Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2

Q: Why is the distance $d$ for the parallel axis theorem equal to $2R$?

The rod has a total length of $2R$, so the distance from its center to its end is $R$. The sphere is attached at the end, and its radius is $R$. Therefore, the distance from the center of the rod to the center of the sphere is $R + R = 2R$.

Q: Why do we multiply the sphere's moment of inertia by 2?

The system consists of *two* identical spheres, one on each end of the rod. Since both are at the same distance from the central axis, they have the same moment of inertia. We multiply by 2 to account for both.

Q: Can I treat the spheres as point masses?

No, because the spheres have a physical radius $R$ that is comparable to the dimensions of the system. Treating them as point masses would ignore their intrinsic moment of inertia ($I_{CM} = \frac{2}{5}MR^2$), leading to an incorrect answer.

Q: What if the axis of rotation was at one end of the rod?

You would need to recalculate the distances for the parallel axis theorem. The rod's inertia would be $\frac{ML^2}{3}$. One sphere would be at distance $R$ (if its center is at the axis, $d=0$), and the other at distance $3R$.

Q: Why is the rod's moment of inertia $\frac{1}{3}MR^2$ and not $\frac{1}{12}MR^2$?

The formula is indeed $\frac{ML^2}{12}$. However, the length of the rod $L$ is given as $2R$. When you substitute $L = 2R$ into the formula, you get $\frac{M(2R)^2}{12} = \frac{4MR^2}{12}$, which simplifies to $\frac{1}{3}MR^2$.

Question

Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Options:

(A) $\frac{137}{15} MR^2$

(B) $\frac{209}{15} MR^2$

(C) $\frac{17}{15} MR^2$

(D) $\frac{152}{15} MR^2$

Sigma Percentile · Accepted Answer

The journey to solving this problem is a beautiful exercise in breaking down a complex system into its fundamental building blocks. Imagine you are an engineer tasked with calculating the rotational inertia of a newly designed dumbbell. The system might look intimidating at first glance, but the principle of superposition is our greatest ally here.

### Deconstructing the System

Our system consists of three distinct components: a central uniform rod and two identical solid spheres attached to its ends. The total moment of inertia about the central axis is simply the sum of the moments of inertia of these individual parts.

$$I_{	ext{total}} = I_{	ext{rod}} + 2 	imes I_{	ext{sphere}}$$

By isolating each component, we can tackle the physics step-by-step without getting overwhelmed.

### The Rod's Contribution

Let's start with the easiest part: the central rod. We know from standard derivations that the moment of inertia of a uniform rod of mass $M$ and length $L$, rotating about an axis through its center and perpendicular to its length, is given by:

$$I = \frac{ML^2}{12}$$

In our specific problem, the length of the rod is given as $2R$. This is where many students make a silly mistake by blindly plugging in $R$ instead of the full length. Substituting $L = 2R$, we get:

$$I_{	ext{rod}} = \frac{M(2R)^2}{12} = \frac{4MR^2}{12} = \frac{1}{3}MR^2$$

That's one piece of the puzzle solved!

### The Spheres and the Parallel Axis Theorem

Now, let's turn our attention to the spheres. If a sphere were rotating about an axis passing directly through its own center of mass, its moment of inertia would be:

$$I_{	ext{CM}} = \frac{2}{5}MR^2$$

However, our spheres are revolving around a distant central axis. This is the perfect scenario to deploy the **Parallel Axis Theorem**, which states:

$$I = I_{	ext{CM}} + Md^2$$

Here, $d$ is the perpendicular distance from the axis of rotation to the center of mass of the sphere. Let's calculate $d$. The rod extends a distance of $R$ from the center to its end. The sphere is attached at this end, and its center is a further distance $R$ away. Therefore, the total distance $d$ is:

$$d = R + R = 2R$$

Now, we substitute this into our theorem:

$$I_{	ext{sphere}} = \frac{2}{5}MR^2 + M(2R)^2$$

$$I_{	ext{sphere}} = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2$$

### Bringing It All Together

We have all the pieces; now it's time to assemble the final equation. Since there are two identical spheres, we must multiply the sphere's inertia by two.

$$I_{	ext{total}} = \frac{1}{3}MR^2 + 2 	imes \left(\frac{22}{5}MR^2ight)$$

$$I_{	ext{total}} = \frac{1}{3}MR^2 + \frac{44}{5}MR^2$$

To add these fractions, we find a common denominator, which is $15$:

$$I_{	ext{total}} = \left(\frac{5 + 132}{15}ight) MR^2 = \frac{137}{15}MR^2$$

And there we have it! By systematically applying fundamental theorems, we've unraveled the mechanics of the system. Always remember, complex physics problems are just a series of simple steps waiting to be executed with precision.

Animated Solution for Physics - Rotational Motion: Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2 R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Select Answer:

Visualized Solution

$Analyzing the System$

$Moment of Inertia of the Rod$

$Parallel Axis Theorem for Sphere$

$Calculating I_{sphere}$

$Total Moment of Inertia$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Moment of Inertia

Deconstructing the System

The Rod's Contribution

The Spheres and the Parallel Axis Theorem

Bringing It All Together

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