Paragraph: 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 $\vec{F}_{\text {rot }}$ 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 inertial frame of reference is $\vec{F}_{\mathrm{rot}}=\vec{F}_{\mathrm{in}}+2 m\left(\vec{v}_{\mathrm{rot}} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega}$, where $\vec{v}_{\text {rot }}$ 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 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 $(\vec{\omega}=\omega \hat{k}) .$ 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.

Question: The net reaction of the disc on the block is

• A

  $-m{\omega }^{2}R\cos \omega t\widehat{j}-mg\widehat{k}$
• B

  $m{\omega }^{2}R\sin \omega t \widehat{j}-mg\widehat{k}$
• Correct answer
  C

  $\frac{1}{2}m{\omega }^{2}R\left(e^{\omega t}-e^{-\omega t}\right) \widehat{j}+mg\widehat{k}$
• D

  $\frac{1}{2}m{\omega }^{2}R \left(e^{2\omega t}-e^{-2\omega t}\right)\widehat{j}+mg\widehat{k}$

Solution

${\overrightarrow{N}}_1=mg\widehat{k}$
${\widehat{N}}_2=2m\left(V_{rot}^{\prime }\times \overrightarrow{\omega }\right) \widehat{j}$
$=2m\left[\frac{\omega R}{4}\left(e^{\omega t}-e^{-\omega t}\right)\right]\omega \widehat{j}$
$=\frac{1}{2}m{\omega }^{2}R\left(e^{\omega t}-e^{-\omega t}\right)\widehat{j}$
Total reaction on block $={\overrightarrow{N}}_1+{\overrightarrow{N}}_2$
$=\frac{1}{2}m{\omega }^{2}R\left(e^{\omega t}-e^{-\omega t}\right) \widehat{j}+mg\widehat{k}$

Hence, the correct answer is option: 'C'.