A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed at a point A o

Q: Why do we use the moment of inertia about the diameter and not the center of the disc?

The axis of rotation $PQ$ is a horizontal chord, which is parallel to a diameter of the disc, not an axis passing through the center and perpendicular to the plane. Therefore, we must use the moment of inertia about the diameter ($mR^2/4$) and apply the parallel axis theorem.

Q: How did we determine the vertical drop of the particle A?

Initially, $A$ is at the highest point, a distance of $R + R/4 = 5R/4$ above the axis $PQ$. When it reaches the lowest point, it has rotated 180 degrees and is now $5R/4$ below the axis. The total vertical drop is the sum of these distances, which is $5R/2$.

Q: Is the linear speed of the center of the disc the same as the particle?

No. Linear speed is given by $v = \omega r$. Since the entire system rotates with the same angular velocity $\omega$, the linear speed depends on the distance $r$ from the axis. The particle is much farther from the axis than the center of the disc, so it moves much faster.

Q: Why is the potential energy of the disc calculated using its center of mass?

For any rigid body, the change in gravitational potential energy can be calculated by treating the entire mass of the body as if it were concentrated at its center of mass.

Q: What if the disc was released from a horizontal position instead?

If released from a horizontal position, the initial heights of the center of mass and the particle above the lowest point would be different. You would need to recalculate the vertical drop for both to find the new loss in potential energy, but the moment of inertia would remain exactly the same.

Question

A uniform circular disc has radius $R$ and mass $m$. A particle, also of mass $m$, is fixed at a point $A$ on the edge of the disc as shown in the figure. The disc can rotate freely about a horizontal chord $PQ$ that is at a distance $R/4$ from the centre $C$ of the disc. The line $AC$ is perpendicular to $PQ$. Initially the disc is held vertical with the point $A$ at its highest position. It is then allowed to fall, so that it starts rotation about $PQ$. Find the linear speed of the particle as it reaches its lowest position.

Sigma Percentile · Accepted Answer

### The Falling Disc: A Symphony of Energy and Inertia

Imagine a beautifully balanced physical system: a uniform circular disc held vertically, but instead of being pivoted at its center, it is hinged along an off-center horizontal chord $PQ$. To make things even more interesting, a small particle of the exact same mass is attached right at the top edge of the disc. When this system is released, gravity takes over, and a fascinating rotational dance begins. Our goal is to find the linear speed of that top particle the moment it swings down to its lowest possible position.

### Analyzing the Setup

Before we dive into equations, let's map out the geometry. The disc has a radius $R$ and mass $m$. The axis of rotation, chord $PQ$, is located at a distance of $R/4$ from the center $C$ of the disc.

The particle $A$ is fixed at the top edge, meaning its distance from the center $C$ is $R$. Since the line $AC$ is perpendicular to $PQ$, the total distance of the particle $A$ from the axis of rotation $PQ$ is simply the sum of these distances: $R + R/4 = 5R/4$.

### The Master Equation

As the system falls, it rotates about the fixed axis $PQ$. Because there are no non-conservative forces (like friction or air resistance) doing work, the total mechanical energy of the system is strictly conserved.

This means the gravitational potential energy lost by the system as it drops will be entirely converted into rotational kinetic energy. Mathematically, we write this as:

$$\Delta U_{	ext{loss}} = \Delta K_{	ext{gain}}$$

### Calculating the Potential Energy Drop

To find the total loss in potential energy, we must track the vertical displacement of both the disc's center of mass and the particle.

Initially, the particle $A$ is at its highest point, a distance of $5R/4$ above the axis $PQ$. When it swings to its lowest point, it will be exactly $5R/4$ below the axis. The total vertical drop for the particle is therefore $2 	imes (5R/4) = 5R/2$. The loss in potential energy for the particle is:

$$\Delta U_A = mg \left( \frac{5R}{2} ight)$$

Similarly, the center of the disc $C$ starts at a distance of $R/4$ above the axis and ends up $R/4$ below it. Its total vertical drop is $2 	imes (R/4) = R/2$. The loss in potential energy for the disc is:

$$\Delta U_C = mg \left( \frac{R}{2} ight)$$

Adding these together, the total potential energy lost by the entire system is:

$$\Delta U_{	ext{total}} = \frac{5mgR}{2} + \frac{mgR}{2} = 3mgR$$

### Unlocking the Moment of Inertia

To find the rotational kinetic energy, we need the total moment of inertia of the system about the axis $PQ$. We will use the parallel axis theorem for both components.

For the disc, the moment of inertia about its own diameter is $mR^2/4$. Shifting this to the parallel axis $PQ$ (a distance of $R/4$ away), we get:

$$I_{	ext{disc}} = \frac{mR^2}{4} + m\left(\frac{R}{4}ight)^2 = \frac{mR^2}{4} + \frac{mR^2}{16} = \frac{5mR^2}{16}$$

For the particle $A$, treating it as a point mass at a distance of $5R/4$ from the axis $PQ$, its moment of inertia is:

$$I_{	ext{particle}} = m\left(\frac{5R}{4}ight)^2 = \frac{25mR^2}{16}$$

The total moment of inertia of the system is the sum of these two:

$$I_{	ext{system}} = \frac{5mR^2}{16} + \frac{25mR^2}{16} = \frac{30mR^2}{16} = \frac{15mR^2}{8}$$

### Final Calculation

Now, we bring it all together. Equating the total potential energy lost to the rotational kinetic energy gained:

$$3mgR = \frac{1}{2} I_{	ext{system}} \omega^2$$

Substituting our calculated moment of inertia:

$$3mgR = \frac{1}{2} \left( \frac{15mR^2}{8} ight) \omega^2$$

Solving for the square of the angular velocity, $\omega^2$:

$$\omega^2 = \frac{3mgR 	imes 16}{15mR^2} = \frac{16g}{5R}$$

Taking the square root gives us the angular velocity $\omega$ at the lowest point:

$$\omega = \sqrt{\frac{16g}{5R}}$$

Finally, the question asks for the linear speed $v$ of the particle $A$. Since the particle is rotating in a circle of radius $r_A = 5R/4$, its linear speed is simply $v = \omega r_A$:

$$v = \sqrt{\frac{16g}{5R}} 	imes \left(\frac{5R}{4}ight) = \sqrt{\frac{16g 	imes 25R^2}{5R 	imes 16}} = \sqrt{5gR}$$

And there we have it! A beautiful interplay of geometry, inertia, and energy conservation leading to a remarkably clean final result.

Visualized Solution

$Visualizing the Setup$

$Energy Conservation Principle$

$Potential Energy Loss: Particle$

$Potential Energy Loss: Disc$

$Total Potential Energy Loss$

$Total Moment of Inertia$

$Moment of Inertia: Disc$

$Moment of Inertia: Particle$

$Summing the Inertia$

$Equating Energies$

$Solving for Angular Velocity$

$Final Linear Speed$

$The Way Forward$

The Sigma Insight: Work and Energy in Rotational Motion

The Falling Disc

Analyzing the Setup

The Master Equation

Calculating the Potential Energy Drop

Unlocking the Moment of Inertia

Final Calculation

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Energy Loss: Particle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Energy Loss: Disc

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

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

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

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

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

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

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

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

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