QJEEAdv23Ph80af: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph80af Solution: A thin circular coin of mass and radius is initially in a…

A coin with a radius of $r = \frac{4}{3} \times 10^{-2} \, \text{m}$ and a mass of $m = 5 \times 10^{-3} \, \text{kg}$ is thrown upwards. An impulse of $J_{\text{lin}} = \sqrt{\frac{\pi}{2}} \times 10^{-2} \, \text{N s}$ is applied perpendicularly to the plane of the coin. This impulse is delivered at a distance of $d = \frac{r}{2}$ from the center of the coin.

Initially at rest, the coin experiences a change in linear momentum due to the applied linear impulse, leading to the equation :

$m v_{\text{cm}} = J_{\text{lin}}$ .

where $m$ is the mass of the coin, $v_{\text{cm}}$ is its center of mass velocity after the impulse, and $J_{\text{lin}}$ is the magnitude of the linear impulse.

After the impulse, the center of mass of the coin ascends with an initial velocity $v_{\text{cm}}$ . Under the influence of gravity, the coin returns to its original position in a time duration expressed by :

$t = \frac{2 v_{\text{cm}}}{g} = \frac{2 J_{\text{lin}}}{mg}$ ,

where $g$ is the acceleration due to gravity, $J_{\text{lin}}$ is the linear impulse applied, and $m$ is the mass of the coin.

The coin undergoes rotation about its axis, and the angular impulse exerted on it around its center is described by the equation :

$J_{\text{ang}} = d J_{\text{lin}} = \left( \frac{r}{2} \right) J_{\text{lin}}$ ,

where $J_{\text{ang}}$ is the angular impulse, $d$ is the distance from the center at which the linear impulse $J_{\text{lin}}$ is applied, and $r$ is the radius of the coin.

The angular impulse leads to a change in the angular momentum of the coin about a horizontal axis passing through its center. This relationship is given by :

$I_{\text{cm}} \omega = L_{\text{ang}} = \left( \frac{r}{2} \right) J_{\text{lin}}$ ,

where $I_{\text{cm}}$ is the moment of inertia of the coin about its center of mass, $\omega$ is the angular velocity after the impulse, and $L_{\text{ang}}$ is the angular momentum resulting from the angular impulse $J_{\text{ang}}$ .

The angular speed of the coin, denoted as $\omega$ , is determined by the equation :

$\omega = \frac{(r / 2) J_{\text{lin}}}{I_{\text{cm}}} = \frac{(r / 2) J_{\text{lin}}}{(1 / 4) m r^2} = \frac{2 J_{\text{lin}}}{m r}$ ,

where $r$ is the radius of the coin, $J_{\text{lin}}$ is the linear impulse applied, $m$ is the mass of the coin, and $I_{\text{cm}}$ is its moment of inertia about the center of mass.

After tossing, the angular speed of the coin remains constant because there is no external torque. The number of rotations $n$ completed by the coin during the time $t$ is calculated as :

$n = \frac{\theta}{2\pi} = \frac{\omega t}{2\pi} = \frac{2 J_{\text{lin}}^2}{\pi m^2 g r}$

Substituting the given values, it becomes :

$n = \frac{2 \left( \frac{\pi}{2} \times 10^{-4} \right)}{\pi \left( 5 \times 10^{-3} \right)^2 \times 10 \times \left( \frac{4}{3} \times 10^{-2} \right)} = 30$ .

where $\theta$ is the total angular displacement, $\omega$ is the angular speed of the coin, $J_{\text{lin}}$ is the linear impulse, $m$ is the mass of the coin, $g$ is the acceleration due to gravity, and $r$ is the radius of the coin.