A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xy-plane. The coin

Q: Why is the moment of inertia $\frac{1}{4}mR^2$ and not $\frac{1}{2}mR^2$?

The coin is spinning about its diameter, not its central perpendicular axis. By the perpendicular axis theorem, $I_z = I_x + I_y$. Since $I_x = I_y$ for a uniform circular coin, $I_x = I_z / 2 = (\frac{1}{2}mR^2) / 2 = \frac{1}{4}mR^2$.

Q: Does gravity affect the angular velocity of the coin?

No. Gravity acts exactly at the center of mass of the coin. Since the torque due to gravity about the center of mass is zero, the angular velocity $\omega$ remains perfectly constant throughout the flight.

Q: Why do we use $T = 2v/g$ for the time of flight?

The coin is tossed vertically upwards. It decelerates at $g$ until it reaches its peak (where $v=0$), taking time $v/g$. It then takes the same amount of time to fall back down, making the total time $2v/g$.

Q: What if the impulse was applied exactly at the center?

If the impulse was applied at the center ($r=0$), the angular impulse would be zero. The coin would simply translate upwards without spinning at all!

Q: Why did we convert the mass and radius to standard SI units (kg and m)?

The impulse is given in Newton-seconds (N-s) and gravity in $m/s^2$, which are standard SI units. To ensure our equations are dimensionally consistent and our final answer is correct, all quantities must be in SI units.

Question

A thin circular coin of mass $5	ext{ gm}$ and radius $4/3	ext{ cm}$ is initially in a horizontal xy-plane. The coin is tossed vertically up ($+z$ direction) by applying an impulse of $\sqrt{\frac{\pi}{2}} 	imes 10^{-2}	ext{ N-s}$ at a distance $2/3	ext{ cm}$ 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 $g = 10	ext{ ms}^{-2}$]

Sigma Percentile · Accepted Answer

The problem of a tossed coin is a classic in physics, but when you add an off-center impulse, it transforms into a beautiful dance of linear and rotational dynamics. Imagine you are holding a coin, and instead of flipping it from the center, you flick it near the edge. It shoots upwards while spinning rapidly. Our goal is to find out exactly how many times it spins before it lands back in your hand.

### The Anatomy of the Toss

Let's break down the initial setup. We have a thin circular coin with a mass of $m = 5 	imes 10^{-3} 	ext{ kg}$ and a radius of $R = \frac{4}{3} 	imes 10^{-2} 	ext{ m}$. It's resting peacefully on the xy-plane.

Suddenly, a sharp vertical impulse $J = \sqrt{\frac{\pi}{2}} 	imes 10^{-2} 	ext{ N-s}$ is applied. But here is the catch: it's not applied at the center. It strikes at a distance $r = \frac{2}{3} 	imes 10^{-2} 	ext{ m}$ from the center.

This off-center hit is the heart of the problem. It does two distinct things simultaneously: it pushes the center of mass upwards, and it twists the coin, causing it to spin. In physics, we can treat these two motions—translation and rotation—completely independently.

### The Dual Nature of Impulse

First, let's look at the linear motion. The upward impulse $J$ gives the coin an initial upward velocity $v$. According to the linear impulse-momentum theorem, the impulse equals the change in linear momentum:
$$J = m v$$

We can easily find the initial velocity by dividing the impulse by the mass:
$$v = \frac{J}{m} = \frac{\sqrt{\frac{\pi}{2}} 	imes 10^{-2}}{5 	imes 10^{-3}}$$

Notice how the powers of ten simplify beautifully. We get:
$$v = \frac{\sqrt{\pi/2} 	imes 10}{5} = 2\sqrt{\frac{\pi}{2}} = \sqrt{2\pi} 	ext{ m/s}$$

