A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed vertical axis

Q: Why does the disc have two types of angular momentum?

Because it is both spinning around its own axis and its center is revolving around point O. Just like the Earth spins on its axis (giving us days) and orbits the Sun (giving us years), the disc has both spin and orbital angular momentum.

Q: Why is the orbital angular velocity of the disc $\Omega$ and not $4\Omega$?

The center of the disc is physically attached to the rod. Therefore, it must revolve around point O at the exact same angular velocity as the rod itself, which is $\Omega$. The $4\Omega$ is strictly its spin rate about its own center.

Q: How did we find the distance of the disc's center from O?

The total length of the rod is $a$. The problem states the disc's center is at a distance of $a/4$ from the *free end*. Therefore, its distance from the pivoted end O is $a - a/4 = 3a/4$.

Q: Can we just use $L = I_{total} \Omega_{total}$ for the whole system?

No, because the different parts of the system are rotating with different angular velocities. The rod rotates at $\Omega$, while the disc spins at $4\Omega$ while orbiting at $\Omega$. You must calculate their angular momenta separately and add them.

Q: What if the disc was fixed and couldn't spin?

If the disc was rigidly fixed to the rod, its spin angular velocity would also be $\Omega$ (it would just rotate as a single rigid body with the rod). In that case, its total angular momentum would simply be its moment of inertia about O multiplied by $\Omega$.

Question

A thin rod of mass $M$ and length $a$ is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass $M$ and of radius $a/4$ is pivoted on this rod with its center at a distance $a/4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point O is $\left(\frac{Ma^2\Omega}{48}\right)n$. The value of n is______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a merry-go-round mounted on top of another merry-go-round! This is exactly what our problem presents. We have a uniform rod rotating about a fixed point O, and sitting on this rod is a disc that is also spinning about its own central axis.

Our ultimate goal is to find the **total angular momentum** of this entire system about point O.

Since angular momentum is an additive property, we can break this complex system down into two simpler parts. The total angular momentum will simply be the sum of the angular momentum of the rod and the angular momentum of the disc.

$$L_{total} = L_{rod} + L_{disc}$$

### The Rod's Contribution

Let's tackle the rod first, as its motion is straightforward. It is a uniform rod of mass $M$ and length $a$, rotating about one of its ends with an angular velocity $\Omega$.

We know from standard rotational mechanics that the moment of inertia of a rod about its end is given by $\frac{Ma^2}{3}$.

Therefore, its angular momentum is simply the product of its moment of inertia and its angular velocity.

$$L_{rod} = I_{rod} \Omega = \left(\frac{Ma^2}{3}\right)\Omega$$

### The Disc's Dual Life

Now comes the tricky part: the disc. The disc is living a dual life. It is spinning about its own center, but simultaneously, its center is revolving around point O because it is attached to the rotating rod.

To find its total angular momentum about O, we must employ the **parallel axis theorem for angular momentum**. This theorem states that the total angular momentum is the sum of its **spin angular momentum** (about its own center of mass) and its **orbital angular momentum** (the angular momentum of its center of mass revolving around O).

$$L_{disc} = L_{spin} + L_{orbital}$$

Let's calculate the spin angular momentum first. The disc has a mass $M$ and a radius of $\frac{a}{4}$. It spins with an angular velocity of $4\Omega$. The moment of inertia of a disc about its central axis is $\frac{MR^2}{2}$.

Substituting the given values, we get:

$$L_{spin} = \frac{1}{2} M \left(\frac{a}{4}\right)^2 (4\Omega) = \frac{1}{2} M \left(\frac{a^2}{16}\right) (4\Omega) = \frac{Ma^2\Omega}{8}$$

### The Orbital Journey

Next, we need the orbital angular momentum. For this, we must determine the exact distance from point O to the center of the disc.

The total length of the rod is $a$, and the problem states that the disc's center is located at a distance of $\frac{a}{4}$ from the free end. Therefore, its distance from the pivoted end O is:

$$r_{cm} = a - \frac{a}{4} = \frac{3a}{4}$$

The center of the disc is physically constrained to revolve around O with the rod's angular velocity, which is $\Omega$. So, its orbital angular momentum is simply $M r_{cm}^2 \Omega$.

$$L_{orbital} = M \left(\frac{3a}{4}\right)^2 \Omega = \frac{9Ma^2\Omega}{16}$$

Adding the spin and orbital components together gives us the total angular momentum of the disc about point O.

$$L_{disc} = \frac{Ma^2\Omega}{8} + \frac{9Ma^2\Omega}{16} = \frac{2Ma^2\Omega + 9Ma^2\Omega}{16} = \frac{11Ma^2\Omega}{16}$$

### The Grand Finale

Finally, let's bring it all together! We add the angular momentum of the rod and the disc to find the total angular momentum of the system.

$$L_{total} = \frac{Ma^2\Omega}{3} + \frac{11Ma^2\Omega}{16}$$

To add these fractions, we find a common denominator, which is $48$.

$$L_{total} = \left(\frac{16 + 33}{48}\right) Ma^2\Omega = \frac{49}{48} Ma^2\Omega$$

The problem states that the total angular momentum is $\left(\frac{Ma^2\Omega}{48}\right)n$. By directly comparing our derived result with this expression, it is crystal clear that the value we are looking for is $49$.

$$n = 49$$

What a beautiful and elegant application of rotational mechanics!

Enter Numerical Value:

Visualized Solution

$S y s t e m$

$L_{t o t a l} = L_{r o d} + L_{d i sc}$

$L_{r o d} = \frac{M a ^{2}}{3} Ω$

$L_{d i sc} = L_{s p in} + L_{or bi t a l}$

$L_{s p in} = \frac{M a ^{2} Ω}{8}$

$r_{c m} = \frac{3 a}{4}$

$L_{or bi t a l} = \frac{9 M a ^{2} Ω}{16}$

$L_{d i sc} = \frac{11 M a ^{2} Ω}{16}$

$L_{t o t a l} = \frac{49 M a ^{2} Ω}{48}$

$n = 49$

The Sigma Insight: Torque and Angular Momentum

Analyzing the Setup

The Rod's Contribution

The Disc's Dual Life

The Orbital Journey

The Grand Finale

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ltotal​=Lrod​+Ldisc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lrod​=3Ma2​Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ldisc​=Lspin​+Lorbital​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lspin​=8Ma2Ω​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rcm​=43a​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lorbital​=169Ma2Ω​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ldisc​=1611Ma2Ω​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ltotal​=4849Ma2Ω​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=49