Question

PARAGRAPH 1

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.

The net reaction of the disc on the block is

Accepted Answer

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

Answer

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

Answer

(C) $-m \omega^{2} R \cos \omega t \hat{j}-m g \hat{k}$

Answer

(D) $m \omega^{2} R \sin \omega t \hat{j}-m g \hat{k}$

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Laws of Motion

Solution

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