A thin circular coin of mass 5 gm and radius $\frac{4}{3}\text{ cm}$ is initially in a horizontal xy-plane. The coin is tossed vertically up (+z-direction) by applying an impulse $\sqrt{\frac{\pi }{2}}\times {10}^{-2}\text{ N-s}$ at a distance $\frac{2}{3}\text{ cm}$ from its center. The coin spins about its diameter and moves along the +z-direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is ______.

The correct answer is 30.

1. Given Parameters:

From the problem description and the attached diagram showing a circular coin lying in the xy-plane:
- Mass of the coin, $m=5\text{ g}=5\times {10}^{-3}\text{ kg}$
- Radius of the coin, $R=\frac{4}{3}\text{ cm}=\frac{4}{3}\times {10}^{-2}\text{ m}$
- Distance of impulse from the center (as indicated by the label $2/3\text{ cm}$ in the image), $r=\frac{2}{3}\text{ cm}=\frac{2}{3}\times {10}^{-2}\text{ m}$
- Linear Impulse applied along +z direction, $J=\sqrt{\frac{\pi }{2}}\times {10}^{-2}\text{ N-s}$
- Acceleration due to gravity, $g=10\text{ m/s}{}^2$ (standard value)

2. Linear Motion of the Coin:

The impulse $J$ provides initial vertical velocity $v$ to the center of mass of the coin in the +z direction:

$J=mv\Rightarrow v=\frac{J}{m}$

Substituting the given values:

$v=\frac{\sqrt{\frac{\pi }{2}}\times {10}^{-2}}{5\times {10}^{-3}}=\frac{\sqrt{\frac{\pi }{2}}}{5\times {10}^{-1}}=2\sqrt{\frac{\pi }{2}}\text{ m/s}=\sqrt{2\pi }\text{ m/s}$

The total time of flight $T$ until the coin returns to its initial horizontal position is given by:

$T=\frac{2v}{g}=\frac{2\sqrt{2\pi }}{10}=\frac{\sqrt{2\pi }}{5}\text{ s}$

3. Rotational Motion of the Coin:

The impulse creates an angular impulse about a diameter of the circular coin.
The angular impulse is given by:

$\tau \Delta t=J\times r=I\omega$

The moment of inertia of a thin circular disc/coin about its diameter is:

$I=\frac{1}{4}m{R}^2$

Therefore, the initial angular velocity $\omega$ is:

$\omega =\frac{J·r}{I}=\frac{J·r}{\frac{1}{4}m{R}^2}=\frac{4Jr}{m{R}^2}$

Substituting $v=\frac{J}{m}$:

$\omega =\frac{4vr}{R^2}$

Plugging in the values of $v$, $r$, and $R$:

$\omega =\frac{4\times \sqrt{2\pi }\times \left(\frac{2}{3}\times {10}^{-2}\right)}{{\left(\frac{4}{3}\times {10}^{-2}\right)}^2}=\frac{\frac{8}{3}\sqrt{2\pi }\times {10}^{-2}}{\frac{16}{9}\times {10}^{-4}}=\frac{3}{2}\times {10}^2\sqrt{2\pi }\text{ rad/s}=150\sqrt{2\pi }\text{ rad/s}$

4. Total Number of Rotations (n):

The total angle rotated by the coin during its time of flight $T$ is:

$\theta =\omega \times T$

Substituting $\omega =150\sqrt{2\pi }$ and $T=\frac{\sqrt{2\pi }}{5}$:

$\theta =\left(150\sqrt{2\pi }\right)\times \left(\frac{\sqrt{2\pi }}{5}\right)=30\times \left(2\pi \right)=60\pi \text{ rad}$

Since 1 complete rotation equals $2\pi$ radians, the number of rotations $n$ is:

$n=\frac{\theta }{2\pi }=\frac{60\pi }{2\pi }=30$