One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment o

Q: Why does the hollow sphere have a larger moment of inertia if they have the exact same mass?

Moment of inertia depends not just on the total mass, but on how that mass is distributed relative to the axis of rotation. In a hollow sphere, all the mass is concentrated at the outer radius, furthest from the axis, making it harder to rotate.

Q: What if the hollow sphere had a smaller mass?

If the masses were different, you would need to plug the specific values into the formulas $I = \frac{2}{5}M_1R^2$ and $I = \frac{2}{3}M_2R^2$ to compare them. You couldn't just rely on the fractions.

Q: Does the axis of rotation matter here?

Yes, absolutely! The formulas $\frac{2}{5}MR^2$ and $\frac{2}{3}MR^2$ are specifically for an axis passing through the center (the diameter). If the axis were a tangent to the sphere, the values would be different (calculated using the parallel axis theorem).

Q: Are these formulas derived using integration?

Yes, they are derived by integrating the moment of inertia of infinitesimally small elements (like disks or rings) over the entire volume or surface of the sphere.

Q: Can I just memorize that 'hollow objects have higher inertia'?

As a general rule of thumb for objects of the same mass and outer dimensions, yes! Because hollow objects have their mass pushed further out from the center, their moment of inertia is generally higher than their solid counterparts.

Question

One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively $I_A$ and $I_B$ such that

Options:

(A) $I_A = I_B$

(B) $I_A > I_B$

(C) $I_A < I_B$

(D) $\frac{I_A}{d_A} = \frac{I_B}{d_B}$ where $d_A$ and $d_B$ are their densities.

Sigma Percentile · Accepted Answer

### The Battle of the Spheres: Solid vs. Hollow

Imagine you are holding two spheres of the exact same size and weight. One is a solid bowling ball, and the other is a hollow metallic shell. If you were to spin both of them like a top, which one would be harder to start spinning? This simple thought experiment lies at the heart of understanding **Moment of Inertia**.

### The Mathematical Breakdown

To solve this, we need to look at the formulas for the moment of inertia of these two shapes about their central diameter.

For the **Solid Sphere (A)**, the mass is distributed uniformly throughout its entire volume. The formula for its moment of inertia is:

$$I_A = \frac{2}{5}MR^2$$

For the **Hollow Sphere (B)**, all of its mass is concentrated entirely on its outer shell. The formula for its moment of inertia is:

$$I_B = \frac{2}{3}MR^2$$

### Comparing the Inertia

Since both spheres have the exact same mass $M$ and the same outer radius $R$, comparing their moments of inertia simply comes down to comparing the numerical fractions in front of $MR^2$.

Let's convert them to decimals for a clearer picture:
- For the solid sphere: $\frac{2}{5} = 0.4$
- For the hollow sphere: $\frac{2}{3} \approx 0.67$

It is mathematically obvious that $0.4 < 0.67$. Therefore, we can confidently conclude that:

$$I_A < I_B$$

### The Physical Intuition

Why does the hollow sphere have a larger moment of inertia? The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends not just on the total mass, but crucially on **how that mass is distributed relative to the axis of rotation**.

In the solid sphere, a significant portion of the mass is located very close to the central axis. Mass close to the axis contributes very little to the moment of inertia.

However, in the hollow sphere, *all* of the mass is pushed as far away from the axis as possible (at distance $R$). Because the moment of inertia scales with the square of the distance from the axis ($r^2$), having all the mass at the maximum distance makes the hollow sphere much harder to spin.

Always remember this golden rule: **For objects of the same mass and outer dimensions, the one with its mass distributed further from the center will always have a higher moment of inertia.**

Animated Solution for Physics - Rotational Motion: One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively $I_{A}$ and $I_{B}$ such that

Select Answer:

Visualized Solution

Visualizing the Spheres

Moment of Inertia of Solid Sphere

Moment of Inertia of Hollow Sphere

Comparing the Fractions

Final Conclusion

Physical Intuition

The Sigma Insight: Moment of Inertia

The Battle of the Spheres

The Mathematical Breakdown

Comparing the Inertia

The Physical Intuition

Similar Questions

Let the moment of inertia of a hollow cylinder of length $30 cm$ (inner radius $10 cm$ and outer radius $20 cm$ ) about its axis be $I$ . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$ , is

Consider two uniform discs of same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ , then the value of $α$ is

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

Mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

From a solid sphere of mass $M$ and radius $R$ , a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is