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 $\sqrt{\frac{\pi}{2}} \times 10^{-2} \mathrm{~N}$-s at a distance $2 / 3 \mathrm{~cm}$ 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 $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Rotational Mechanics

A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal

x y

-plane. The coin is tossed vertically up (

+ z

direction) by applying an impulse of

\frac{π}{2} \times 1 0^{- 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 ms^{- 2}

]

Answer: 30

From impulse-momentum:

J = m v

, so

v = \frac{J}{m} = \frac{π /2 \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 3}} = 2 π

m/s.

Time of flight:

t = \frac{2 v}{g} = \frac{2 2 π}{10} = \frac{2 π}{5}

s.

Angular impulse:

J r = I_{c} ω

, where

I_{c} = \frac{1}{4} m R^{2}

(moment of inertia about diameter).

ω = \frac{J r}{I _{c}} = \frac{J r}{\frac{1}{4} m R ^{2}} = \frac{4 J r}{m R ^{2}}

Substituting values:

ω = \frac{4 \times π /2 \times 1 0 ^{- 2} \times ( 2/3 ) \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 3} \times ( 4/3 ) ^{2} \times 1 0 ^{- 4}} = 150 2 π

rad/s.

Number of rotations:

N = \frac{ω t}{2 π} = \frac{150 2 π \times 2 π /5}{2 π} = \frac{150 \times 2 π}{5 \times 2 π} = 30

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer