Physics 1 · Course Topics
Rotational Motion and Rigid Body Dynamics
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In 30 seconds
Every linear quantity has a rotational analog. Position x becomes angle θ. Velocity v becomes angular velocity ω. Acceleration a becomes angular acceleration α. Force F becomes torque τ. Mass m becomes moment of inertia I. And Newton's second law F = ma becomes τ= Iα.
The table below summarizes the complete analogy between linear and rotational quantities — a powerful tool for reasoning about rotational problems. Once you understand this mapping, you can often translate a linear problem into its rotational counterpart and apply the same logic.
Linear ↔ Rotational Analogy Table
| Linear Quantity | Symbol | SI Unit | Rotational Quantity | Symbol | SI Unit |
|---|---|---|---|---|---|
| Position | x | m | Angular position | θ | rad |
| Displacement | Δx | m | Angular displacement | Δθ | rad |
| Velocity | v = dx/dt | m/s | Angular velocity | ω= dθ/dt | rad/s |
| Acceleration | a = dv/dt | m/s² | Angular acceleration | α= dω/dt | rad/s² |
| Mass | m | kg | Moment of inertia | I | kg·m² |
| Force | F | N | Torque | τ | N·m |
| Newton's 2nd Law | F = ma | — | Rotational 2nd Law | τ= Iα | — |
| Momentum | p = mv | kg·m/s | Angular momentum | L = Iω | kg·m²/s |
| Kinetic energy | K = 12mv2 | J | Rotational KE | Krot = 12Iω2 | J |
| Work | W = FΔx | J | Rotational work | W = τΔθ | J |
| Power | P = Fv | W | Rotational power | P = τω | W |
| Impulse | J = FΔt | N·s | Angular impulse | ΔL = τΔt | N·m·s |
Key Assumptions of Rigid Body Rotation
Throughout this topic we assume:
- Rigid body: The object does not deform — distances between any two points in the body remain fixed. This means we never worry about internal vibrations, bending, or stretching.
- Fixed axis of rotation: The axis around which the object rotates does not move or tilt (unless we explicitly consider rolling, where the axis translates but maintains its orientation). The kinematics and dynamics simplify enormously under this assumption.
- Ideal constraints: When we say "rolling without slipping," we assume perfect static friction that prevents sliding — but does no work (the point of contact is instantaneously at rest relative to the surface).
- Point of application matters: Unlike linear forces, where you can slide a force vector along its line of action, the point where a force is applied to a rotating body determines the torque it produces.
These assumptions hold well for most everyday rotating objects — bicycle wheels, flywheels, doors, and rolling balls. They break down for fluids, soft bodies, or objects rotating around shifting axes, which require more advanced treatments.
ELI-10: Explain It Like I'm 10
Think of a seesaw. Sitting farther from the center makes you "heavier" in terms of rotating the seesaw — even if you weigh the same. That is torque: force times distance from the pivot. A spinning ice skater who pulls their arms in speeds up because their "spread-out-ness" (moment of inertia) decreases while their "spin-energy" stays the same. Rotation follows the same logic as straight-line motion, just wrapped around a circle.
Why this matters
Most real-world motion involves rotation — wheels, gears, turbines, planets, and molecules all spin. Rotational dynamics mirrors linear dynamics with a parallel set of quantities (angle ↔ position, torque ↔ force, moment of inertia ↔ mass). Understanding rotation is essential for engineering, astronomy, and quantum mechanics.
The college version
Big Picture
Every linear quantity has a rotational analog. Position x becomes angle θ. Velocity v becomes angular velocity ω. Acceleration a becomes angular acceleration α. Force F becomes torque τ. Mass m becomes moment of inertia I. And Newton's second law F = ma becomes τ= Iα.
The table below summarizes the complete analogy between linear and rotational quantities — a powerful tool for reasoning about rotational problems. Once you understand this mapping, you can often translate a linear problem into its rotational counterpart and apply the same logic.
