Q.A circular disc has mass M and radius R. The moment of inertia about its diameter is
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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.
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=∑miri2
For a collection of point masses, rotational inertia I is the sum of each mass mi multiplied by the square of its perpendicular distance ri 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 Iring=MR2.
- A solid disk has mass spread evenly from center to edge. Its rotational inertia is Idisk=21MR2.
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=121ML2.
- A rod spun about one end: I=31ML2.
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.
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 | mR2 | Baseline |
This tests the standard moment-of-inertia result for a uniform circular disc about an axis lying in its own plane, through its centre (a diameter). …
Moment of inertia of a disc about its diameter = MR^2/4 (option c).
For a uniform circular disc of mass M and radius R, the moment of inertia about an axis perpendicular to its plane, through its centre, is I_z = MR^2/2 (this is the standard result obtained by integrating r^2 dm over the disc).
By symmetry, the disc's two diameters (say the x-axis and y-axis, both lying in the plane of the disc, through the centre) are equivalent, so I_x = I_y = I_diameter.
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Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set ANNUAL1 markQ.Moment of inertia of a uniform circular ring about a diameter of a ring is ............. .
›Reveal solutionSolution
Using the perpendicular axis theorem on a ring's axis (I=MR2) and its two diameters gives Idiameter=21MR2.
For a uniform circular ring of mass M and radius R, the moment of inertia about the axis perpendicular to its plane, through its centre, is Iz=MR2 (all the mass lies at distance R from that axis).
By the perpendicular axis theorem, for any two mutually perpendicular diameters x and y lying in the plane of the ring:
Iz=Ix+Iy …
- CBSE 2026Set ANNUAL1 markMCQQ.(C) The M.I. of a Solid sphere of mass 5 kg and radius 1 m about its diameter is :(a) 1/2 Kg m^2(b) 2 Kg m^2(c) 1/5 Kg m^2(d) 5 Kg m^2.
›Reveal solutionSolution
Using I = (2/5) M R² for a solid sphere about its diameter, with M = 5 kg and R = 1 m, gives I = 2 kg m².
The standard formula for the moment of inertia of a uniform solid sphere about an axis through its diameter (centre) is:
I = (2/5) M R²
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- CBSE 2026Set ANNUAL1 markMCQQ.(D) The moment of Inertia of a body, depends upon :(a) Distribution of Mass from axis of Rotation.(b) Angular velocity of the Body.(c) Angular acceleration of the Body.(d) None of the above.
›Reveal solutionSolution
Since I = Σ m r², moment of inertia depends purely on how mass is distributed relative to the chosen axis, not on how fast the body happens to be rotating.
Moment of Inertia is defined as I = Σ mi ri², where ri is the perpendicular distance of each mass element from the axis of rotation.
This makes I depend entirely on:
- the total mass of the body, and
- how that mass is distributed with respect to the particular axis chosen (mass farther from the axis contributes more, since r² appears in the formula). …
- CBSE 2026Set ANNUAL1 markMCQQ.The moment of inertia of a circular disc of mass M and radius R about its diameter is(a) MR^2(b) MR^2/4(c) 3MR^2/4(d) MR^2/2
›Reveal solutionSolution
Moment of inertia of a disc about a diameter is MR^2/4. Answer (B).
About the central axis perpendicular to the disc, I_z = ½ MR^2.
…
- CBSE 2026Set ANNUAL1 markMCQQ.If the mass of a body is M and moment of inertia about an axis is I, then the radius of gyration about that axis is(a) MI^2(b) I^2/M(c) √(I/M)(d) √(M/I)
›Reveal solutionSolution
Radius of gyration k = sqrt(I/M). Answer (C).
The radius of gyration k is the distance from the axis at which the whole mass could be concentrated to give the same moment of inert …
- CBSE 2025Set ANNUAL1 markMCQQ.What is the moment of inertia of a cylindrical shell of mass M and radius R about its own axis?(a) MR^2(b) (1/2) MR^2(c) 2MR^2(d) (2/3) MR^2
›Reveal solutionSolution
A cylindrical (hollow) shell has its entire mass concentrated at a fixed distance R from the axis, so its moment of inertia is simply MR^2, exactly like a ring or hoop.
