A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal 𝑥𝑦-plane. The coin is tossed vertically up (+𝑧 direction) by applying an impulse of √ 𝜋 2 × 10−2 N-s at a distance 2/3 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 ____. [Given: The acceleration due to gravity 𝑔 = 10 m s −2 ]

To solve this problem, we need to analyze both the translational and rotational motion of the coin. Here's a step-by-step breakdown:

1. Given Data:

• Mass of the coin, $m=5\text{gm}=5\times 10^{-3}\text{kg}$

• Radius of the coin, $r=\frac{4}{3}\text{cm}=\frac{4}{3}\times 10^{-2}\text{m}$

• Impulse applied, $J=\sqrt{\frac{\pi }{2}}\times 10^{-2}\text{N-s}$

• Distance from the center where impulse is applied, $d=\frac{2}{3}\text{cm}=\frac{2}{3}\times 10^{-2}\text{m}$

• Acceleration due to gravity, $g=10{\text{m/s}}^2$

2. Translational Motion:

The impulse $J$ gives the coin an initial velocity $v_0$ in the $+z$ direction. The relationship between impulse and momentum is:

$$J=m{v}_0⟹v_0=\frac{J}{m}=\frac{\sqrt{\frac{\pi }{2}}\times 10^{-2}}{5\times 10^{-3}}=\sqrt{\frac{\pi }{2}}\times 2\text{m/s}$$

The time $t$ it takes for the coin to go up and come back down to its initial position is given by the kinematic equation:

$$v=v_0-gt$$

At the highest point, $v=0$, so:

$$t_{\text{up}}=\frac{v_0}{g}=\frac{\sqrt{\frac{\pi }{2}}\times 2}{10}=\frac{\sqrt{\frac{\pi }{2}}\times 2}{10}=\frac{\sqrt{\frac{\pi }{2}}}{5}\text{s}$$

The total time for the round trip is:

$$t_{\text{total}}=2t_{\text{up}}=\frac{2\sqrt{\frac{\pi }{2}}}{5}\text{s}$$

3. Rotational Motion:

The impulse $J$ applied at a distance $d$ from the center imparts an angular momentum $L$ to the coin:

$$L=J\cdot d=\sqrt{\frac{\pi }{2}}\times 10^{-2}\times \frac{2}{3}\times 10^{-2}=\frac{2}{3}\sqrt{\frac{\pi }{2}}\times 10^{-4}{\text{kg m}}^2\mathrm{/}\text{s}$$

The moment of inertia $I$ of the coin about its diameter is:

$$I=\frac{1}{4}m{r}^2=\frac{1}{4}\times 5\times 10^{-3}\times {(\frac{4}{3}\times 10^{-2})}^2=\frac{1}{4}\times 5\times 10^{-3}\times \frac{16}{9}\times 10^{-4}=\frac{5\times 16}{4\times 9}\times 10^{-7}=\frac{20}{9}\times 10^{-7}{\text{kg m}}^2$$

The angular velocity $\omega$ is:

$$\omega =\frac{L}{I}=\frac{\frac{2}{3}\sqrt{\frac{\pi }{2}}\times 10^{-4}}{\frac{20}{9}\times 10^{-7}}=\frac{2}{3}\times \frac{9}{20}\times \sqrt{\frac{\pi }{2}}\times 10^3=\frac{3}{10}\sqrt{\frac{\pi }{2}}\times 10^3\text{rad/s}$$

The number of rotations $n$ the coin completes in time $t_{\text{total}}$ is:

$$n=\frac{\omega \cdot t_{\text{total}}}{2\pi }=\frac{\frac{3}{10}\sqrt{\frac{\pi }{2}}\times 10^3\times \frac{2\sqrt{\frac{\pi }{2}}}{5}}{2\pi }=\frac{3}{10}\times 10^3\times \frac{2}{5}\times \frac{\pi }{2\pi }=\frac{3}{10}\times 10^3\times \frac{2}{5}\times \frac{1}{2}=\frac{3}{10}\times 10^3\times \frac{1}{5}=\frac{3}{50}\times 10^3=60$$

4. Final Answer:

The coin completes $\boxed{{\displaystyle 60}}$ rotations by the time it returns to its initial position.