Motion Dual Reading Pack The Dynamics of Rotation
Advanced Comparative Mechanics
A rigorous conceptual exploration into the physical intuition and mathematical symmetries that govern translational and rotational motion. This volume examine inertia distribution, the mechanical advantage of torque, the energetics of non-slipping contact, and momentum preservation.
Kinematic Symmetry Inertia Quadratic Torque Efficiency Conservation Systems
Technical Resource 04-P
01
Kinematic Symmetry
1. The Displacement Bridge
The radian is a dimensionless unit defined by the ratio of arc length (\(s\)) to radius (\(r\)). This definition allows us to translate angular motion into physical ground distance. The bridge equation \(s = r\theta\) reveals a physical truth: every point on a rotating disk completes the same turn, but the physical path covered depends entirely on the distance from the axle.
s = r θ
2. The Motion Mirror
Every translational variable has a rotational "twin." Displacement (\(s\)) maps to \(\theta\), velocity (\(v\)) maps to \(\omega\), and acceleration (\(a\)) maps to \(\alpha\). Because the bridge equations link these variables by a constant factor (\(r\)), the calculus of motion remains identical across both worlds.
Translational (m, m/s)
v = u + at
s = ut + ½at²
v² = u² + 2as
Rotational (rad, rad/s)
ω = ω0 + αt
θ = ω0t + ½αt²
ω² = ω0² + 2αθ
02
Mass & Inertia
Linear Mass vs. Rotational Resistance
In translation, mass (\(m\)) is an intrinsic property—a 10kg block resists change the same way regardless of the direction of the force. In rotation, resistance to change depends not only on how much mass exists, but how far it is from the axis. This is the Moment of Inertia (\(I\)) .
I = ∑ m r²
Radius compounds resistance quadratically
The r² Compounding Effect
Radius is squared because distance amplifies two distinct physical factors simultaneously: the physical speed of that mass (for any spin rate) and the mechanical disadvantage of trying to stop it. Moving mass twice as far from the axis doesn't just double the difficulty of rotation—it quadruples it.
Shape Factor (k)
Solid Sphere: 0.4 MR²
Solid Disk: 0.5 MR²
Hollow Hoop: 1.0 MR²
03
Torque Dynamics
The Architecture of the Twist
Torque (\(\tau\)) is the rotational analogue of Force. However, force only tells you how hard you push. Torque tells you how effective that push is at creating rotation. A torque is a vector quantity defined by the cross product \(\vec{\tau} = \vec{r} \times \vec{F}\).
τ = r F sin θ
Twist Effectiveness Formula
τnet = I α
Lever Arm (Moment Arm)
The Line of Action is the infinite line passing through the force vector. The Lever Arm is the perpendicular distance from the pivot to this line. Increasing the lever arm increases the "turning advantage."
The Sine Component
Only the tangential component of force creates rotation. If you push directly toward the axis (\(\theta=0\)), sine is zero, resulting in zero torque. Torque is maximized when the force is applied perpendicularly (\(90^\circ\)).
04
The Energy Budget
Why Conserve Energy?
In a rolling system (without slipping), total mechanical energy is conserved because static friction does no work . The point of contact between the wheel and the floor is momentarily at rest relative to the surface. Since work is Force \(\times\) Displacement, and there is no displacement at the point of application, friction acts as a constraint, not an energy sink.
"Potential energy is a finite currency. A rolling object must pay a 'spin tax' (Krot), leaving less for forward speed (Ktrans)."
mgh = ½mv² + ½Iω²
Angular Impulse
A torque applied over a time interval changes total spin-momentum: \(\tau \Delta t = \Delta L\). This builds up the rotation speed.
Angular Work
When a torque rotates an object through an angle, energy is transferred: \(W = \tau \theta = \Delta K_{rot}\). This work changes the object's rotational energy.
05
Momentum Systems
1. The Skater Paradox
Angular Momentum (\(L = I\omega\)) is perfectly conserved in isolated systems. When a skater tucks their arms in, their Moment of Inertia (\(I\)) drops. Since \(L\) is locked, the angular velocity \(\omega\) must skyrocket.