Moment of inertia I = Integral of r^2 dm, where r is the perpendicular distance of each mass element from the axis.
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- CBSE 2025Set ANNUAL1 markQ.Define radius of gyration.
›Reveal solutionSolution
Radius of gyration is a way of representing how the mass of an extended body is distributed relative to an axis, using a single equivalent 'point-mass' distance.
For a rigid body of total mass M and moment of inertia I about a given axis, the radius of gyration k is defined as the distance from the axis at which, if the entire mass M of the body were concentrated as a single point mass, the moment of inertia about that axis would be the same as the actual moment of inertia of the body.
That is, I=Mk2, so k=I/M.
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- CBSE 2024Set ANNUAL1 markMCQQ.Mass and moment of inertia about diameter of a solid sphere and a thin spherical shell is same. The ratio between their radii is (A) 3:5 (B) 5:3 (C) √3:√5 (D) √5:√3
›Reveal solutionSolution
Equal I and equal M for a solid sphere and shell gives radii ratio 5:3.
Moment of inertia about a diameter: solid sphere I1=52MR12; thin spherical shell I2=32MR22.
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- CBSE 2024Set ANNUAL1 markMCQQ.What is the unit of the radius of gyration?(a) kg-m^2(b) kg(c) m(d) kg-m
›Reveal solutionSolution
Radius of gyration k is defined by I = M k^2, so k = sqrt(I/M), which works out to a length — its SI unit is the metre.
Moment of inertia I has units kg·m^2. Setting I = Mk^2 and solving for k:
k = sqrt(I/M), units = sqrt(kg·m^2 / kg) = sqrt(m^2) = m
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- CBSE 2024Set ANNUAL1 markMCQQ.What is the moment of inertia of a solid cylinder of mass M and radius R about its own axis?(a) MR^2(b) (1/2) MR^2(c) (3/2) MR^2(d) (5/2) MR^2
›Reveal solutionSolution
The moment of inertia of a solid cylinder (or disc) of mass M and radius R about its own central axis is (1/2) MR^2 — a standard result derived by integrating r^2 dm over circular mass shells.
Model the solid cylinder as a stack of thin circular discs. Integrating I = ∫ r^2 dm over the cross-section (using dm = (2M/R^2) r dr for a uniform disc of radius R) gives:
I = (1/2) M R^2
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- CBSE 2023Set ANNUAL1 markMCQQ.Analogue of mass in rotational motion is(1) moment of inertia(2) torque(3) radius of gyration(4) angular momentum
›Reveal solutionSolution
Moment of inertia (I) plays the same role in rotational dynamics that mass (m) plays in linear dynamics - both quantify inertia, i.e. resistance to a change in motion.
In linear motion, Newton's second law is F = ma - mass m measures how much a body resists linear acceleration for a given force.
In rotational motion, the analogous law is tau = I alpha - moment of inertia I measures how much a body resists angular acceleration for a given torque. It depends not just on the mass of the body but on how that mass is distributed relative to the axis of rotation (I = sum of m_i r_i^2).
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- CBSE 2023Set ANNUAL1 markMCQQ.M is mass and R is radius of a circular ring. The moment of Inertia about an axis passing through the centre and perpendicular to the plane is:(a) MR²(b) (1/2) MR²(c) (2/5) MR²(d) (2/3) MR²
›Reveal solutionSolution
Every mass element of a ring is at distance R from the central perpendicular axis, so I=MR2 exactly, with no fractional factor.
Moment of inertia about an axis is I=∑miri2, where ri is each mass element's perpendicular distance from the axis. For a thin circular ring of mass M and radius R, every particle of the ring lies exactly on the circumference, at distance R from the centre, and the axis considered passes through the centre perpendicular to the plane of the ring. Hence every mass element has the same ri=R:
I=∑miR2=R2∑mi=MR2 …
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