Linear ↔ Rotational Analogy Table
| Linear Quantity | Symbol | SI Unit | Rotational Quantity | Symbol | SI Unit |
|---|---|---|---|---|---|
| Position | x | m | Angular position | θ | rad |
| Displacement | Δx | m | Angular displacement | Δθ | rad |
| Velocity | v = dx/dt | m/s | Angular velocity | ω= dθ/dt | rad/s |
| Acceleration | a = dv/dt | m/s² | Angular acceleration | α= dω/dt | rad/s² |
| Mass | m | kg | Moment of inertia | I | kg·m² |
| Force | F | N | Torque | τ | N·m |
| Newton's 2nd Law | F = ma | — | Rotational 2nd Law | τ= Iα | — |
| Momentum | p = mv | kg·m/s | Angular momentum | L = Iω | kg·m²/s |
| Kinetic energy | K = 12mv2 | J | Rotational KE | Krot = 12Iω2 | J |
| Work | W = FΔx | J | Rotational work | W = τΔθ | J |
| Power | P = Fv | W | Rotational power | P = τω | W |
| Impulse | J = FΔt | N·s | Angular impulse | ΔL = τΔt | N·m·s |
Key Assumptions of Rigid Body Rotation
Throughout this topic we assume:
- Rigid body: The object does not deform — distances between any two points in the body remain fixed. This means we never worry about internal vibrations, bending, or stretching.
- Fixed axis of rotation: The axis around which the object rotates does not move or tilt (unless we explicitly consider rolling, where the axis translates but maintains its orientation). The kinematics and dynamics simplify enormously under this assumption.
- Ideal constraints: When we say "rolling without slipping," we assume perfect static friction that prevents sliding — but does no work (the point of contact is instantaneously at rest relative to the surface).
- Point of application matters: Unlike linear forces, where you can slide a force vector along its line of action, the point where a force is applied to a rotating body determines the torque it produces.
These assumptions hold well for most everyday rotating objects — bicycle wheels, flywheels, doors, and rolling balls. They break down for fluids, soft bodies, or objects rotating around shifting axes, which require more advanced treatments.
ELI-10: Explain It Like I'm 10
Think of a seesaw. Sitting farther from the center makes you "heavier" in terms of rotating the seesaw — even if you weigh the same. That is torque: force times distance from the pivot. A spinning ice skater who pulls their arms in speeds up because their "spread-out-ness" (moment of inertia) decreases while their "spin-energy" stays the same. Rotation follows the same logic as straight-line motion, just wrapped around a circle.
5.1 Rotational Kinematics
Core Idea
Rotational kinematics describes angular position, velocity, and acceleration using the same structure as linear kinematics. The equations are identical in form — just with angular symbols replacing linear ones.
Physics and Mathematics
- Angular position θ: angle relative to a reference line. SI unit: radian (rad). 2π rad = 360°.
- Angular displacement Δθ= θf - θi.
- Angular velocity ω= dθdt. SI unit: rad/s.
- Angular acceleration α= dωdt. SI unit: rad/s².
Relation to linear quantities (for a point at distance r from the axis): s = rθ, vt = rω, at = rα
Where vt is tangential speed and at is tangential acceleration.
Constant angular acceleration equations mirror the linear kinematic equations:
ωf = ωi + αt Δθ= ωi t + 12αt2 ωf2 = ωi2 + 2αΔθ
The full analogy between the linear and rotational kinematic equations is:
| Linear Kinematics | Rotational Kinematics |
|---|---|
| vf = vi + at | ωf = ωi + αt |
| Δx = vi t + 12at2 | Δθ= ωi t + 12αt2 |
| vf2 = vi2 + 2aΔx | ωf2 = ωi2 + 2αΔθ |
| Δx = 12(vi + vf)t | Δθ= 12(ωi + ωf)t |
Centripetal acceleration. Any point moving in a circle at constant speed still accelerates — toward the center:
ac = v2r = rω2
This is NOT the same as tangential acceleration at = rα. Tangential acceleration changes the speed of rotation; centripetal acceleration keeps the object on the circular path. They are perpendicular to each other.
ELI-10: Explain It Like I'm 10
Imagine a merry-go-round. One kid sits near the center; another sits at the edge. They both complete one circle in the same time — they have the same angular speed. But the kid at the edge has a much faster linear speed (they travel a bigger circle in the same time). Rotational speed is the same for everyone on the ride; linear speed depends on how far out you are.
5.2 Moment of Inertia
Core Idea
Moment of inertia I is the rotational analog of mass. It measures how hard it is to change an object's angular velocity. Unlike mass, which is a fixed property of an object, moment of inertia depends on both the mass AND how that mass is distributed relative to the axis of rotation.
Physics and Mathematics
For a point mass: I = mr2
For a system of point masses: I = ∑mi ri2
For a continuous body: I = ∫r2 dm
The key insight: mass farther from the axis contributes disproportionately more to the moment of inertia because of the r2 dependence. Doubling the distance quadruples the contribution of that mass element.
Common moments of inertia (about center of mass):
- Solid sphere: I = 25MR2
- Solid cylinder/disk: I = 12MR2
- Thin rod (center): I = 112ML2
- Thin rod (end): I = 13ML2
- Hoop: I = MR2
- Spherical shell: I = 23MR2
Parallel-Axis Theorem: I = Icm + Md2, where d is the distance from the center of mass to the new axis. This is why a rod rotated about its end has a larger I (13ML2) than the same rod rotated about its center (112ML2): mass is, on average, farther from the end-axis.
Proportional Reasoning
Understanding how changes in I or ω affect other quantities is essential for qualitative reasoning:
Scenario 1: Halving the moment of inertia. If an isolated spinning object's moment of inertia is suddenly halved (I → I/2), conservation of angular momentum (L = Iω= constant) forces the angular velocity to double (ω → 2ω). What about kinetic energy?
Krot = 12Iω2 → 12(I2)(2ω)2 = 12 · I2 · 4ω2 = Iω2
The kinetic energy doubles! Where does the extra energy come from? The skater must do work to pull their arms inward against the centrifugal force. This internal work is what increases the rotational kinetic energy.
Scenario 2: Same torque, different I. If the same net torque τ is applied to object A (IA) and object B (IB), then:
αA = τIA, αB = τIB
If IB = 2IA, object B experiences half the angular acceleration. This is the rotational version of "heavier objects accelerate less for the same force."
Scenario 3: Same ω, different I. If two objects spin at the same angular velocity, the one with the larger moment of inertia has:
- Larger angular momentum (L = Iω)
- Larger rotational kinetic energy (K = 12Iω2)
This is why flywheels used for energy storage are designed with most of their mass at the rim — maximizing I for a given mass and size.
ELI-10: Explain It Like I'm 10
A figure skater spinning with arms out spins slowly. When she pulls her arms in, she spins faster. Her mass has not changed, but her moment of inertia has — mass closer to the axis means smaller I. Since angular momentum is conserved, smaller I means larger ω. An object's "resistance to spinning" depends not just on how heavy it is, but on how far the weight is from the spin axis.
5.3 Torque
Core Idea
Torque is the rotational analog of force — it causes angular acceleration.
Physics and Mathematics
τ = r × F τ= rFsinθ
Where:
- τ = torque (N·m)
- r = distance from axis to point of force application (m)
- F = magnitude of force (N)
- θ = angle between r and F
The direction of torque follows the right-hand rule. Positive torque typically causes counterclockwise rotation.
Lever arm (moment arm) r ⊥ = rsinθ is the perpendicular distance from the axis to the line of action of the force. τ= r ⊥ F.
Worked Example: Wrench
Problem: A 50 N force is applied perpendicular to a 0.30 m wrench handle. Find the torque.
Solution: τ= rFsin90°= (0.30)(50) = 15 N·m
If the force were applied at 30° to the handle, τ= (0.30)(50)sin30°= 7.5 N·m — half as effective.
ELI-10: Explain It Like I'm 10
You use a wrench to loosen a bolt. Pushing near the bolt head does almost nothing — you need to push at the end of the handle. Pushing straight along the handle does nothing — you need to push sideways, perpendicular. Torque is the twisting effectiveness of a force: it depends on how hard you push, how far from the pivot you push, and the angle.
5.4 Newton's Second Law for Rotation
Core Idea
The net torque on an object equals its moment of inertia times its angular acceleration.
∑τ= Iα
Where:
- ∑τ = net torque (N·m)
- I = moment of inertia (kg·m²)
- α = angular acceleration (rad/s²)
This is the direct rotational analog of ∑F = ma. Just as net force causes linear acceleration proportional to 1/m, net torque causes angular acceleration proportional to 1/I.
Worked Example: Flywheel Acceleration
Problem: A solid cylindrical flywheel of mass M = 10 kg and radius R = 0.25 m is free to rotate about its central axis. A constant tangential force of F = 20 N is applied at the rim. Find (a) the torque, (b) the moment of inertia, (c) the angular acceleration, and (d) the angular velocity after 4.0 seconds starting from rest.
Solution:
(a) Torque. The force is tangential (θ= 90° at the rim r = R): τ= RFsin90°= (0.25 m)(20 N)(1) = 5.0 N·m
(b) Moment of inertia. For a solid cylinder about its central axis: I = 12MR2 = 12(10 kg)(0.25 m)2 = 12(10)(0.0625) = 0.3125 kg·m2
(c) Angular acceleration: α= τI = 5.00.3125 = 16 rad/s2
(d) Angular velocity after 4.0 s. From rest (ωi = 0) with constant α: ωf = ωi + αt = 0 + (16)(4.0) = 64 rad/s
In revolutions per second: 642π ≈ 10.2 rev/s — the flywheel spins up to over 10 revolutions per second in just 4 seconds.
Check: If the same flywheel were a hoop (I = MR2 = 0.625 kg·m2, double the solid cylinder's I), the same torque would produce α= 8.0 rad/s2 — half the acceleration. This demonstrates proportional reasoning: doubling I halves α for the same torque.
ELI-10: Explain It Like I'm 10
This is just F=ma for spinning. Net twist = (resistance to twisting) × (how fast the spinning changes). A heavy merry-go-round (big I) needs a bigger push (torque) to speed up than a light one.
5.5 Rotational Kinetic Energy and Rolling
Core Idea
A rotating object has kinetic energy: Krot = 12Iω2
Rolling without slipping: The object both translates and rotates. Ktotal = 12mvcm2 + 12Iω2
The no-slip condition links them: vcm = Rω
For a rolling object on an incline, the acceleration depends on the shape's moment of inertia: a = gsinθ1 + I/(mR2)
A solid sphere (I = 25mR2) reaches the bottom faster than a hoop (I = mR2) because less of the initial potential energy goes into rotation.
Derivation Insight
Why does a depend on I? When an object rolls down an incline without slipping, static friction provides the torque that causes rotation. The energy released by gravity (mgh) must be split between translational kinetic energy (12mv2) and rotational kinetic energy (12Iω2). Using v = Rω:
mgh = 12mv2 + 12I(vR)2 = 12mv2(1 + ImR2)
Solving for v: v = 2gh1 + I/(mR2)
The denominator (1 + I/(mR2)) is always greater than 1, so a rolling object is always slower than a frictionless sliding block (v = 2gh). The larger I/(mR2), the more energy is diverted into rotation and the slower the object rolls.
Worked Example: Solid Sphere vs. Hoop Down an Incline
Problem: A solid sphere (I = 25mR2) and a thin hoop (I = mR2), both of mass m = 2.0 kg and radius R = 0.10 m, roll without slipping from rest down a ramp inclined at θ= 30°, starting from a height h = 1.5 m. Find (a) the speed of each at the bottom, (b) the acceleration of each down the ramp, and (c) the time each takes to reach the bottom.
Solution:
(a) Speed at the bottom. From energy conservation: mgh = 12mv2 + 12Iω2 with ω= v/R.
For the solid sphere, I/(mR2) = 25 = 0.40: vsphere = 2gh1 + 0.40 = 2(9.8)(1.5)1.40 = 29.41.40 = 21.0 ≈ 4.58 m/s
For the hoop, I/(mR2) = 1.0: vhoop = 2gh1 + 1.0 = 29.42.0 = 14.7 ≈ 3.83 m/s
The solid sphere is about 20% faster — because only 40% of its energy goes into rotation vs. 50% for the hoop.
(b) Acceleration down the ramp. Using a = gsinθ1 + I/(mR2):
For the solid sphere: asphere = (9.8)sin30°1.40 = 4.91.40 ≈ 3.50 m/s2
For the hoop: ahoop = 4.92.0 = 2.45 m/s2
(c) Time to reach the bottom. The ramp length is L = h / sinθ= 1.5 / 0.5 = 3.0 m. From rest with constant acceleration: L = 12at2 ⇒ t = 2L/a.
For the solid sphere: tsphere = 2(3.0)3.50 = 1.714 ≈ 1.31 s
For the hoop: thoop = 2(3.0)2.45 = 2.449 ≈ 1.56 s
The solid sphere wins by about a quarter-second. This is the quantitative version of "solid balls beat hollow ones down a ramp."
