Question

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-inertial frame of reference. The relationship between the force $\vec{F}_{rot}$ experienced by a particle of mass $m$ moving on the rotating disc and the force $\vec{F}_{in}$ experienced by the particle in an inertial frame of reference is

$$\vec{F}_{rot} = \vec{F}_{in} + 2m(\vec{v}_{rot} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega}.$$

where $\vec{v}_{rot}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the centre of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis ($\vec{\omega} = \omega\hat{k}$). A small block of mass $m$ is gently placed in the slot at $\vec{r} = (R/2)\hat{i}$ at $t=0$ and is constrained to move only along the slot.

The distance $r$ of the block at time $t$ is:

• A. $\displaystyle \frac{R}{4}\bigl(e^{\omega t}+e^{-2\omega t}\bigr)$
• B. $\displaystyle \frac{R}{2}\cos 2\omega t$
• C. $\displaystyle \frac{R}{2}\cos \omega t$
• D. $\displaystyle \frac{R}{4}\bigl(e^{\omega t}+e^{-\omega t}\bigr)$

Answer 1

Answer: (D)

$\displaystyle \frac{R}{4}\bigl(e^{\omega t}+e^{-\omega t}\bigr)$

Answer 2

Equation of motion in the rotating frame
We use the effective forces:

• Coriolis force yields no $x$-component since $\vec{v}_{rot}\parallel\hat{i}$.
• Centrifugal force gives

$$(\vec{\omega}\times \vec{r})\times \vec{\omega} = \omega^2\,x\,\hat{i}.$$

Thus along the slot,

$$m\,\ddot x = m\,\omega^2\,x \quad\Longrightarrow\quad \ddot x - \omega^2 x = 0.$$

General solution

$$x(t)=Ae^{\omega t}+Be^{-\omega t}.$$

Initial conditions

$$x(0)=\tfrac{R}{2}\ \Rightarrow\ A+B=\tfrac{R}{2},\qquad \dot x(0)=0\ \Rightarrow\ A=B.$$

Hence $A=B=\tfrac{R}{4}$ and

$$x(t)=\frac{R}{4}\bigl(e^{\omega t}+e^{-\omega t}\bigr).$$