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## Physics on a turntable

A turntable (merry-go-round, roundabot, carousel) is a simple example of non-inertial reference frame. I discuss some simple examples of motion in the inertial as well as rotating frame.

### Introduction

In this article I will discuss simple physical phenomena in a horizontally rotating two dimensional system. A physical realization of such a system are: a turntable for playing music records, a merry-go-round, a roundabot, a carousel, a horizontal centrifuge, etc. The view of such a system from above is:

The turnable rotates with the constant angular velocity $\omega$ with respect to an invarint system. I denote the variables in the invariant system by capital letters and the variables in the rotating system by small letters. The relation between these variables is: $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos \omega t & \sin \omega t \\ - \sin \omega t & \cos \omega t \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} \tag {1}$$

The equation of motion in the rotating frame are: $$m \mathbf {\ddot r} = \mathbf F - 2 m \pmb \omega \times \mathbf {\dot r} - m \pmb \omega \times (\pmb \omega \times \mathbf r) \tag {2}$$ or in components $\mathbf r = (x, y)$ : $$m \begin{bmatrix} \ddot x \\\ddot y \end{bmatrix} = \mathbf F - 2 m \omega \begin{bmatrix} - \dot y \\ \dot x \end{bmatrix} + m \omega^2 \begin{bmatrix} x \\ y \end{bmatrix} \tag {3}$$ On the right hand of this equation is $\mathbf F$, the force outside the rotating system. i.e. in the inertial system, followed by the Coriolis force orthogonal to the movement in the rotating system and the centrifugal force that acts radially from the rotation center outside:

### Particle resting in the inertial system

Consider a particle with mass $m$ that is at rest in the inertial system, say at the position $(R, 0)$. It is $\mathbf F = 0$ and we have \begin{align} X(t) & = R \\ \tag {4} Y(t) & = 0 \end{align} In the rotating system we would need to solve the system of differential equations (3). Since we have obtained easily the solution (4) in the inertial system we can just use the eq. (1) to transform it into the rotating system. We obtain: \begin{align} x(t) & = R \cos(- \omega t) \\ \tag {5} y(t) & = R \sin(- \omega t) \end{align} By differentiating eq.(5) we can indeed confirm that these $x(t), y(t)$ satisfy the eq. (3).

A particle that is at rest in the inertial system appears as rotating with angular velocity $- \omega$ in the rotation system that rotates with $\omega$.

### Ball in a groove

Consider a turntable with a radial groove. The turntable rotates with a constant angular velocity $\omega$. A ball with mass $m$ is released inside the groove at a distance $x_0$ from the center. What is the movement of the ball?

We adopt a rotating coordinate system with the x-axis along the groove. The only effective force is the centrifugal force because the gravity is compensated by the turntable and the Coriolis force by the groove. The ball moves radially according to $$m \frac{d^2 x}{dt^2} = m \omega^2 x$$ The solution of this differential equation is \begin{align} x(t) & = \frac{x_0}{2} (e^{\omega t} + e^{- \omega t}) \\ v(t) & = \frac{x_0}{2} \omega (e^{\omega t} - e^{- \omega t}) \end{align}

From the formulas above we can compute the velocity $v$ as a function of the distance $x$: $$v^2 = \omega^2 (x^2 - x_0^2)$$

How can the distance and the velocity increase exponentially? Well, it cannot. When no energy is supplied to the system then the initial rotational energy $$E_{rot} = \frac{1}{2} m x_0^2 \omega_0 = \frac{1}{2} m x(t)^2 \omega(t)$$ remains constant and consequently $\omega$ must decrease as $x(t)$ increases.