Internal Work Insight: Kinetic energy increases during the tuck because the skater's muscles perform chemical work to pull mass inward against centrifugal force. Internal work increases energy, but preserves momentum.
2. The Clay-Rod Collision
Moving objects possess Orbital Angular Momentum relative to a pivot point (\(L = mvr\)). When clay strikes a rod and sticks, the linear momentum is "captured" by the pivot and converted into rotation.
Initial Linear Momentum \(\rightarrow\) Final Rotational Momentum
mvr = (Irod + mr²)ωf
I1 ω1 = I2 ω2
Isolated System Momentum Conservation
Motion Analysis Worksheet Motion Analysis Worksheet
Part A: Mapping & Kinematics (Q1-Q5)
Name:
Date:
01 Symmetry Map
Linear Angular SI Unit (Angular) Displacement (s) Velocity (v) Inertia / Mass (m) Newton's 2nd Law Kinetic Energy
2. A centrifuge (r=0.15m) rest to 1,200 rad/s in 30s. Calculate angular acceleration (\(\alpha\)) and final tangential speed (\(v\)).
3. A wheel rotates from rest through 50 radians in 4 seconds. Find \(\alpha\) and the linear distance swept by the rim (r=0.5m).
4. A motor at 3,000 RPM brakes at -150 rad/s². How long until it stops?
5. A clock hand of length 0.2m moves at constant speed. Find its angular velocity (ω) and tip speed (v).
Part B: Mass, Inertia & Torque (Q6-Q10)
6. Point Mass: Calculate the inertia of a 0.5kg ball attached to a 2m string pivoted at the end.
7. Pivot Shift: Calculate the inertia of a 2kg, 1.0m rod pivoted at its end (\(I = \frac{1}{3}ML^2\)).
8. Shape Factor: A 2kg disk and 2kg hoop (r=0.4m) start a race. Calculate both Moments of Inertia.
9. Wrench: Apply 60N at 45° to a 0.5m wrench. Calculate the net Torque produced.
10. Net Torque: A 2m rod is pivoted at center. 10N up on left (90°), 15N down on right (30°). Find \(\tau_{net}\).
Part C: The Energy Budget (Q11-Q15)
11. Angular Impulse: A torque of 25 Nm is applied for 4s. Calculate the change in momentum (\(\Delta L\)).
12. Angular Work: A motor turns a flywheel through 10 revs with 5 Nm torque. Find the work done.
13. Energy Partition: Calculate the rotational KE of a 10kg disk (r=0.2m) rolling at 5 m/s.
14. Speed Ratio: A solid sphere and a disk race. What is the ratio of their final speeds (\(v_{sph}/v_{disk}\))?
15. Potential to Rotation: A 2kg block falls 3m, turning a massless pulley. Find final \(\Delta K_{rot}\).
Part D: Conservation & Challenges (Q16-Q20)
Motion Mashup Answer Key Answer Key: Q1-Q10
Technical Solutions Part A & B
Teacher Key
1. Symmetry: Displacement (θ, rad, s=rθ) | Velocity (ω, rad/s, v=rω) | Accel (α, rad/s², a=rα) | Inertia (I, kgm², I=∑mr²) | N2L (τ=Iα, Nm, τ=rFsinθ). K: ½Iω² (Joules).
