Question

A frame of reference that is accelerated with respect to an internal 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 ${\vec{F}}_{\text{rot }}$ experienced by a particle of mass $m$ moving on the rotating disc and the force ${\vec{F}}_{\text{in }}$ experienced by the particle in an internal frame of reference is
${\vec{F}}_{rot}={\vec{F}}_{in}+2m\left({\vec{v}}_{rot}\times \vec{\omega }\right)+m(\vec{\omega }\times \vec{r})\times \vec{\omega }$
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 $\vec{r}=\left(\frac{R}{2}\right)\hat{i}$ at $t=0$ and is constrained to move only along the slot.

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

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

Answer 1

Answer: (C)

Force on block along slot $=m{\omega }^{2}r$
$=ma=m\left(\frac{vdv}{dr}\right)$
$\underset{0}{\overset{V}{\int }}vdv=\underset{\frac{R}{2}}{\overset{r}{\int }}{\omega }^{2}rdr$
$\frac{v^2}{2}=\frac{{\omega }^2}{2}\left(r^2-\frac{R^2}{4}\right)$
$\Rightarrow v=\omega \sqrt{r^2-\frac{R^2}{4}}=\frac{dr}{dt}$
$\Rightarrow \underset{\frac{R}{4}}{\overset{r}{\int }}\frac{dr}{\sqrt{r^2-\frac{R^2}{4}}}=\underset{0}{\overset{t}{\int }}\omega dt$
$\ln \left(\frac{\frac{r+\sqrt{r^2-\frac{R^2}{4}}}{R}}{2}\right)-\ln <br/>\left(\frac{\frac{\frac{R}{2}+\sqrt{\frac{R^2}{4}-\frac{R^2}{4}}}{R}}{2}\right)=\omega t$
$\Rightarrow r+\sqrt{r^2-\frac{R^2}{4}}=\frac{R}{2}{e}^{\omega t}$
$\Rightarrow r^2-\frac{R^2}{4}=\frac{R^2}{4}{e}^{2\omega t}+r^2-2r\frac{R}{2}{e}^{\omega t}$
$\Rightarrow r=\frac{\frac{R^2}{4}{e}^{2\omega t}+\frac{R^2}{4}}{R{e}^{\omega t}}=\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$