Question

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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 ${\overrightarrow{F}}_{\text{rot}}$ experience by a particle of mass $m$ moving on the rotating disc and the force ${\overrightarrow{F}}_{\text{in}}$ experienced by the particle in an inertial frame of reference is
${\overrightarrow{F}}_{\text{rot}}={\overrightarrow{F}}_{\text{in}}+2m({\overrightarrow{v}}_{\text{rot}}\times \overrightarrow{\omega })+m(\overrightarrow{\omega }\times \overrightarrow{r})\times \overrightarrow{\omega },$

where ${\overrightarrow{v}}_{\text{rot}}$ is the velocity of the particle in the rotating frame of reference and $\overrightarrow{r}$ 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 counterclockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x-$axis along the slot, the $y-$ axis perpendicular to the slot and the $z-$ axis along the rotation axis $\overrightarrow{\omega =\omega \overrightarrow{k}}$. A small block of mass $m$ is gently placed in the slot at $\overrightarrow{r}=\left(R/2\right)\overrightarrow{i}$ 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. $\frac{R}{4}(e^{\omega t}+e^{-\omega t})$
• D. $\frac{R}{4}(e^{2\omega t}+e^{-2\omega t})$

Answer 1

The correct option is C $\frac{R}{4}(e^{\omega t}+e^{-\omega t})$

Force experienced by the particle in the inertial frame of reference ${\overrightarrow{F}}_{in}$ will have the components in the direction perpendicular to direction of ${\overrightarrow{F}}_{rot}$.
Considering the rotating frame of reference for the particle, the velocity of particle will be, ${\overrightarrow{v}}_{rot}=\frac{dr}{dt}\hat{i}$ and $\overrightarrow{\omega }=\omega \hat{k}$
But the cross product $({\overrightarrow{v}}_{rot}\times \overrightarrow{\omega })$ will be in the direction perpendicular to the direction of ${\overrightarrow{F}}_{rot}$ which is along the slot i.e $(+xor\hat{i})$, since the particle will accelerate only in the +ve direction along the slot.
Now, $m(\overrightarrow{\omega }\times \overrightarrow{r})\times \overrightarrow{\omega }=m(\omega \hat{k}\times r\hat{i})\times \omega \hat{k}=m{\omega }^{2}r$
Also ${\overrightarrow{F}}_{rot}=m\overrightarrow{a}=ma\hat{i}$
Now,
${\overrightarrow{F}}_{\text{rot}}={\overrightarrow{F}}_{\text{in}}+2m({\overrightarrow{v}}_{\text{rot}}\times \overrightarrow{\omega })+m(\overrightarrow{\omega }\times \overrightarrow{r})\times \overrightarrow{\omega }$
As discussed first two terms i.e ${\overrightarrow{F}}_{in}$ and $2m({\overrightarrow{v}}_{rot}\times \overrightarrow{\omega })$ will be in the direction perpendicular to that of ${\overrightarrow{F}}_{rot}$, therefore neglecting those we are left with
$mr{\omega }^2=ma=m\frac{d^{2}r}{d{t}^2}$
$\frac{d^{2}r}{d{t}^2}={\omega }^{2}r$
Now substituting all the given options in the differential equation it was found that only option C will satisfy the equation and the initial condition i.e $r=R/2$ at $t=0$
Hence option C is the correct answer.