Rod moment of inertia about one end
A 2 kg rod 1.5 m long, pivoted at one end, has a moment of inertia of 1.5 kg·m²: enter mass and length and the one-third factor is applied for you, with no parallel-axis step. Free in your browser.
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Run — free
Worked example
- Mass
- 2
- Length
- 1.5
- Moment of inertia
- 1.5
- Axis
- through one end, perpendicular to the thin rod length
The moment of inertia of a uniform rod about one end is a standard result in rigid-body dynamics and first-year physics: for a thin rod of mass m and length L, the inertia about an axis through either tip and perpendicular to the length is one third m L squared. Students, tutors, and engineers need that closed form without re-deriving the integral or mis-applying the parallel-axis theorem every time. This calculator accepts mass and length in consistent units, multiplies them with the exact one-third factor, and returns the scalar moment of inertia together with the formula and axis description. Zero mass or zero length correctly yields zero inertia.
Formula: I = (1/3) m L² about one end
A uniform thin rod has constant linear density along its length L and total mass m. The mass moment of inertia about an axis is the integral of r squared dm for every mass element. When the axis passes through the centre of mass and is perpendicular to the rod, the familiar result is I_cm equals one twelfth m L squared.
Shifting the same axis to one end by the parallel-axis theorem adds m times (L over two) squared: I_end equals I_cm plus m (L/2)², which simplifies to one twelfth m L² plus one fourth m L², or one third m L². The axis is fixed through one tip, perpendicular to the rod length—the usual classroom setup for a rod pivoted at a hinge, a door modelled as a slender bar, or a physical pendulum rod.
The thin-rod idealisation ignores thickness and end caps; if your body is a thick bar or a plate, use a full solid-body inertia table instead. Units stay consistent: if m is in kilograms and L in metres, I is in kilogram metres squared; any other coherent pair (grams and centimetres, slugs and feet) yields I in mass times length squared. The calculator does not convert units; it multiplies the numbers you send.
What you can do with it
Homework and lab checks
Verify I = (1/3) m L² for a rod pivoted at one end against textbook answers before coupling into torque or energy problems.
Exam and LMS auto-grading
Mark rod-about-end inertia items with a deterministic oracle that rejects negative mass or length and returns a stable scalar.
Simple mechanism models
Supply end-pivot inertia for gates, robotic links, and physical pendulums in notebooks without a full solid-body CAD export.
FAQ
What is the moment of inertia of a uniform rod about one end?
For a thin uniform rod of mass m and length L, about an axis through one end and perpendicular to the rod, I = (1/3) m L².
How does this differ from inertia about the centre?
About the centre (perpendicular to the rod) I_cm = (1/12) m L². About one end, the parallel-axis theorem adds m (L/2)², giving I_end = (1/3) m L²—the formula this calculator uses.
What happens if mass or length is negative?
Negative mass or length is not physical for this model and is not billed.
Which units should I use?
Any consistent pair: kilograms and metres yield kg·m².
Can I model a door or a gate as a rod about its hinge?
Yes, if it is slender: the hinge is the end axis, perpendicular to the length. The thin-rod model ignores thickness and end caps, so for a thick bar or a plate use a full solid-body inertia table instead.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/mech/moment-of-inertia-rod-end \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"mass":2,"length":1.5}'const res = await fetch("https://api.kit.forhosting.com/mech/moment-of-inertia-rod-end", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"mass": 2,
"length": 1.5
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/mech/moment-of-inertia-rod-end",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"mass": 2,
"length": 1.5
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/mech/moment-of-inertia-rod-end", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"mass":2,"length":1.5}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"mass":2,"length":1.5}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/mech/moment-of-inertia-rod-end", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"mass": 2,
"length": 1.5
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "mech.moment_of_inertia_rod_end",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
Negative mass or length fails with invalid_input instead of a silent nonsense number.
Formula: I = (1/3) m L² about one end
That is the identity this endpoint evaluates: I equals (1/3) m L².
Input contract, response fields, and errors
Example: mass equals two kilograms and length equals one and a half metres gives I equals (1/3)×2×(1.5)² equals one and a half kilogram metres squared. POST a JSON object with two required finite numbers: mass (m) and length (L). Each field may be a JSON number or a numeric string; missing fields, empty strings, NaN, and Infinity fail with invalid_input naming the field. Zero is allowed and returns moment_of_inertia equal to zero, which is mathematically correct for a zero-mass or zero-length idealisation. On success the body echoes the cleaned mass and length (rounded to twelve decimal places for stable goldens), the scalar moment_of_inertia, the formula string I = (1/3) m L^2, a short axis description, and a units object that reminds callers of kg·m² conventions. Pair the response with your own unit labels or graphics if you need a diagram of the rod and pivot; this endpoint only computes the scalar inertia about the end axis and does not emit plots or multi-axis tensors.
Where a rod-end inertia API earns its keep
Classrooms and tutoring platforms use a dedicated rod moment of inertia about one end calculator so learners can check the one-third formula without confusing it with the centre-axis one-twelfth result that many multi-shape tools default to. Exam generators and LMS plugins need a deterministic oracle that rejects negative mass or length and never invents an answer when the parallel-axis step was forgotten. Failed validation does not consume balance. Engineering notebooks that model gates, robotic links, and physical pendulums often need only this closed form before coupling I into τ equals I α or energy (1/2) I ω²; calling a tiny parse API avoids pulling a full CAD inertia pipeline for a two-parameter ideal rod. Keep requests small, pin the rounded moment_of_inertia in tests, and you get reproducible I values for homework, lab write-ups, and edge workers without standing up symbolic math infrastructure of your own.
Technical questions
What does the rod moment of inertia end calculator cost?
The browser tool is free. API requests use prepaid KIT balance at $0.002 per successful call; invalid input is not billed.