Interactive physics simulator
Moment of Inertia
Explore how shape and mass distribution resist rotation. Spin discrete masses on a central rod, race cylinders and spheres down an incline, and study the parallel axis theorem on physical swinging pendulums.
Moment of Inertia Lab
Adjust sliders, select your graph option, and click Simulate to start the physics engine.
Live Telemetry
- Left Inertia (I₁)
- 0.000 kg·m²
- Right Inertia (I₂)
- 0.000 kg·m²
- Total Inertia (I)
- 0.000 kg·m²
- Angular Accel (α)
- 0.00 rad/s²
- Current Leader
- Solid Sphere
- COM Inertia (Icm)
- 0.000 kg·m²
- Parallel Shift (Md²)
- 0.000 kg·m²
- Total Inertia (I)
- 0.000 kg·m²
- Oscillation Period
- 0.00 s
- System State
- Ready
Introduction to Moment of Inertia
Just as linear inertia measures how strongly an object resists changes to its straight-line velocity, moment of inertia (also called rotational inertia or angular mass) measures how strongly an object resists changes to its rotation speed. In linear mechanics, inertia is determined solely by mass. In rotational mechanics, however, resistance depends not only on *how much* mass is present, but also on *where* that mass is located relative to the rotation pivot axis.
An object whose mass is concentrated far from its axis of rotation will have a much larger moment of inertia than an object of identical mass whose material is packed tightly around the pivot. This is why a tightrope walker carries a long, heavy pole to maximize their moment of inertia, slowing down any accidental tilting falls and giving them more time to restore balance.
Key Rotational Mechanics Concepts
1. Mathematical Formulations
For a collection of individual point masses (discrete system) rotating about a single axis, the moment of inertia (I) is calculated by multiplying each mass (mi) by the square of its perpendicular distance (ri) to the axis and summing them:
For continuous solid shapes, this summation turns into an integral: I = ∫ r² dm. This leads to standard shape coefficients where moment of inertia is written in terms of total mass (M) and outer radius (R):
Here, β is the **inertia shape factor**. Common factors include:
- Thin Ring / Hoop: β = 1.0 (all mass at outer edge)
- Hollow Sphere: β ≈ 0.67 (mass distributed along shell)
- Solid Cylinder / Disk: β = 0.5 (mass distributed evenly throughout volume)
- Solid Sphere: β = 0.4 (mass concentrated closer to center)
2. Parallel Axis Theorem
If you know the moment of inertia of a body about an axis passing through its center of mass (Icm), the Parallel Axis Theorem allows you to calculate its moment of inertia about any parallel axis displaced by a perpendicular distance d:
This relation shows that the minimum moment of inertia for any object always occurs about the axis that passes directly through its center of mass (where d = 0). Displacing the pivot axis always increases the rotational resistance quadratically.
3. Radius of Gyration
The radius of gyration (k) is the distance from the rotation axis to a single point where the entire mass of the object could be concentrated without altering its moment of inertia:
Solved Numerical Examples
Compare the moment of inertia of: (a) a solid cylinder (disk) of mass 4.0 kg and radius 0.25 meters rotating about its central axis, and (b) a thin cylindrical ring (hoop) of the same mass and radius rotating about the same axis.
View Step-by-Step Solution
- Given: Mass M = 4.0 kg, radius R = 0.25 m.
- (a) Calculate Moment of Inertia for the Solid Cylinder:
The formula for a solid cylinder/disk is: Idisk = ½ · M · R². - Substitute known values: Idisk = 0.5 × 4.0 kg × (0.25 m)².
- Solve: Idisk = 2.0 × 0.0625 = 0.125 kg·m².
- (b) Calculate Moment of Inertia for the Thin Ring:
The formula for a thin ring/hoop is: Iring = M · R² (since all mass is at the outer rim). - Substitute known values: Iring = 4.0 kg × (0.25 m)².
- Solve: Iring = 4.0 × 0.0625 = 0.250 kg·m².
- Results: (a) The solid cylinder has an inertia of 0.125 kg·m². (b) The thin ring has an inertia of 0.250 kg·m² (exactly double). This means the ring is twice as difficult to angularly accelerate under the same torque.
A dumbbell system consists of two small spheres of mass 3.0 kg each, mounted on the ends of a light rod of negligible mass. Find the moment of inertia of the system about a central axis perpendicular to the rod if: (a) the spheres are positioned at a distance of 0.80 meters from the axis, and (b) they are slid inward to a distance of 0.40 meters from the axis.
View Step-by-Step Solution
- Given: Mass m₁ = 3.0 kg, mass m₂ = 3.0 kg.
- Recall the discrete moment of inertia formula: I = Σ mi · ri² = m₁ · r₁² + m₂ · r₂².
- (a) Spheres at r = 0.80 m:
Substitute known values: I = (3.0 kg × (0.80 m)²) + (3.0 kg × (0.80 m)²). - Calculate: I = (3.0 × 0.64) + (3.0 × 0.64) = 1.92 + 1.92 = 3.84 kg·m².
- (b) Spheres at r = 0.40 m:
Substitute known values: I = (3.0 kg × (0.40 m)²) + (3.0 kg × (0.40 m)²). - Calculate: I = (3.0 × 0.16) + (3.0 × 0.16) = 0.48 + 0.48 = 0.96 kg·m².