Now, let's tackle the spin. The impulse creates a torque about the center of mass, which imparts an angular momentum. The angular impulse is simply the linear impulse multiplied by the lever arm distance $r$. This must equal the change in angular momentum, which is the moment of inertia $I_c$ times the angular velocity $\omega$:
$$J \cdot r = I_c \omega$$

Since the coin is spinning about its diameter, we use the perpendicular axis theorem to find its moment of inertia. For a uniform circular disc, the moment of inertia about its diameter is:
$$I_c = \frac{1}{4} m R^2$$

Let's substitute this into our angular impulse equation to find $\omega$:
$$\omega = \frac{J \cdot r}{\frac{1}{4} m R^2}$$

Here is where the magic happens. We know that $J = m v$. Also, if you look closely at the given values, the distance $r$ ($2/3$ cm) is exactly half of the radius $R$ ($4/3$ cm). So, $r = R/2$. Substituting these makes our calculation incredibly elegant:
$$\omega = \frac{(m v) \cdot (R/2)}{\frac{1}{4} m R^2} = \frac{2v}{R}$$

The mass cancels out entirely! Plugging in our value of $v$ and the radius $R$, we get:
$$\omega = \frac{2 \sqrt{2\pi}}{\frac{4}{3} 	imes 10^{-2}} = 150\sqrt{2\pi} 	ext{ rad/s}$$

The coin is spinning at a blistering rate!

### The Flight of the Coin

Once the coin leaves the finger, it is in free fall. Gravity pulls it down, slowing its ascent until it stops momentarily at its peak, and then it falls back down.

The total time of flight $T$ for an object thrown vertically upwards with velocity $v$ is given by basic kinematics:
$$T = \frac{2v}{g}$$

Substituting our velocity and the acceleration due to gravity ($g = 10 	ext{ m/s}^2$):
$$T = \frac{2\sqrt{2\pi}}{10} = \frac{\sqrt{2\pi}}{5} 	ext{ s}$$

This is the exact duration the coin spends in the air, performing its aerial acrobatics.

### The Final Spin Count

While the coin is flying up and down, it is also spinning at a constant angular velocity $\omega$. Why constant? Because gravity acts exactly at the center of mass, creating zero torque.

The total angle $	heta$ it rotates through is simply the angular velocity multiplied by the time of flight:
$$	heta = \omega T$$

Let's multiply our two expressions:
$$	heta = \left(150\sqrt{2\pi}ight) \left(\frac{\sqrt{2\pi}}{5}ight)$$

The square roots of $2\pi$ multiply together to give exactly $2\pi$. The numbers simplify perfectly:
$$	heta = 30 	imes 2\pi = 60\pi 	ext{ rad}$$

We want to find the number of full rotations $n$. Since one full rotation is $2\pi$ radians, we divide the total angle by $2\pi$:
$$n = \frac{	heta}{2\pi} = \frac{60\pi}{2\pi} = 30$$

The coin completes exactly 30 rotations before returning to its starting position. A beautiful, clean integer answer born from a complex interplay of linear and rotational dynamics!

Enter Numerical Value:

Visualized Solution

$I ni t ia l S e t u p$

$L in e a r I m p u l se - M o m e n t u m$

$C a l c u l a t in g L in e a r V e l oc i t y$

$L in e a r V e l oc i t y R es u l t$

$A n g u l a r I m p u l se - M o m e n t u m$

$C a l c u l a t in g A n g u l a r V e l oc i t y$

$A n g u l a r V e l oc i t y R es u l t$

$T im e o f F l i g h t$

$T o t a l A n g l e R o t a t e d$

$N u mb er o f R o t a t i o n s$

$T h e W a y F or w a r d$

The Sigma Insight: Dynamics of Rigid Body Rotation

The Anatomy of the Toss

The Dual Nature of Impulse

The Flight of the Coin

The Final Spin Count

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Linear Impulse−Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Linear Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Linear Velocity Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Impulse−Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Angular Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Velocity Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Flight

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Angle Rotated

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Rotations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward