Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about a

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Q. Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

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Concept understanding — Rotational Inertia Comparison

Rotational Inertia Comparison: From Intuition to Precision

Imagine pushing a shopping cart that's nearly empty, then pushing the same cart loaded with bricks. The loaded cart is harder to get moving — it resists changes to its motion more. That resistance is inertia, and it depends only on how much mass is there.

Now imagine spinning a bicycle wheel. If you hold the axle and try to tilt the spinning wheel, it fights you. But here's the twist: a lightweight wheel that's large in diameter can be harder to spin or stop than a heavy wheel that's small in diameter, even if the heavy wheel has more mass. Why? Because rotational inertia depends not just on how much mass, but on where that mass is placed relative to the axis of rotation.

Note

Rotational inertia (also called moment of inertia) is the rotational equivalent of mass. It measures how difficult it is to change an object's rotational motion — to start it spinning, stop it, or change its spin speed.

The Core Idea: Mass × Distance²

The precise statement is this:

I = \sum m_i r_i^2

For a collection of point masses, rotational inertia $I$ is the sum of each mass $m_i$ multiplied by the square of its perpendicular distance $r_i$ from the axis of rotation.

That square is crucial. Doubling the distance from the axis quadruples the rotational inertia. A mass far from the axis contributes much more to rotational inertia than the same mass close to the axis.

Why Comparison Matters

When you compare two objects, you're asking: Which is harder to spin? The answer depends on both mass and shape.

Example 1: A ring vs. a disk of the same mass and radius

A ring has all its mass at the outer edge ( $r = R$ for all mass). Its rotational inertia is $I_{\text{ring}} = MR^2$ .
A solid disk has mass spread evenly from center to edge. Its rotational inertia is $I_{\text{disk}} = \frac{1}{2}MR^2$ .

The ring has twice the rotational inertia of the disk. Same mass, same radius — but the ring is harder to spin because its mass is concentrated farther from the axis.

Example 2: A long rod vs. a short rod of the same mass

A rod spun about its center: $I = \frac{1}{12}ML^2$ .
A rod spun about one end: $I = \frac{1}{3}ML^2$ .

The same rod, same mass — but spinning it about the end is four times harder than spinning it about the center. The mass is, on average, farther from the axis.

Watch out

A common mistake is to think that rotational inertia depends only on mass. It does not. Two objects with the same mass can have wildly different rotational inertias depending on how their mass is distributed.

The Intuition Behind the Square

Why distance squared? Think of a spinning object. A mass far from the axis has to travel a longer path in the same time — it has a higher linear speed for the same angular speed. To change that speed (to accelerate or decelerate the rotation), you need to apply a force over that longer distance. The square comes from the geometry: the work required scales with distance, and the lever-arm effect also scales with distance. The two factors multiply.

A Quick Comparison Table

Object	Axis location	Rotational inertia $I$	Relative difficulty to spin
Point mass $m$ at distance $R$	Through point	$mR^2$	Baseline

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A solid sphere has a smaller moment of inertia than a hollow cylinder of the same mass and radius because, in the sphere, a greater proportion of its mass is distributed closer to the axis of rotation, resulting in less resistance to angular acceleration.

The moment of inertia, often called rotational inertia, is a fundamental concept in rotational dynamics. It quantifies an object's resistance to changes in its rotational motion, much like mass quantifies resistance to changes in linear motion. The key factors determining moment of inertia are the object's total mass and, crucially, how that mass is distributed relative to the axis of rotation. The further the mass is, on average, from the axis, the greater the moment of inertia.

To understand why a solid sphere has a smaller moment of inertia than a hollow cylinder of the same mass and radius about their axes of symmetry, we need to compare how mass is distributed in each object.

Recall the Moment of Inertia Formulas: For a solid sphere of mass $M$ and radius $R$ , rotating about an axis passing through its center (which is its axis of symmetry), the moment of inertia is given by:

I_{sphere} = \frac{2}{5}MR^2

For a thin-walled hollow cylinder of mass $M$ and radius $R$, rotating about its central axis (its axis of symmetry), the moment of inertia is given by:

I_{hollow\_cylinder} = MR^2

> [!IMPORTANT]
> These formulas are standard results derived from integration, considering the mass distribution.

2. Compare the Coefficients:

Let's look at the coefficients multiplying $MR^2$ in both formulas:

* For the solid sphere: $\frac{2}{5} = 0.4$

* For the hollow cylinder: $1$

Clearly, $0.4 < 1$ . This mathematical comparison directly shows that $I_{sphere} < I_{hollow\_cylinder}$ for the same mass $M$ and radius $R$ .

Understand the Physical Reason (Mass Distribution): The difference in these coefficients arises from how the mass is distributed in each object relative to the axis of rotation.
- Hollow Cylinder: In a thin-walled hollow cylinder, all of its mass $M$ is concentrated at the maximum radius $R$ from the axis of rotation. There is no mass closer to the axis than $R$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$\dfrac{7MR^2}{16}$
›Reveal solutionSolution
Take the left semicircular half of the disc ( $MR^2/8$ about $AB$ ) and subtract the two removed half-circles of radius $R/2$ (together $MR^2/64$ ). The result is $\mathbf{7MR^2/64}$ — option (B).

The concept first

Two ideas make this a short calculation instead of a nasty integral.

1. Moment of inertia is additive. $I$ is a sum (integral) of $dm\,r^2$ , so if a body is a whole minus a hole, then $I_{body} = I_{whole} - I_{hole}$ , provided both are taken about the same axis. That lets us handle an awkward hatched shape with two standard results.

2. Symmetry halves the work. The axis $AB$ is a vertical diameter, and the disc is symmetric about it. So the left half and right half contribute equally: each is $\tfrac12 I_{disc}$ . The same symmetry argument applies to each small circle, because its centre lies on $AB$ — that is the crucial geometric fact in the figure. $AB$ passes through the centre of each small circle, so $AB$ is a diameter of the small circle too, and its left half contributes half of the small circle's $I$ .

Standard result used: for a uniform disc of mass $m$ , radius $r$ , about a diameter, $I = \tfrac14 m r^2$ .

Step-by-step

Surface mass density.

$\sigma = \frac{M}{\pi R^2}$

Whole disc about the diameter $AB$ .

$I_{disc} = \frac{1}{4}MR^2$

Left half of the big disc about $AB$ . By symmetry about $AB$ ,

$I_{left\ half} = \frac12 \times \frac{MR^2}{4} = \frac{MR^2}{8}$

Mass of one small circle (radius $R/2$ ):

$m = \sigma \times \pi\left(\frac{R}{2}\right)^2 = \frac{M}{\pi R^2}\times \frac{\pi R^2}{4} = \frac{M}{4}$

One whole small circle about $AB$ . $AB$ passes through its centre, so $AB$ is a diameter of that circle: …
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$\dfrac{16}{27}$
›Reveal solutionSolution
For a head-on elastic collision between two spheres, the fraction of kinetic energy transferred from the smaller to the larger depends only on their mass ratio. Here the masses are proportional to the cubes of the radii, giving a mass ratio of 1:8, and the fraction transferred is $\frac{16}{81}$ . The correct option is (D).

The key idea is that in a one-dimensional elastic collision, the velocities after impact are completely determined by conservation of momentum and conservation of kinetic energy. The fraction of energy transferred from the projectile to the target depends only on the mass ratio. Since both spheres are made of steel, their masses are proportional to their volumes, i.e., to the cube of their radii.

Why this approach works:

We don’t need to compute forces or time of contact — elastic collisions have a neat algebraic result: the final velocity of the initially stationary target is $v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i}$ . The kinetic energy transferred is then simply $\frac{1}{2} m_2 v_{2f}^2$ , and the fraction is that divided by the initial kinetic energy $\frac{1}{2} m_1 v_{1i}^2$ . The result simplifies to a function of the mass ratio alone.

Find the masses. The smaller sphere has diameter 2 cm, so its radius is $r_1 = 1$ cm. The larger sphere has radius $r_2 = 2$ cm. Since both are steel (same density $\rho$ ), mass is proportional to volume:

$m_1 = \rho \cdot \frac{4}{3}\pi (1)^3, \quad m_2 = \rho \cdot \frac{4}{3}\pi (2)^3$
Hence
$\frac{m_1}{m_2} = \frac{1^3}{2^3} = \frac{1}{8}.$
So $m_2 = 8 m_1$ .

Set up the collision. Let the smaller sphere (mass $m_1$ ) move initially with speed $u$ toward the larger sphere (mass $m_2 = 8m_1$ ) at rest. After an elastic head-on collision, the velocity of the larger sphere is given by the standard formula:

$v_2 = \frac{2 m_1}{m_1 + m_2} u.$
Substituting $m_2 = 8 m_1$ :
$v_2 = \frac{2 m_1}{m_1 + 8 m_1} u = \frac{2 m_1}{9 m_1} u = \frac{2}{9} u.$

Compute the kinetic energy transferred. Initial kinetic energy of the smaller sphere:
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$Q. Two solid spheres of radii 10 cm and 5 cm are in contact with each other. If the density of the material of the smaller sphere is twice the density of the material of the larger sphere, then the distance of the centre of mass of the system of the two spheres from the centre of the smaller sphere is (A) 6 cm (B) 9 cm (C) 12 cm (D) 8 cm$
›Reveal solutionSolution
The centre of mass lies along the line joining the centres, weighted by mass.

Using the smaller sphere’s centre as origin, the COM distance from it is 9 cm → option (B).

Concept & Intuition

The centre of mass of two point masses (or uniform spheres, which behave as point masses at their centres) is the weighted average of their positions, where the weights are the masses.

Here the spheres touch, so the distance between their centres is the sum of their radii: $10 + 5 = 15$ cm.

The smaller sphere has twice the density of the larger, so even though its volume is smaller, its mass is not simply proportional to volume — we must compute each mass using density × volume.

Step-by-step solution

Set up coordinates

Place the centre of the smaller sphere at $x = 0$ and the centre of the larger sphere at $x = 15$ cm (since they touch, the distance between centres is $10 + 5 = 15$ cm).

Find the masses

Let the density of the larger sphere be $\rho$ . Then the density of the smaller sphere is $2\rho$ .

Volume of a sphere: $V = \frac{4}{3}\pi r^3$ .

Larger sphere: radius $R = 10$ cm → $V_L = \frac{4}{3}\pi (10)^3 = \frac{4000\pi}{3}$ cm³. Mass $M_L = \rho \cdot V_L = \frac{4000\pi\rho}{3}$ .

Smaller sphere: radius $r = 5$ cm → $V_S = \frac{4}{3}\pi (5)^3 = \frac{500\pi}{3}$ cm³. Mass $M_S = 2\rho \cdot V_S = 2\rho \cdot \frac{500\pi}{3} = \frac{1000\pi\rho}{3}$ .

Apply the centre of mass formula

For points along a line, the COM coordinate (with origin at the smaller sphere’s centre) is:

$x_{\text{COM}} = \frac{M_S \cdot 0 + M_L \cdot 15}{M_S + M_L}$
Substitute the masses:
$x_{\text{COM}} = \frac{0 + \frac{4000\pi\rho}{3} \cdot 15}{\frac{1000\pi\rho}{3} + \frac{4000\pi\rho}{3}}$
The factor $\frac{\pi\rho}{3}$ cancels from numerator and denominator:
$x_{\text{COM}} = \frac{4000 \times 15}{1000 + 4000} = \frac{60000}{5000} = 12 \text{ cm}$

Interpret the result …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$6.53\, \mathrm{ms}^{-2}$
›Reveal solutionSolution
When a solid cylinder unwinds a rope by falling, it undergoes both translational and rotational motion. By applying Newton's second law for both types of motion and relating the linear and angular accelerations, we find the acceleration of the cylinder to be $\frac{2}{3}g$ . The calculated acceleration is $\boxed{6.53\, \mathrm{ms}^{-2}}$ .

The problem describes a solid cylinder with a rope wound around it, which is allowed to unwind as the cylinder falls. This setup involves two types of motion simultaneously: the cylinder's center of mass moves downwards (translational motion), and the cylinder rotates about its axis (rotational motion). To find the acceleration, we need to apply Newton's second law for both translational and rotational dynamics and then combine these equations.

Here's the underlying concept:

Translational Motion: The net force acting on the cylinder determines its linear acceleration. The forces involved are gravity pulling it down and the tension in the rope pulling it up.

Rotational Motion: The net torque acting on the cylinder determines its angular acceleration. The tension in the rope creates a torque about the cylinder's central axis, causing it to rotate.

Constraint: Since the rope unwinds without slipping, there's a direct relationship between the linear acceleration of the cylinder's center of mass and its angular acceleration.

Let's denote the mass of the cylinder as $M$ , its radius as $R$ , its linear acceleration as $a$ , and its angular acceleration as $\alpha$ . The acceleration due to gravity is $g$ .

Analyze Translational Motion: The forces acting on the cylinder are its weight, $Mg$ , acting downwards, and the tension in the rope, $T$ , acting upwards. Applying Newton's second law for translational motion in the vertical direction:

$F_{net} = Ma$ $Mg - T = Ma \quad (1)$
Here, we've taken the downward direction as positive.

2. Analyze Rotational Motion:

The tension $T$ in the rope creates a torque about the cylinder's central axis. This torque causes the cylinder to rotate. The moment of inertia for a solid cylinder about its central axis is $I = \frac{1}{2}MR^2$ .

The torque $\tau$ is given by the force multiplied by the perpendicular distance from the axis of rotation, which is the radius $R$ .
$\tau = TR$
Applying Newton's second law for rotational motion:
$\tau_{net} = I\alpha$ $TR = \left(\frac{1}{2}MR^2\right)\alpha \quad (2)$

Relate Linear and Angular Acceleration: Since the rope unwinds without slipping, the linear acceleration of the point on the cylinder's circumference in contact with the rope is the same as the linear acceleration of the rope itself (which is the acceleration of the cylinder's center of mass). This gives us the kinematic relationship:

$a = R\alpha$
From this, we can express $\alpha$ in terms of $a$:
$\alpha = \frac{a}{R} \quad (3)$

Solve the System of Equations: Now we substitute equation (3) into equation (2):

$TR = \left(\frac{1}{2}MR^2\right)\left(\frac{a}{R}\right)$ $TR = \frac{1}{2}MRa$
We can cancel $R$ from both sides (assuming $R \neq 0$): …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$450\,\mathrm{ms}^{-1}$
›Reveal solutionSolution

Conserving angular momentum about the hinge: the bullet's $L=m v (L/2)=3.04\ \text{kg m}^2\text{s}^{-1}$ divided by the total moment of inertia $I=\tfrac13 M_dL^2+m(L/2)^2=4.86\ \text{kg m}^2$ gives $\omega\approx 0.625\ \text{rad s}^{-1}$ .

Angular momentum of the bullet about the hinge (embeds at the centre, $r=L/2=0.45\ \text{m}$ ):
$L_{\text{bullet}}=m\,v\,r=(0.015)(450)(0.45)=3.0375\ \text{kg m}^2\text{s}^{-1}.$
Moment of inertia after embedding. Door hinged at one edge (mass $M_d=18\ \text{kg}$ , width $L=0.9\ \text{m}$ ):

$I_{\text{door}}=\tfrac13 M_d L^{2}=\tfrac13(18)(0.9)^{2}=4.86\ \text{kg m}^2,$ …
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$1 : \sqrt{3}$
›Reveal solutionSolution
The radius of gyration about an axis through one end of a thin uniform rod is $K = L/\sqrt{3}$ , so the ratio $K : L = 1 : \sqrt{3}$ . The correct option is (A).

The key idea is that the radius of gyration $K$ is defined by $I = M K^2$ , where $I$ is the moment of inertia about the given axis. For a thin uniform rod of length $L$ and mass $M$ , the moment of inertia about an axis perpendicular to the rod through one end is $\frac{1}{3} M L^2$ . Equating gives $K^2 = L^2/3$ , so $K = L/\sqrt{3}$ . The ratio $K:L$ is therefore $1:\sqrt{3}$ .

Recall the definition of radius of gyration. For any rigid body, the radius of gyration $K$ about a given axis is the distance from the axis at which the entire mass $M$ could be concentrated to produce the same moment of inertia. Mathematically,

$I = M K^2 \quad \Rightarrow \quad K = \sqrt{\frac{I}{M}}.$
So finding $K$ reduces to finding the moment of inertia $I$ about the specified axis.

Identify the moment of inertia for the rod. A thin uniform rod of length $L$ and mass $M$ has a well-known moment of inertia about an axis perpendicular to its length through its center:

$I_{\text{center}} = \frac{1}{12} M L^2.$
But here the axis passes through one end, not the center. Use the parallel axis theorem:
$I_{\text{end}} = I_{\text{center}} + M d^2,$
where $d$ is the distance from the center to the new axis. For a rod, the center is at $L/2$ from either end, so $d = L/2$ .
$I_{\text{end}} = \frac{1}{12} M L^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + \frac{1}{4} M L^2.$
Combine the fractions: $\frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$ . Hence
$I_{\text{end}} = \frac{1}{3} M L^2.$

Relate to the radius of gyration. Set $I_{\text{end}} = M K^2$ :
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$1 : 3$
›Reveal solutionSolution
The radius of gyration $K$ about an axis through one end of a thin uniform rod is $\frac{L}{\sqrt{3}}$ , so the ratio $K : L$ is $1 : \sqrt{3}$ .

The radius of gyration $K$ is defined as the distance from the axis at which the entire mass of the body could be concentrated to give the same moment of inertia. In other words, if $I$ is the moment of inertia about the given axis and $M$ is the mass, then $I = M K^2$ . So $K = \sqrt{I/M}$ .

For a thin uniform rod of length $L$ and mass $M$ , the moment of inertia about an axis perpendicular to its length and passing through one end is a standard result. Why? Because the rod is a continuous distribution of mass, and we can integrate $r^2 \, dm$ from the axis. The centre of mass is at $L/2$ , and the parallel axis theorem lets us shift from the centre to the end.

Let’s work it out step by step.

Moment of inertia about the centre For a thin rod about a perpendicular axis through its centre, the moment of inertia is

$I_{\text{cm}} = \frac{1}{12} M L^2.$
This comes from integrating $x^2 \, dm$ from $-L/2$ to $L/2$ , where $dm = (M/L) dx$ .

Shift the axis to one end using the parallel axis theorem The parallel axis theorem says: $I = I_{\text{cm}} + M d^2$ , where $d$ is the distance between the two parallel axes. Here, the axis through the end is a distance $d = L/2$ from the centre. So

$I_{\text{end}} = \frac{1}{12} M L^2 + M \left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + \frac{1}{4} M L^2.$

Combine the fractions $\frac{1}{4} = \frac{3}{12}$ , so

$I_{\text{end}} = \frac{1}{12} M L^2 + \frac{3}{12} M L^2 = \frac{4}{12} M L^2 = \frac{1}{3} M L^2.$

Find the radius of gyration By definition, $I = M K^2$ , so …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$r_1$
›Reveal solutionSolution
Masses scale as $r^3$ (same material). Taking the contact point as origin, the centre of mass lies at $\dfrac{r_2^4 - r_1^4}{r_1^3 + r_2^3}$ from it. Option (D).

Same material ⇒ same density, so each mass is proportional to volume: $m_1 \propto r_1^3$ , $m_2 \propto r_2^3$ .

Set up coordinates. Put the point of contact at the origin. The centre of sphere 1 lies at $x = -r_1$ and the centre of sphere 2 at $x = +r_2$ (their centres are one radius away from the contact point, on opposite sides).

Centre of mass. …
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Q.Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Concept understanding — Rotational Inertia Comparison

Rotational Inertia Comparison: From Intuition to Precision

The Core Idea: Mass × Distance²

Why Comparison Matters

The Intuition Behind the Square

A Quick Comparison Table

Concept: Moment of Inertia Depends on How Mass Is Distributed Relative to the Axis

The concept first

Step-by-step

Q.Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

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