2. Centrifuge: α = 1200 / 30 = 40 rad/s² . v = (0.15)(1200) = 180 m/s .
3. Wheel: α = 2θ / t² = 100 / 16 = 6.25 rad/s² . s = (0.5)(50) = 25 meters .
4. Drill: 3000 RPM = 100π rad/s. t = Δω/α = 100π / 150 ≈ 2.09 seconds .
5. Clock: ω = 2π / 60 = 0.105 rad/s. v = (0.2)(0.105) = 0.021 m/s .
6. Point Mass: I = mr² = (0.5)(2)² = 2.0 kg·m² .
7. End Pivot: I = 1/3(2)(1)² = 0.67 kg·m² .
8. Shape: Disk I = 0.5(2)(0.16) = 0.16. Hoop I = (2)(0.16) = 0.32 kg·m².
9. Wrench: τ = rFsinθ = (0.5)(60)sin(45) = 21.2 N·m .
10. Net Torque: τ1 = 1(10)sin90 = 10 (CCW). τ2 = 1(15)sin30 = 7.5 (CW). τnet = 2.5 N·m (CCW) .
Answer Key: Q11-Q20
Technical Solutions Part C & D
11. Impulse: ΔL = τΔt = (25)(4) = 100 kg·m²/s .
12. Work: W = τθ = (5)(10 × 2π) = 100π J (~314.2 J) .
13. Rolling KE: Krot = 0.5(0.5MR²)(v/R)² = 0.25mv² = 0.25(10)(25) = 62.5 J .
14. Budget: v = √(2gh/(1+k)) = √(200/1.4) ≈ 11.95 m/s .
15. Ratio: Hoop k=1. Krot/Ktot = k/(1+k) = 1/2 = 50% .
16. Skater: (4)(3) = (1.5)ωf → ωf = 8.0 rad/s . KE up because muscle work.
17. Collision: Lin = (0.5)(10)(1) = 5. Itot = 1/3(3)(1) + 0.5(1) = 1.5. ωf = 5/1.5 = 3.33 rad/s .
18. Collapse: L1=L2 → MR²ω1 = M(R/2)²ω2 → ω2 = 4ω1. Period Ratio = 1/4 .
19. Paradox: L conserved because zero external torque. Energy up because internal work is done.
20. Challenge: a = g(m1 - m2) / (m1 + m2 + 0.5M). Atwood logic with pulley inertia term.
Motion Mashup Slides The Unified Map
Translational and Rotational Duality
The Basics: Linear vs. Angular
Displacement
s = r θ
m vs. rad
Velocity
v = r ω
m/s vs. rad/s
Acceleration
a = r α
m/s² vs. rad/s²
Moment of Inertia: Mass Distribution
The Formula
I = ∑ mr²
Resistance to rotation depends on distance from the pivot .
The r² impact
Moving mass twice as far makes the object four times harder to stop.
Shape Coefficients
Sphere (0.4) vs. Disk (0.5) vs. Hoop (1.0).
Torque: The Law of Spin
The Turning Force
τ = r F sin θ
τ = I α
The Lever Arm (r)
Distance from the pivot increases torque exponentially.
The Angle (θ)
Only the perpendicular component causes rotation.
Rotational Kinetic Energy
Energy in Spin
Kr = ½ Iω²
Mirrors linear energy (½ mv²)
Mechanical Energy
E = Kt + Kr + Ug
Rolling objects "spend" potential energy on spin. The winner is the object that rotates with the least resistance.
Conservation: Momentum & Collisions
The Skater Effect
I1ω1 = I2ω2
Reducing \(I\) forces \(\omega\) to increase to keep momentum constant.
Angular Collisions
mvr = (Irod + mr²)ω
Linear objects contribute orbital angular momentum relative to a pivot.
Motion Summary Cornell Notes Page Reading Notes: Kinematics
Part 1: The Bridges & Mirror Equations
Name:
Date:
Conceptual Cues
Geometric role of the radian.
Tangential speed vs. spin rate.
Translational to Rotational Mirror.
Elaborate Reading Notes
I. The Radian Foundation
Elaborate on why the radian is a 'dimensionless necessity' for cross-domain physics:
II. The Tangential velocity Link
Explain why the rim of a fan is physically faster than the hub at the same spin rate:
Practice A
A wheel (r=0.5m) reaches 1,200 rad/s in 30s. Find linear distance s.
Summary
Summarize how the radius acts as a scale factor between the two domains.
Reading Notes: Resistance
Part 2: Mass vs. Inertia & Torque Dynamics
Conceptual Cues
Why does mass distribution matter?
The "Quadratic" nature of r.
Line of Action vs Lever Arm.
Torque Efficiency Components.
