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 net reaction of the disc on the block is

• A. $m{\omega }^{2}Rsin\omega t\hat{j}-mg\hat{k}$
• B. $\frac{1}{2}m{\omega }^{2}R(e^{\omega t}-e^{-\omega t})\hat{j}+mg\hat{k}$
• C. $\frac{1}{2}m{\omega }^{2}R(e^{2\omega t}-e^{-2\omega t})\hat{j}+mg\hat{k}$
• D. $-m{\omega }^{2}R\cos \omega t\hat{j}-mg\hat{k}$

Answer 1

The correct option is B $\frac{1}{2}m{\omega }^{2}R(e^{\omega t}-e^{-\omega t})\hat{j}+mg\hat{k}$

${\overrightarrow{F}}_{rot}={\overrightarrow{F}}_{in}+2m(v_{rot}\hat{i}\times \omega \hat{k})+m(\omega \hat{k}\times r\hat{i})\times \omega \hat{k}$
Since ${\overrightarrow{F}}_{rot}$ will be in +ve x-direction i.e along $\hat{i}$ in the rotating frame of reference and $(\omega \hat{k}\times r\hat{i})\times \omega \hat{k}$ will also be in the same direction whereas ${\overrightarrow{F}}_{in}$ will be in direction perpendicular to x-direction since the reactions acting on the particle due to circular disc and weight of the particle will be in y- direction and z-direction respectively .
Hence,
$0={\overrightarrow{F}}_{in}+2m{v}_{rot}\omega (-\hat{j}){\overrightarrow{F}}_{in}=2m{v}_{rot}\omega \hat{j}r=\frac{R}{4}(e^{\omega t}+e^{-\omega t})v_r=\frac{dr}{dt}=\frac{R\omega }{4}(e^{\omega t}-e^{-\omega t}){\overrightarrow{F}}_{in}=\frac{2m{\omega }^{2}R}{4}(e^{\omega t}-e^{-\omega t})\hat{j}$
There wil be a reaction due to the weight of the particle in +ve z-direction given by $mg\hat{k}$
Hence net reaction force on the particle will be
${\overrightarrow{F}}_{reaction}=\frac{m{\omega }^{2}R}{2}(e^{\omega t}-e^{-\omega t})\hat{j}+mg\hat{k}$