- Observe that halving the distance reduces the moment of inertia by a factor of four (from 3.84 to 0.96) because inertia scales quadratically with radius (r²).
- Results: (a) Inertia at 0.80 m is 3.84 kg·m². (b) Inertia at 0.40 m is 0.96 kg·m².
A uniform door has a mass of 18.0 kg and a width of 0.90 meters. Its moment of inertia about a vertical axis running through its center of mass is given by I<sub>cm</sub> = 1.215 kg·m². Using the Parallel Axis Theorem, calculate the door's moment of inertia when rotating about its vertical hinges on the edge (distance d = 0.45 meters from the center of mass).
View Step-by-Step Solution
- Given: Mass M = 18.0 kg, width W = 0.90 m, Icm = 1.215 kg·m², pivot displacement offset d = 0.45 m.
- Recall the Parallel Axis Theorem: I = Icm + M · d².
- Substitute known values into the equation:
I = 1.215 kg·m² + [18.0 kg × (0.45 m)²]. - Calculate the parallel displacement shift (M · d²):
M · d² = 18.0 × 0.2025 = 3.645 kg·m². - Add the components together:
I = 1.215 + 3.645 = 4.860 kg·m². - Note: This is equivalent to using the end-pivot rod formula (I = ⅓ M L²) directly:
I = ⅓ × 18.0 × (0.90)² = 6.0 × 0.81 = 4.860 kg·m², verifying the theorem. - Result: The moment of inertia about the hinges is 4.860 kg·m².
Conceptual Practice
What is the physical significance of "moment of inertia" (rotational inertia), and how does it compare to mass?
Show Explanation
In linear kinematics, **mass** (m) is the measure of inertia—it resists translational acceleration. In rotational kinematics, the **moment of inertia** (I) is the measure of rotational inertia—it resists angular acceleration. While mass is an intrinsic property that remains constant regardless of shape, the moment of inertia depends heavily on both the total mass and *how* that mass is distributed relative to the axis of rotation.
Why does a solid sphere roll down an incline faster than a hollow hoop (ring) of the same mass and radius?
Show Explanation
When rolling down a hill, gravitational potential energy converts into both **translational** and **rotational** kinetic energy. The ring distributes all of its mass at its outer radius, yielding a larger moment of inertia ($I = MR^2$, inertia factor $eta = 1.0$), which resists rolling. The solid sphere distributes its mass throughout its volume, yielding a much lower moment of inertia ($I = 0.4 MR^2$, $eta = 0.4$). Consequently, the sphere requires less energy to rotate, leaving more energy to convert into translational speed.
State the Parallel Axis Theorem. Under what conditions can it be applied?
Show Explanation
The Parallel Axis Theorem states that the moment of inertia ($I$) of a body about any axis is equal to its moment of inertia about a parallel axis passing through its center of mass ($I_{cm}$) plus the product of the body's mass ($M$) and the square of the perpendicular distance ($d$) between the two axes: $I = I_{cm} + Md^2$. It can only be applied when the two rotation axes are parallel and one of them passes directly through the body's center of mass.
What is the "radius of gyration" of a rotating body, and how is it defined mathematically?
Show Explanation
The **radius of gyration** ($k$) is the radial distance from the rotation axis to a point where the entire mass of the body could be concentrated without changing its moment of inertia. Mathematically, it is defined as $k = sqrt{I / M}$ or $I = M k^2$. It provides a convenient way to compare the mass distribution of differently shaped bodies.
Why does a figure skater spin faster when they pull their arms and legs close to their rotation axis?
Show Explanation
This is a classic demonstration of the **Conservation of Angular Momentum** ($L = I cdot omega$). In the absence of external torques, $L$ must remain constant. By pulling their limbs inward, the skater decreases their mass distribution radius ($r$), which dramatically reduces their moment of inertia ($I$). To conserve angular momentum, the angular velocity ($omega$) must increase proportionally, causing them to spin much faster.
Frequently Asked Questions
What is moment of inertia?
The moment of inertia is a measure of an object's resistance to changes in its rotation speed. It is determined by the mass of the object and how that mass is distributed relative to the axis of rotation.
What is the SI unit of moment of inertia?
The SI unit of moment of inertia is the kilogram-meter squared (kg·m²).
How does mass distribution affect moment of inertia?
Since the moment of inertia formula is I = Σ m·r², distance has a squared relationship. Spreading mass further from the axis of rotation increases the moment of inertia exponentially, making the object much harder to spin or stop.
What is the difference between mass and moment of inertia?
Mass is a measure of translational inertia (resisting straight-line acceleration) and remains constant. Moment of inertia is a measure of rotational inertia (resisting rotational acceleration) and changes depending on the axis of rotation and shape.
What is the Parallel Axis Theorem?
It is a formula used to calculate the moment of inertia about any axis parallel to an axis through the center of mass: I = I_cm + Md².
Why do hollow objects roll slower than solid ones down hills?
Hollow objects (like rings or hollow balls) have their mass distributed further from their centers, giving them a larger moment of inertia. This means they convert a larger fraction of their energy into rotation, leaving less energy for straight-line speed.
Can an object have negative moment of inertia?
No. Mass (M) and squared distance (r²) are always positive quantities, meaning the moment of inertia is always positive.