A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Tw

Q: Why do we use the velocity relative to the ground and not relative to the platform?

Angular momentum must be conserved in an inertial frame of reference. The ground is an inertial frame, whereas the rotating platform is a non-inertial frame. Using the velocity relative to the ground ensures our conservation equation is valid.

Q: Why is the moment of inertia of the platform $MR^2/2$ and not $MR^2$?

The platform is a solid circular disc, not a hollow ring. For a uniform solid disc rotating about its central axis, the mass is distributed throughout its area, leading to a moment of inertia of $\frac{1}{2}MR^2$.

Q: Do we need to include the mass of the guns in the moment of inertia?

The problem explicitly states that the spring toy-guns are "massless". Therefore, they do not contribute to the moment of inertia of the platform.

Q: What if the balls were fired in the same direction instead of opposite directions?

If both balls were fired in the same direction (e.g., both "up" in the diagram), their angular momenta about the center would cancel each other out ($mvr - mvr = 0$). The platform would not rotate at all!

Q: Why is the initial angular momentum zero?

Before the guns are fired, the platform and the balls are completely at rest. Since nothing is moving or rotating, the total initial angular momentum of the system is zero.

Question

A horizontal circular platform of radius $0.5	ext{ m}$ and mass $0.45	ext{ kg}$ is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass $0.05	ext{ kg}$ are attached to the platform at a distance $0.25	ext{ m}$ from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of $9	ext{ ms}^{-1}$ with respect to the ground. The rotational speed of the platform in $	ext{rad s}^{-1}$ after the balls leave the platform is

Sigma Percentile · Accepted Answer

Imagine you are standing in a physics lab, looking down at a perfectly smooth, horizontal circular platform. It's free to rotate about its central axis, like a merry-go-round. On this platform, we have a fascinating setup: two toy guns, each loaded with a steel ball, positioned symmetrically along a diameter.

This is a classic scenario that tests our understanding of one of the most profound principles in physics: the conservation of angular momentum. But before we dive into the math, let's really visualize what's happening.

### Analyzing the Setup

We have a platform of mass $M = 0.45 	ext{ kg}$ and radius $R = 0.5 	ext{ m}$. It's initially at rest. Mounted on it are two steel balls, each of mass $m = 0.05 	ext{ kg}$, located at a distance $r = 0.25 	ext{ m}$ from the center.

Suddenly, the guns fire! The balls are shot horizontally and perpendicular to the diameter, in opposite directions. They leave the platform with a speed $v = 9 	ext{ m/s}$ relative to the ground.

Why is this detail—"relative to the ground"—so crucial? Because angular momentum must be calculated from a single, inertial frame of reference. By giving us the velocity relative to the ground, the problem saves us from the headache of dealing with relative velocities and non-inertial frames.

When the balls are fired, they carry away angular momentum. But wait, the system as a whole (platform + balls) experiences no external torque in the horizontal plane. The forces of the springs are purely internal. What does this mean? It means the total angular momentum of the system must remain exactly as it was before the guns fired: zero!

### The Master Equation

To keep the total angular momentum at zero, the platform must start rotating in the opposite direction to the balls. Let's set up our master equation.

The initial angular momentum is:

$L_i = 0$

The final angular momentum consists of two parts: the angular momentum of the balls, and the angular momentum of the platform.

$L_f = L_{	ext{balls}} + L_{	ext{platform}}$

Let's calculate the angular momentum of the balls. Each ball has mass $m$, velocity $v$, and is fired at a perpendicular distance $r$ from the center. The angular momentum of a single ball is $mvr$. Since both balls are fired in a way that causes rotation in the same sense (say, counter-clockwise), their angular momenta add up:

$L_{	ext{balls}} = mvr + mvr = 2mvr$

Now, what about the platform? It's a solid uniform disc, so its moment of inertia $I$ about the central axis is:

$I = \frac{MR^2}{2}$

If the platform rotates with an angular velocity $\omega$, its angular momentum is $I\omega$. Since it must rotate in the opposite direction to cancel the balls' angular momentum, we can write:

$L_{	ext{platform}} = -I\omega = -\frac{MR^2}{2}\omega$

Equating the total final angular momentum to the initial angular momentum (zero), we get:

$0 = 2mvr - \frac{MR^2}{2}\omega$

This is our master equation! It beautifully captures the physics of the situation. The angular momentum "stolen" by the balls is perfectly balanced by the angular momentum "gained" by the platform.

### Final Calculation

Now, let's rearrange our master equation to solve for the angular velocity $\omega$, which is what we're looking for:

$\omega = \frac{4mvr}{MR^2}$

This expression tells us a lot. The rotational speed is directly proportional to the mass and velocity of the balls, and inversely proportional to the mass and the square of the radius of the platform. It makes perfect intuitive sense!

Let's plug in the numbers provided in the problem:
- $m = 0.05 	ext{ kg}$
- $v = 9 	ext{ m/s}$
- $r = 0.25 	ext{ m}$
- $M = 0.45 	ext{ kg}$
- $R = 0.5 	ext{ m}$

Substituting these into our equation:

$\omega = \frac{4(0.05)(9)(0.25)}{0.45(0.5)^2}$

Let's simplify the numerator first. $4 	imes 0.25$ is exactly $1$. So we are left with $0.05 	imes 9$, which is $0.45$.

Numerator $= 0.45$

Now for the denominator. $0.5$ squared is $0.25$. So the denominator is $0.45 	imes 0.25$.

Denominator $= 0.45 	imes 0.25$

Putting it all together:

$\omega = \frac{0.45}{0.45 	imes 0.25}$

Notice how beautifully the $0.45$ cancels out from the numerator and the denominator! This is the hallmark of a well-designed JEE problem. We are left with:

$\omega = \frac{1}{0.25}$

Since $0.25$ is $1/4$, dividing by it gives us exactly $4$.

$\omega = 4 	ext{ rad/s}$

And there we have it! The platform will rotate at a speed of $4 	ext{ rad/s}$.

This problem is a fantastic reminder of the power of conservation laws. By simply trusting that angular momentum is conserved, we bypassed any complex force or torque calculations and arrived straight at the answer. Always look for these symmetries and conservation principles—they are your best friends in physics!

Enter Numerical Value:

Visualized Solution

$Initial Setup$

$Conservation of Angular Momentum$

$Initial and Final States$

$Angular Momentum of Balls$

$Angular Momentum of Platform$

$Equating Angular Momenta$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup

The Master Equation

Final Calculation

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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; } Conservation of Angular Momentum

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

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

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

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

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

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

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