ELI-10: Explain It Like I'm 10
Roll a solid ball and a hollow ball (same size and weight) down a ramp. The solid ball wins. Why? Both start with the same stored gravitational energy. But the hollow ball has more of its mass far from the center, so more energy goes into spinning it — leaving less for forward speed. The solid ball converts more energy into forward motion.
5.6 Angular Momentum
Core Idea
Angular momentum is the rotational analog of linear momentum.
L = Iω
For a point mass: L = m v r sinθ (magnitude), or L = r × p.
Conservation of angular momentum: In the absence of net external torque, L is constant. This explains why spinning skaters speed up when pulling in their arms, why neutron stars spin extremely fast, and why Earth's axis stays (roughly) stable.
Limiting Cases and Extreme Scenarios
Angular momentum conservation produces dramatic effects when I changes by orders of magnitude:
Case 1: Stellar collapse. When a massive star's core collapses from roughly the Sun's radius (R⊙ ≈ 7 × 108 m) to a neutron star radius ( ∼ 104 m), the radius shrinks by a factor of ∼ 7 × 104. Since I ∝ R2, the moment of inertia drops by a factor of ∼ 5 × 109. Conservation of L = Iω forces ω to increase by the same factor. A star rotating once every 30 days can become a neutron star spinning hundreds of times per second. These are pulsars — cosmic lighthouses whose extreme regularity makes them some of the most precise clocks in the universe.
Case 2: Skater's limit. How fast can a figure skater spin? There is a practical limit: even if the skater could reduce I to nearly zero by pulling all mass to the axis, real skaters are limited by (a) how tightly they can pull in their limbs and (b) the friction in the ice and air resistance, which exert small but non-zero external torques. A typical skater reduces their I by roughly a factor of 2–3, corresponding to a spin rate increase from about 2 rev/s to 5–6 rev/s.
Case 3: Zero external torque. In deep space, far from any gravitating body, an object set spinning will maintain both its angular speed AND the direction of its angular momentum vector forever. This is the principle behind reaction wheels and gyroscopes used in spacecraft attitude control — spinning up an internal wheel in one direction causes the spacecraft to rotate in the opposite direction to conserve total angular momentum.
Case 4: The collapsing cloud. A diffuse interstellar gas cloud slowly rotating as it contracts under gravity will spin faster and faster. This is why galaxies, stars, and planetary systems are disk-shaped — the rotation flattens the material into a plane perpendicular to the angular momentum vector. Our solar system is a prime example: all planets orbit in roughly the same plane and in the same direction because they formed from a single rotating protoplanetary disk.
Where conservation breaks down. Angular momentum is only conserved when net external torque is zero. Common situations where it is NOT conserved:
- A door pushed by your hand (external torque from the hinges and your hand changes L)
- A spinning top slowing down (friction at the tip exerts external torque)
- A car braking while turning (friction from the road applies torque)
ELI-10: Explain It Like I'm 10
Angular momentum is "spinning momentum." Once something is spinning in empty space, it keeps spinning forever unless something twists it. A star collapsing into a neutron star shrinks to a tiny size — its moment of inertia drops enormously, so its spin rate increases enormously (like the skater). This is conservation of angular momentum on a cosmic scale.
Topic Summary
- Rotational kinematics mirrors linear kinematics: θ, ω, α replace x, v, a. s = rθ, v = rω, at = rα.
- Centripetal acceleration ac = v2/r = rω2 keeps objects on circular paths; it is perpendicular to tangential acceleration.
- Moment of inertia I depends on mass distribution relative to the axis. I = ∑mi ri2. Mass farther from the axis contributes more (quadratically).
- Torque τ= rFsinθ is the rotational effect of a force. The lever arm r ⊥ = rsinθ is the perpendicular distance from axis to force line.
- Newton's second law for rotation: ∑τ= Iα. Same torque produces less angular acceleration for larger I.
- Rotational kinetic energy: Krot = 12Iω2.
- Rolling without slipping: vcm = Rω; total K = 12mv2 + 12Iω2. The acceleration down an incline is a = gsinθ1 + I/(mR2) — objects with larger I/(mR2) roll slower.
- Angular momentum L = Iω is conserved without net external torque. This principle explains phenomena from spinning skaters to pulsars to galaxy formation.
- Assumptions: Rigid body, fixed axis, perfect rolling constraints. These hold for most everyday rotating systems.
Essential Equations
| Equation | Name |
|---|---|
| vt = rω | Tangential velocity |
| at = rα | Tangential acceleration |
| ac = v2r = rω2 | Centripetal acceleration |
| τ= rFsinθ= r ⊥ F | Torque |
| ∑τ= Iα | Rotational Newton's Second Law |
| I = ∑mi ri2 | Moment of inertia (discrete) |
| I = Icm + Md2 | Parallel-axis theorem |
| Krot = 12Iω2 | Rotational kinetic energy |
| Ktotal = 12mv2 + 12Iω2 | Rolling kinetic energy |
| a = gsinθ1 + I/(mR2) | Rolling acceleration on incline |
| L = Iω | Angular momentum |
| ΔL = τΔt | Angular impulse |
Concept Check
- A solid disk and a hoop of equal mass and radius roll down an incline. Which reaches the bottom first? Why?
- Why does a tightrope walker carry a long pole?
- If the Earth's radius suddenly shrank by half while keeping the same mass, how would the length of a day change?
- You push on a door near the hinges vs. near the handle. Same force. Which produces more torque? Why?
- A spinning figure skater pulls her arms in. Her angular velocity increases. Does her kinetic energy increase? If so, where does the extra energy come from?
- A solid sphere and a hollow sphere of equal mass and radius roll down the same incline. By what factor does the solid sphere's acceleration exceed the hollow sphere's? (Hint: Isolid = 25mR2, Ihollow = 23mR2.)
Open Educational References
- OpenStax, College Physics, Chapter 10: Rotational Motion and Angular Momentum
- OpenStax, University Physics, Volume 1, Chapters 10–11: Rotation, Angular Momentum

Eli explains
The same idea, in plain words
Explain it like I’m 10
ELI-10: Explain It Like I'm 10
Think of a seesaw. Sitting farther from the center makes you "heavier" in terms of rotating the seesaw — even if you weigh the same. That is torque: force times distance from the pivot. A spinning ice skater who pulls their arms in speeds up because their "spread-out-ness" (moment of inertia) decreases while their "spin-energy" stays the same. Rotation follows the same logic as straight-line motion, just wrapped around a circle.
ELI-10: Explain It Like I'm 10
Imagine a merry-go-round. One kid sits near the center; another sits at the edge. They both complete one circle in the same time — they have the same angular speed. But the kid at the edge has a much faster linear speed (they travel a bigger circle in the same time). Rotational speed is the same for everyone on the ride; linear speed depends on how far out you are.
ELI-10: Explain It Like I'm 10
A figure skater spinning with arms out spins slowly. When she pulls her arms in, she spins faster. Her mass has not changed, but her moment of inertia has — mass closer to the axis means smaller I. Since angular momentum is conserved, smaller I means larger ω. An object's "resistance to spinning" depends not just on how heavy it is, but on how far the weight is from the spin axis.
ELI-10: Explain It Like I'm 10
You use a wrench to loosen a bolt. Pushing near the bolt head does almost nothing — you need to push at the end of the handle. Pushing straight along the handle does nothing — you need to push sideways, perpendicular. Torque is the twisting effectiveness of a force: it depends on how hard you push, how far from the pivot you push, and the angle.
ELI-10: Explain It Like I'm 10
This is just F=ma for spinning. Net twist = (resistance to twisting) × (how fast the spinning changes). A heavy merry-go-round (big I) needs a bigger push (torque) to speed up than a light one.
ELI-10: Explain It Like I'm 10
Roll a solid ball and a hollow ball (same size and weight) down a ramp. The solid ball wins. Why? Both start with the same stored gravitational energy. But the hollow ball has more of its mass far from the center, so more energy goes into spinning it — leaving less for forward speed. The solid ball converts more energy into forward motion.
ELI-10: Explain It Like I'm 10
Angular momentum is "spinning momentum." Once something is spinning in empty space, it keeps spinning forever unless something twists it. A star collapsing into a neutron star shrinks to a tiny size — its moment of inertia drops enormously, so its spin rate increases enormously (like the skater). This is conservation of angular momentum on a cosmic scale.
