future reference, what is the command? And thank you for the advice on the parentheses, unfortunately my lecturer never wrote it out with them (which lead to some confusion). $\endgroup$ – Daniel Blay Sep 9 '11 at 5:27
  • $\begingroup$ \widehat{}. \hat{} should also work. $\endgroup$ – Jonathan Gleason Sep 10 '11 at 2:10

My hint is to solve the problem using polar coodriantes along with the handy equations founds here. We have that $$ \mathbf{a}=\ddot{\mathbf{r}}=\left( \ddot{r}-r\dot{\theta}^2\right) \widehat{\mathbf{r}}+\left( 2\dot{r}\dot{\theta}+r\ddot{\theta}\right) \widehat{\mathbf{\theta}}=-\Omega ^2r\widehat{\mathbf{z}}\times \widehat{\mathbf{\theta}}-2\Omega \widehat{\mathbf{z}}\times \left( \dot{r}\widehat{\mathbf{r}}+r\dot{\theta}\widehat{\mathbf{\theta}}\right) $$ Thus, $$ \left( \ddot{r}-r\dot{\theta}^2\right) \widehat{\mathbf{r}}+\left( 2\dot{r}\dot{\theta}+r\ddot{\theta}\right) \widehat{\mathbf{\theta}}=\left( \Omega ^2r+2\Omega r\dot{\theta}\right) \widehat{\mathbf{r}}-2\Omega \dot{r}\widehat{\mathbf{\theta}}. $$ Thus, $$ \ddot{r}-r\dot{\theta}^2=\Omega ^2r+2\Omega r\dot{\theta} $$ and $$ 2\dot{r}\dot{\theta}+r\ddot{\theta}=-2\Omega \dot{r}. $$ This might seem difficult to solve, but you easily show that $\dot{\theta}=-\Omega$ (just think about how $\theta$ and $\Omega$ are defiend), and so these just reduce to $$ \ddot{r}=0. $$

Think about it. This should make a lot of physical sense.

  • $\begingroup$ Thank you. Ultimately I asked my lecturer and he clarified he just wanted me to derive the differential equations and then check them with the given solutions. The ODEs were indeed correct. $\endgroup$ – Daniel Blay Sep 9 '11 at 5:32

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