Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. Weights near the middle
2. The same weights, far out
Quick check: two equal small masses sit on a spinning rod, one 1 m from the axis and one 2 m from the axis. The far one's share of I is:
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3. The baseball bat trick
Read the story as text
Hold a baseball bat by the thin handle and swing it quickly back and forth. It is hard work. Now flip it and hold it by the thick barrel end, and swing again. Suddenly it feels light and easy.
The bat has the same mass both times. Nothing about it changed except where you hold it.
So what makes something hard to start (or stop) spinning, if it is not just its mass?
Quick check: a 2.0 kg ball sits 0.50 m from the axis. Its rotational inertia is:
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4. Check yourself
Think of your answer first, then tap to see it.
a) A 1.0 kg weight on a spinning rod is moved from 0.10 m to 0.30 m from the axis. How many times bigger is its share of I?
Show answer
3 times as far, so r² is 9 times bigger: 1.0(0.10)² = 0.010 kg·m² becomes 1.0(0.30)² = 0.090 kg·m².
b) A hoop and a solid disk have the same mass and the same radius. Which is harder to spin up?
Show answer
The hoop. All its mass is at the rim, far from the axis (I = MR²). The disk has mass close in too (I = ½MR²).
c) Why is a bat easier to swing when you hold it by the thick barrel?
Show answer
The heavy barrel is then close to your hands (the axis), so I is small. Held by the handle, the heavy part is far from the axis and I is big.
Already know this?
Three quick questions. Get all three right first time and you can skip ahead.
A small 3.0 kg ball sits 2.0 m from the axis. Its rotational inertia is:
A thin hoop and a solid disk have the same mass and radius. Which has the bigger I about its centre?
You move a small mass twice as far from the axis. Its share of I becomes:
Rotational inertia I measures how hard it is to change an object's spin. It depends on mass and, even more, on how far that mass is from the axis.
Mass far from the axis counts more
In pictures
Weights near the middle: the rod spins up easily.
The same weights far out: the same push spins it up much less.
The bat: heavy barrel far from your hands is hard to swing; heavy barrel at your hands is easy.
It is not just how much mass, but how far from the axis it sits. Far mass counts much more.
Below is the same idea in words, and then with numbers.
In straight-line motion, mass m tells you how hard it is to change velocity. In rotation, the same job is done by rotational inertia I (also called moment of inertia). For a set of small masses, each at distance r from the axis:
I = Σ m r² (units kg·m²)
The r is squared. Move a mass twice as far from the axis and its share of I becomes four times bigger. That is the bat: held by the handle, the heavy barrel is far from your hands (big I). Held by the barrel, most of the mass is near your hands (small I).
For solid shapes, adding up all the tiny pieces gives these standard results (mass M, radius R, length L). You do not need to derive them, but you should know the pattern: the more mass sits far out, the bigger the number in front.
Object (axis)
I
Why
Point mass at distance r
m r²
All mass at r
Thin hoop or ring (centre)
M R²
All mass at the rim
Solid disk or cylinder (centre)
½ M R²
Mass spread from centre to rim
Hollow sphere (centre)
⅔ M R²
Thin shell, but not all at distance R from the axis
Solid sphere (centre)
⅖ M R²
Much mass close to the axis
Thin rod (centre)
¹⁄₁₂ M L²
Mass on both sides, max distance L/2
Thin rod (one end)
⅓ M L²
Mass reaches out a full L
Parallel axis idea. If you know Icm about an axis through the centre of mass, then about any parallel axis a distance d away:
I = Icm + M d²
The centre of mass axis always gives the smallest I. Check with the rod: ¹⁄₁₂ML² + M(L/2)² = ¹⁄₁₂ML² + ³⁄₁₂ML² = ⅓ML². ✓
With numbers: a dumbbell
Two 2.0 kg balls sit at the ends of a light 1.0 m rod (mass of rod ignored).
Show all steps as text
Axis through the middle: each ball is 0.50 m away. I = 2.0(0.50)² + 2.0(0.50)² = 1.0 kg·m².Add m r² for each mass.
Axis through one ball: that ball has r = 0, the other r = 1.0 m. I = 0 + 2.0(1.0)² = 2.0 kg·m².Same object, different axis, different I.
Parallel axis check: I = Icm + Md² = 1.0 + (4.0)(0.50)² = 1.0 + 1.0 = 2.0 kg·m². ✓M = 4.0 kg total, axes 0.50 m apart.
Slide the balls in to 0.25 m from the middle: I = 2 × 2.0 × (0.25)² = 0.25 kg·m², a quarter of before.Half the distance, r² is a quarter.
Masses on a rod
Two equal masses slide along a light rod (the hub adds a tiny 0.0005 kg·m²). A string on the hub applies a steady torque. Slide the masses in and out and watch how fast the rod spins up: α = τ / I (more in 5.6).
Worked examples
Three beads on a rod basic
A light rod spins about one end. Beads: 1.0 kg at 0.20 m, 2.0 kg at 0.50 m, 0.50 kg at 1.0 m. Find I.
Show solutionShow all steps as text
1.0(0.20)² = 0.040; 2.0(0.50)² = 0.50; 0.50(1.0)² = 0.50 (kg·m²).m r² for each bead.
I = 0.040 + 0.50 + 0.50 = 1.04 kg·m².The lightest bead adds as much as the heaviest because it is twice as far out.
Disk vs hoop medium
A solid disk and a thin hoop each have M = 2.0 kg and R = 0.30 m. Find I for each about the centre.
Show solutionShow all steps as text
Disk: I = ½MR² = ½(2.0)(0.30)² = 0.090 kg·m².Mass spread all the way in.
Hoop: I = MR² = (2.0)(0.090) = 0.18 kg·m², twice the disk.All its mass is at the rim.
Rod about the centre and the end medium
A 1.2 m rod of mass 0.50 kg. Find I about its centre and about one end.
Show solutionShow all steps as text
Centre: ¹⁄₁₂ML² = (0.50)(1.44)/12 = 0.060 kg·m².L² = 1.44 m².
A rod has Irod = 0.020 kg·m² about its centre. Two 0.50 kg clamps sit 0.10 m from the centre. A student moves them to 0.20 m and predicts the total I will become 4 times bigger. Find the actual factor and explain.
Show solutionShow all steps as text
Before: I = 0.020 + 2(0.50)(0.10)² = 0.020 + 0.010 = 0.030 kg·m².Rod plus two point masses.
After: I = 0.020 + 2(0.50)(0.20)² = 0.020 + 0.040 = 0.060 kg·m².Only the clamps' part changes.
Factor = 0.060 / 0.030 = 2, not 4.The clamps' share went up 4 times (0.010 → 0.040), but the rod's 0.020 did not change. "Doubling r quadruples I" is true only for the part of the mass that moved.
Practice
The SI unit of rotational inertia is:
Show answer
I = Σmr², so kg × m² = kg·m².
A hoop and a solid disk have the same mass and radius. The same torque acts on each. Which speeds up its spin more slowly?
Show answer
The hoop: I = MR², twice the disk's ½MR². Bigger I means smaller α for the same torque.
A small mass on a light rod is moved from 0.2 m to 0.4 m from the axis. Its rotational inertia becomes:
Show answer
I = mr². Doubling r multiplies I by 2² = 4.
A uniform rod is spun about one end, then about its centre. Iend / Icentre is:
Show answer
(⅓ML²) / (¹⁄₁₂ML²) = 12/3 = 4.
A solid disk (mass M, radius R) spins about an axis perpendicular to it through a point on its rim. Its rotational inertia is:
Show answer
Parallel axis: I = ½MR² + MR² = ³⁄₂MR² (the axes are R apart).
Two solid disks have the same mass. Disk B has twice the radius of disk A. IB / IA is:
Show answer
I = ½MR²: same M, R doubled, so I is 4 times bigger.
Short answer. A tightrope walker carries a long, heavy pole. Use rotational inertia to explain how it helps.
Show answer
The pole puts a lot of mass far from the walker's axis of tipping, so the system's rotational inertia is large. A small push or wobble gives a torque τ, and the angular acceleration α = τ / I is small. The walker tips slowly and has time to correct.
Short answer. Four 0.25 kg balls sit at the corners of a light square frame with sides 0.40 m. Find I about an axis perpendicular to the square through its centre.
Show answer
Each corner is half a diagonal from the centre: r = ½ × 0.40√2 = 0.283 m, so r² = 0.080 m².
I = 4 × 0.25 × 0.080 = 0.080 kg·m².
AP question types for this topic
Four short free response questions, one of each AP Physics 1 type: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. Try each one on paper before you open the worked answer.
1. Mathematical Routines (6 points)
Joel builds a dumbbell: two small balls, each of mass m, fixed to the ends of a light rod of length L. He spins it first about the rod's centre, then about one end.
Derive expressions for the rotational inertia about the centre, Ic, and about one end, Ie, in terms of m and L.
m = 0.50 kg and L = 0.60 m. Calculate both values.
Calculate Ie/Ic, and check Ie using the parallel axis theorem.
Show worked answer and scoring
(a) For point masses I = Σmr². About the centre each ball is L/2 away: Ic = 2m(L/2)² = ½mL². About one end, one ball is on the axis and the other is L away: Ie = mL².
(b) Ic = ½ × 0.50 × 0.60² = 0.090 kg·m². Ie = 0.50 × 0.60² = 0.18 kg·m².
(c) Ratio = 0.18/0.090 = 2. Parallel axis: Ie = Ic + Md² = 0.090 + 1.0 × 0.30² = 0.18 kg·m² ✓ (total mass 1.0 kg, centre 0.30 m from the end).
Scoring (6 points): (a) 2 points: 1 point for each expression. (b) 2 points: 1 point for each value with units. (c) 2 points: 1 point for the ratio 2, 1 point for a correct parallel axis check.
2. Translation Between Representations (6 points)
Lena slides a 0.20 kg bead along a light rod that spins about one end. She treats the bead as a point mass at distance r from the axis.
ABCD
In words: Four graphs of rotational inertia I against r. A is a straight line from the origin. B starts at the origin and curves upward more and more steeply. C is a flat line. D rises steeply at first and then levels off.
Which graph (A to D) shows I against r? Justify.
What should Lena put on the horizontal axis to make the graph a straight line, and what is its slope?
Calculate I at r = 0.30 m and r = 0.60 m, and explain how the ratio matches your graph.
Show worked answer and scoring
(a)B. I = mr², a parabola through the origin that gets steeper as r grows. A would be I ∝ r, D would be I ∝ √r.
(b) Plot I against r²: a straight line through the origin with slope = m = 0.20 kg.
(c) I = 0.20 × 0.30² = 0.018 kg·m²; I = 0.20 × 0.60² = 0.072 kg·m². Doubling r makes I four times larger, the upward curve of B.
Scoring (6 points): (a) 2 points: 1 point for B, 1 point for I = mr². (b) 2 points: 1 point for r², 1 point for slope = m. (c) 2 points: 1 point for both values, 1 point for the factor of 4 link.
3. Experimental Design and Analysis (6 points)
Kwame clamps two 0.10 kg masses on a light cross bar of a turntable, both at distance r from the axis. A string around the spindle applies a constant torque of 0.050 N·m, and a rotary motion sensor reads the angular acceleration α.
r (m)
0.05
0.10
0.15
0.20
α (rad/s²)
11.1
8.3
5.9
4.2
In words: r of 0.05, 0.10, 0.15 and 0.20 metres gave alpha of 11.1, 8.3, 5.9 and 4.2 radians per second squared.
Describe how he could make sure the torque stays the same for every run.
What should he graph to get a straight line? What do the slope and intercept mean?
Use the data to find the slope and the rotational inertia of the empty turntable.
Show worked answer and scoring
(a) Use the same hanging mass and the same spindle radius each run, keep the string horizontal off the spindle, and measure α over the same short time so the string tension barely changes. Repeat each run.
(b) I = τ/α and I = I0 + 2mr². Graph τ/α against r²: a straight line with slope = 2m (0.20 kg) and intercept = I0, the turntable alone.
Scoring (6 points): (a) 2 points: 1 point for the same hanging mass and spindle radius, 1 point for repeats. (b) 2 points: 1 point for τ/α against r², 1 point for slope 2m and intercept I0. (c) 2 points: 1 point for the slope near 0.20 kg, 1 point for I0 near 0.0040 kg·m².
Omar has a thin hoop and a solid disk, each of mass 2.0 kg and radius 0.25 m. He claims: “Same mass, same radius, so they have the same rotational inertia about their centres.”
Explain in words why Omar is wrong, and which object has the larger rotational inertia.
Calculate both rotational inertias.
Find the radius a 2.0 kg solid disk would need to match the hoop's rotational inertia, and connect the answer to (a).
Show worked answer and scoring
(a) Rotational inertia depends on where the mass is, not just how much. All the hoop's mass is at the rim, but much of the disk's mass is near the centre, so the hoop has more.
(b) Hoop: I = MR² = 2.0 × 0.25² = 0.125 kg·m². Disk: I = ½MR² = 0.0625 kg·m², half as much.
(c) ½M R′² = MR² gives R′ = √2 R = 1.41 × 0.25 ≈ 0.35 m. The disk must spread its mass out farther to match the hoop, which is the idea in (a): mass far from the axis counts more.
Scoring (6 points): (a) 2 points: 1 point for mass distribution mattering, 1 point for hoop larger. (b) 2 points: 1 point for each value. (c) 2 points: 1 point for 0.35 m, 1 point for the link to (a).
Common mistakes
The mistake: "same mass means same rotational inertia".
Why it is wrong: I depends on where the mass is relative to the axis (r²), not just how much.
How to spot it: compare where the mass sits: rim (hoop) beats spread out (disk) beats packed in (sphere).
The mistake: measuring r from the end of the object, or using the diameter.
Why it is wrong: r is the distance from the axis of rotation to the mass.
How to spot it: mark the axis first, then measure every r from it.
The mistake: treating I as a fixed property, like mass.
Why it is wrong: the same object has different I about different axes (rod: ¹⁄₁₂ML² vs ⅓ML²).
How to spot it: the question names a new axis or a new pivot point.
The mistake: forgetting to square r (so doubling r "doubles" I).
Why it is wrong: I = Σmr², so distance matters as its square.
How to spot it: ratio questions: a factor of 2 in r should give 4 in I (for that mass).