A slide-by-slide walkthrough of all 6 topics in Energy and Momentum of Rotating Systems. Each slide covers the big idea, key definitions, and a worked example.
Just like a moving object has KE = ½mv², a spinning object has rotational KE: K_rot = ½Iω². For a rolling ball, BOTH forms exist at once — K_total = ½mv² + ½Iω². The same energy conservation rules apply, just with an extra rotational term.
Key Properties
Formula
K_rot = ½Iω². Always ≥ 0.
Same Form As
K = ½mv², with I↔m and ω↔v.
For Rolling Objects
K_total = K_trans + K_rot.
Units
Joules (J).
Example
A spinning disk
A disk with rotational inertia I = 2 kg·m² spins at ω = 4 rad/s. What is its rotational KE?
K_rot = ½Iω² = ½(2)(4)² = ½(2)(16) = 16 J
If you doubled ω to 8 rad/s, K_rot would be 64 J — FOUR times larger (ω is squared, same as v in linear KE).
Topic 6.2
Torque and Work
Torque does work, just like force does
The Big Idea
When a torque acts over an angular displacement, it does work: W = τΔθ (with Δθ in radians). The rotational work-energy theorem says W_net = ΔK_rot. Rotational power: P = τω. Same logic as Unit 3 for linear motion.
Key Equations
Rotational Work
W = τΔθ.
Work-Energy Theorem
W_net = ½Iω_f² − ½Iω_i².
Rotational Power
P = τω.
Compare to Linear
W = Fd; P = Fv.
Example
A motor on a wheel
A motor applies a constant torque of 5 N·m to a wheel, spinning it through 4 radians. How much work does the motor do?
W = τΔθ = (5)(4) = 20 J
If the wheel is rotating at ω = 10 rad/s, the motor's power output is P = τω = (5)(10) = 50 W.
Topic 6.3
Angular Momentum and Angular Impulse
L = Iω, and torque-time changes it
The Big Idea
Angular momentum L = Iω (for a rigid body). For a point mass moving in a straight line, L = mvr_⊥. The angular impulse-momentum theorem says J_ang = τΔt = ΔL — a torque applied over time changes angular momentum. This is the rotational analog of J = Δp from Unit 4.
Key Equations
Rigid Body
L = Iω. Vector quantity.
Point Mass
L = mvr_⊥.
Angular Impulse
J_ang = τΔt.
Impulse-Momentum Thm
J_ang = ΔL.
Example
A torque slowing down a wheel
A wheel with I = 3 kg·m² is spinning at ω_i = 6 rad/s. A constant friction torque of −2 N·m acts on it for 3 seconds. What is its final angular velocity?
J_ang = τΔt = (−2)(3) = −6 kg·m²/s
By J_ang = ΔL = I(ω_f − ω_i): −6 = 3(ω_f − 6). So ω_f − 6 = −2, and ω_f = 4 rad/s. The wheel slowed down by 2 rad/s.
Topic 6.4
Conservation of Angular Momentum
When no external torque acts, L stays constant
The Big Idea
If a system has no external TORQUE acting on it, its total angular momentum stays constant: L_i = L_f. The most important principle of Unit 6. If something inside the system changes its rotational inertia (like a skater pulling in her arms), the angular velocity adjusts so that Iω stays the same. Internal forces always come in equal-and-opposite pairs and can't change L.
Classic Applications
Figure Skater
Pull in arms → smaller I → faster ω.
Neutron Star
Star collapses → tiny I → very fast rotation.
Child on Merry-Go-Round
Child jumps on → L_child + L_merry conserved.
Diver Tucking
Tuck → smaller I → more flips per second.
Example
Figure skater pulling in arms
A skater spins at ω_1 = 2 rad/s with arms out (I_1 = 5 kg·m²). She pulls her arms in (I_2 = 1.5 kg·m²). What's her new angular velocity?
Conservation: I_1·ω_1 = I_2·ω_2
(5)(2) = (1.5)(ω_2)
10 = 1.5·ω_2 → ω_2 ≈ 6.7 rad/s
She spins about 3.3× faster. (Her K_rot also increased — energy came from her muscles doing internal work.)
Topic 6.5
Rolling
Translation AND rotation at the same time
The Big Idea
A rolling object has both translational AND rotational motion. The rolling-without-slipping condition is v_cm = rω, which links them. Total KE: ½mv² + ½Iω². For an object rolling down a ramp, use energy conservation: mgh = ½mv² + ½Iω². Objects with HIGHER rotational inertia (per unit mass) put more energy into spinning, so they go slower at the bottom.
The Ramp Race
Sliding Block
No rotation. Fastest down the ramp.
Solid Sphere
I = ⅖mR². Fast.
Solid Disk
I = ½mR². Slower.
Hoop
I = mR². Slowest of the rolling objects.
Example
A hoop rolling down a ramp
A hoop (I = mR²) of mass m and radius R rolls from rest down a ramp of height h. Find its speed at the bottom.
Compare to a sliding block: v = √(2gh). The hoop is slower by a factor of √2 because half its energy goes to rotation.
Topic 6.6
Motion of Orbiting Satellites
Gravity provides the centripetal force
The Big Idea
A satellite in orbit is constantly falling toward Earth — but moving fast enough sideways that it always "misses." Gravity provides the centripetal force: F_grav = F_centripetal, so GMm/r² = mv²/r, which gives v_orbit = √(GM/r). The satellite's mass cancels — orbital speed depends only on the central mass and orbital radius.
Key Facts
Orbital Speed
v_orbit = √(GM/r). Mass-independent.
Orbital Period
T = 2πr/v_orbit. Longer at bigger r.
Kepler's 3rd Law
T² ∝ r³. Doubling r → period × 2.83.
Closer = Faster
ISS at low orbit: ~90 min. Moon: ~28 days.
Example
Satellite orbital speed
A satellite orbits Earth in a circular orbit of radius r = 7,000 km = 7 × 10⁶ m. Find its orbital speed. (Use GM_Earth = 4 × 10¹⁴ m³/s².)