Equation of Motion in a Non-Inertial (Rotating) Frame

Question

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?

Answer 1

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.