ELI-10 Final Recap
Rotation is straight-line motion wrapped around a circle. Everything you learned about position, velocity, acceleration, force, mass, and momentum has a spinning twin. The "spread-out-ness" of mass — moment of inertia — replaces mass. Torque — twist — replaces force. Angular momentum replaces regular momentum.
The same conservation rules apply: no external twist means spinning momentum stays the same. That is why skaters spin faster when they pull in, why planets orbit smoothly, and why collapsing stars become the fastest-spinning objects in the universe. Rolling things are a mix: they move forward AND spin. A rolling object's shape matters — solid balls beat hollow ones down a ramp because they spend less energy on spinning and more on moving. And just as heavier objects resist changes in linear motion, objects with mass farther from the axis resist changes in rotation — the ultimate reason a tightrope walker's long pole makes balancing possible.
Worked example
Worked Example: Wrench
Problem: A 50 N force is applied perpendicular to a 0.30 m wrench handle. Find the torque.
Solution: τ= rFsin90°= (0.30)(50) = 15 N·m
If the force were applied at 30° to the handle, τ= (0.30)(50)sin30°= 7.5 N·m — half as effective.
Worked Example: Flywheel Acceleration
Problem: A solid cylindrical flywheel of mass M = 10 kg and radius R = 0.25 m is free to rotate about its central axis. A constant tangential force of F = 20 N is applied at the rim. Find (a) the torque, (b) the moment of inertia, (c) the angular acceleration, and (d) the angular velocity after 4.0 seconds starting from rest.
Solution:
(a) Torque. The force is tangential (θ= 90° at the rim r = R): τ= RFsin90°= (0.25 m)(20 N)(1) = 5.0 N·m
(b) Moment of inertia. For a solid cylinder about its central axis: I = 12MR2 = 12(10 kg)(0.25 m)2 = 12(10)(0.0625) = 0.3125 kg·m2
(c) Angular acceleration: α= τI = 5.00.3125 = 16 rad/s2
(d) Angular velocity after 4.0 s. From rest (ωi = 0) with constant α: ωf = ωi + αt = 0 + (16)(4.0) = 64 rad/s
In revolutions per second: 642π ≈ 10.2 rev/s — the flywheel spins up to over 10 revolutions per second in just 4 seconds.
Check: If the same flywheel were a hoop (I = MR2 = 0.625 kg·m2, double the solid cylinder's I), the same torque would produce α= 8.0 rad/s2 — half the acceleration. This demonstrates proportional reasoning: doubling I halves α for the same torque.
Worked Example: Solid Sphere vs. Hoop Down an Incline
Problem: A solid sphere (I = 25mR2) and a thin hoop (I = mR2), both of mass m = 2.0 kg and radius R = 0.10 m, roll without slipping from rest down a ramp inclined at θ= 30°, starting from a height h = 1.5 m. Find (a) the speed of each at the bottom, (b) the acceleration of each down the ramp, and (c) the time each takes to reach the bottom.
Solution:
(a) Speed at the bottom. From energy conservation: mgh = 12mv2 + 12Iω2 with ω= v/R.
For the solid sphere, I/(mR2) = 25 = 0.40: vsphere = 2gh1 + 0.40 = 2(9.8)(1.5)1.40 = 29.41.40 = 21.0 ≈ 4.58 m/s
For the hoop, I/(mR2) = 1.0: vhoop = 2gh1 + 1.0 = 29.42.0 = 14.7 ≈ 3.83 m/s
The solid sphere is about 20% faster — because only 40% of its energy goes into rotation vs. 50% for the hoop.
(b) Acceleration down the ramp. Using a = gsinθ1 + I/(mR2):
For the solid sphere: asphere = (9.8)sin30°1.40 = 4.91.40 ≈ 3.50 m/s2
For the hoop: ahoop = 4.92.0 = 2.45 m/s2
(c) Time to reach the bottom. The ramp length is L = h / sinθ= 1.5 / 0.5 = 3.0 m. From rest with constant acceleration: L = 12at2 ⇒ t = 2L/a.
For the solid sphere: tsphere = 2(3.0)3.50 = 1.714 ≈ 1.31 s
For the hoop: thoop = 2(3.0)2.45 = 2.449 ≈ 1.56 s
The solid sphere wins by about a quarter-second. This is the quantitative version of "solid balls beat hollow ones down a ramp."
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