Elaborate Reading Notes
III. Mass vs. Rotational Inertia (I)
Contrast linear mass with the Moment of Inertia (\(I = \sum mr^2\)):
IV. Torque: The Twist Architecture
Elaborate on the Lever Arm distance and the role of the Sine component:
Dynamics Practice
Find torque for 60N applied at 45° to a 0.5m wrench.
Summary
Explain how shape and lever arm combine to determine angular acceleration.
Reading Notes: Conservation
Part 3: Energy Budgets & Momentum Preservation
Conceptual Cues
The "Finite Budget" of rolling motion.
Why static friction does no work.
Skater Work vs. Momentum balance.
Impulse vs. Angular Work.
Reading Elaboration & Synthesis
V. The Rolling Paradox Paradox
Explain the energy split and why friction is a constraint, not a loss:
VI. Angular Impulse & Skater Effect
Detail how internal work changes KE but preserves isolated momentum:
Momentum Practice
I=4.0 spins at 3 rad/s. Tucks arms to I=1.5. Find new speed.
Summary
Synthesize the budget paradox: Why does shape matter more than mass for speed?
Motion Cornell Notes Teacher Key Cornell Key: Foundations
Elaborate Reading Summary
Teacher Reference
Conceptual Cues
Radian: Dimensionless ratio (m/m) allowing cross-domain mapping without arbitrary conversion factors.
Rim Speed: Outer points cover longer arcs in the same time interval, requiring physical speed increase.
Mirroring: Bridge equations link variables linearly, resulting in mirrored motion calculus.
Elaborate Reading Summary
I. The Radian Foundation
The radian is defined as \(\theta = s/r\). Since meters cancel, it's a natural geometric unit. Displacement Bridge: \(s = r\theta\). Intuition: Every turn swept results in more linear coverage for objects at larger radii.
II. Velocity & Rim Speed
Angular velocity (\(\omega\)) is uniform for a rigid spinning body. Tangential speed (\(v = r\omega\)) is not. Fan rim points have more linear kinetic energy than the hub due to their physical path length.
Practice A Solution
α = (15 - 0) / 5 = 3 rad/s²
θ = ω0t + ½αt² = 0.5(3)(25) = 37.5 rad
s = rθ = 0.4 × 37.5 = 15 meters
Teacher Key: Dynamics
Part 2: Resistance & Impulse
Conceptual Cues
r² Impact: Resistance scales quadratically. Radius compounds both local velocity and lever arm magnification.
Impulse vs. Torque: Torque causes acceleration; Impulse (\(\tau \Delta t\)) changes total momentum (L).
Work: torque acting through an angle transfers energy (\(W = \tau \theta\)).
Elaborate Reading Summary
III. Moment of Inertia (I)
\(I = \sum mr^2\). Distance squared matters because a rim point is moving faster (energy increases as \(r^2\)) AND needs more torque to stop. Shape constant k determines efficiency. Hoop (1.0) is hardest to move.
IV. Angular Impulse & Work
Impulse: \(\tau \Delta t = \Delta L\). Prolonged torque builds up momentum. Work: \(\tau \theta = \Delta K_{rot}\). Rotating a wheel requires energy transfer proportional to the twist magnitude and distance.
Practice B Solution
τ = r F \sin(45°) = (0.3)(20)(0.707)
τ = 6 \times 0.707 = 4.24 N·m
Teacher Key: Conservation
Part 3: Budgeting & Systems
Conceptual Cues
Rolling Paradox: Potential energy pays a "spin tax." Concentrated mass saves energy for speed.
Skater: Muscles do work (KE up), but no external torque exists (L constant).
Clay-Rod: Linear L = mvr is converted to system spin momentum.
Elaborate Reading Summary
Motion Rotation Quiz Assessment AP Physics C: Rotational Dynamics Quiz
Assessment 04-P: Angular Mechanics & Conservation
Student:
Score: ______ / 50
Section 1: Multiple Choice (2 pts each)
1. A wheel of radius \(R\) starts from rest and rotates with a constant angular acceleration \(\alpha\). After time \(t\), the linear acceleration of a point on the rim is:
A) \(R\alpha\)
B) \(\sqrt{(R\alpha)^2 + (R\alpha^2t^2)^2}\)
C) \(R\alpha^2t\)
D) \(\alpha t\)
2. A uniform rod of mass \(M\) and length \(L\) is pivoted at its center. A point mass \(M\) is attached to one end. The new moment of inertia is:
A) \(1/3 ML^2\)
B) \(7/12 ML^2\)
C) \(1/12 ML^2 + 1/4 ML^2\)
D) \(1/3 ML^2 + 1/4 ML^2\)
3. A force \(\vec{F}\) is applied to a body at a position \(\vec{r}\) from the pivot. The torque is zero if:
A) \(\vec{r}\) and \(\vec{F}\) are perpendicular.
B) \(\vec{r}\) and \(\vec{F}\) are parallel.
C) The force is applied at the center of mass.
D) The object is already spinning at constant \(\omega\).
4. A solid sphere and a hollow hoop of the same mass and radius roll without slipping down an incline. At the bottom:
A) They have the same translational kinetic energy.
B) The hoop has more translational kinetic energy.
C) The sphere has more translational kinetic energy.
D) The hoop has more total mechanical energy.
5. An ice skater spinning with arms out pulls them in. Which of the following is true?
A) \(L\) is conserved; \(K\) is conserved.
B) \(L\) increases; \(K\) is conserved.
C) \(L\) is conserved; \(K\) increases.
D) \(L\) decreases; \(K\) decreases.
6. The work done by a constant torque \(\tau\) acting through an angle \(\theta\) is:
A) \(\tau \omega\)
B) \(1/2 I\omega^2\)
C) \(\tau \theta\)
D) \(I \alpha \theta\)
7. A disk of mass \(M\) and radius \(R\) rolls without slipping. The ratio of its rotational KE to its total KE is:
A) 1/2
B) 1/3
C) 2/3
D) 1/4
8. A particle moves in a straight line with constant velocity. Its angular momentum relative to a fixed point not on the line is:
A) Zero
B) Increasing
C) Constant and non-zero
D) Decreasing
9. Which shape has the highest Moment of Inertia for the same \(M\) and \(R\)?
A) Solid Sphere
Motion Rotation Quiz Key Rotation Quiz: Teacher Key
Solutions for Assessment 04-P
AP Physics C Level
Section 1: Multiple Choice Answers
1. B (atot = √[at² + ac²]; at = Rα, ac = v²/R = Rα²t²)
2. A (I = Irod,cm + M(L/2)² = 1/12 ML² + 1/4 ML² = 1/3 ML²)
3. B (Torque is the cross product; τ = rFsinθ. If parallel, θ=0, sinθ=0)
4. C (Sphere k=0.4, Hoop k=1.0. Lower k means higher v, so more linear KE)
5. C (Internal work increases KE; zero external torque preserves momentum L)
6. C (Work is the rotational equivalent of Force × distance)
7. B (Krot/Ktot = k/(1+k) = 0.5/1.5 = 1/3)
8. C (L = mvr sinφ. r sinφ is the constant dist from the point to the path)
9. D (Hollow Hoop: all mass at maximum radius R, k = 1.0)
10. B (τ = Iα. If τ and I are constant, α must be constant)
Section 2: Quantitative Solutions
11. Multi-Force Dynamics
τ1 = (10 N)(0.5 m) = 5.0 Nm (CW)18 rad/s²
12. Derivation & Ratio
mgh = ½mv² + ½(kMR²)(v/R)² → mgh = ½mv²(1+k)1.19
13. Conservation & Work
L1 = L2 → (4.0)(2.0) = (1.0)ω2 → ω2 = 8.0 rad/s 24 Joules
14. Angular Collision
Lin = mvr = (0.5)(20)(1.0) = 10.0 kgm²/s6.67 rad/s
15. Pulley System
mg - T = ma | τ = TR = Iα → T = (1/2 MR²)(α/R) = 1/2 Mag / (1 + M/2m)
