Unit 6: Energy & Momentum of Rotating Systems · First Principles Physics
← PREVIEW MODE →   Delete the <div class="fpp-setup-banner">...</div> block before pasting into Squarespace (or leave it, harmless).
AP Physics 1 · Unit 06 of 08

Energy & Momentum of Rotating Systems.

The figure skater pulls in her arms and speeds up. A planet orbits closer to the sun and whips faster through perihelion. A rolling wheel stores half its energy in rotation. This unit is why the universe behaves the way it does.

Exam Weight · 5–8% Class Periods · ~8–14 CED Topics · 6.1–6.5 Difficulty · Moderate-High
01
The lesson

The skater, the wheel, and why the universe conserves angular momentum

You've watched a figure skater do a spin — starts with arms extended, pulls them in, suddenly accelerates into a blur so fast the audience gasps, then extends her arms again to slow down. No one is touching her. No ice-rink motors are speeding her up. And yet she controls her rotation rate with millimeter-precise movements of her own body. The question is simple: what physical law is this?

It's conservation of angular momentum, and it's the centerpiece of Unit 6. We'll also cover rotational kinetic energy — the energy a spinning object stores by virtue of its motion — and the special case of rolling without slipping, where an object carries both translational and rotational kinetic energy at once.

If Unit 5 was "Newton's laws with Greek letters," Unit 6 is "Units 3 and 4 with Greek letters." Everything you learned about energy and momentum has a rotational twin.

Rotational kinetic energy

A spinning disk has energy. You can feel this — grab a spinning bike wheel by the axle and you can feel the resistance to changing its orientation. That energy didn't come from translation (the wheel's center of mass may not even be moving). It's rotational kinetic energy:

$$K_{\text{rot}} = \tfrac{1}{2} I \omega^2$$

Compare this to the translational version \(K = \tfrac{1}{2}mv^2\). The analogy is clean: mass becomes rotational inertia, velocity becomes angular velocity, the ½ stays the same, and the result is measured in joules (same units as any other energy). You can use rotational KE in energy conservation problems exactly the same way you use translational KE.

Worked example 1 · rotational KE

A flywheel storing energy

A solid cylindrical flywheel of mass 50 kg and radius 0.4 m spins at 100 rad/s. How much kinetic energy does it store?

Calculate rotational inertia

\(I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(50)(0.16) = 4.0\) kg·m².

Apply \(K = \tfrac{1}{2}I\omega^2\)

\(K = \tfrac{1}{2}(4.0)(100)^2 = 20{,}000\) J = 20 kJ.

Energy stored: 20 kJ (about enough to lift a 100 kg person up 20 meters).

Rolling without slipping: the combined case

A ball rolls down a ramp. Its center of mass translates — it's moving. But the ball is also rotating. Both motions exist simultaneously. The total kinetic energy is the sum:

$$K_{\text{total}} = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2$$

When the ball rolls without slipping, the linear speed of the center of mass and the angular velocity are linked by \(v = R\omega\). This means some of the gravitational PE released as the ball rolls down the ramp goes into translation, and some goes into rotation. For the same reason, a rolling ball accelerates down a ramp slower than a frictionless sliding block would — a portion of the gravitational energy is diverted into spinning the ball up.

The rolling race

A solid sphere (\(I = \tfrac{2}{5}MR^2\)), a solid cylinder (\(I = \tfrac{1}{2}MR^2\)), and a hoop (\(I = MR^2\)) are released from the top of a ramp. Which wins the race to the bottom? Always the sphere. Why? The sphere has the least rotational inertia, so less energy goes into rotation and more goes into translation. Mass and radius don't matter — the race depends only on the shape factor \(\beta\) where \(I = \beta MR^2\): smaller \(\beta\) means faster finisher.

For rolling without slipping, static friction acts (the contact point doesn't slide) but does no work because the contact point has zero velocity at the instant of contact. Mechanical energy is conserved. You can use energy conservation straight from top to bottom of a ramp.

Worked example 2 · rolling energy

A solid sphere rolls down a ramp

A solid sphere of mass \(M\) and radius \(R\) starts from rest at the top of a ramp of height \(h\) and rolls without slipping to the bottom. What is its speed \(v\) at the bottom?

Apply energy conservation

\(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2\).

Use \(I = \tfrac{2}{5}MR^2\) and rolling constraint \(\omega = v/R\)

\(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}(\tfrac{2}{5}MR^2)(v/R)^2 = \tfrac{1}{2}Mv^2 + \tfrac{1}{5}Mv^2 = \tfrac{7}{10}Mv^2\).

Solving: \(v = \sqrt{10gh/7}\).

\(v = \sqrt{10gh/7}\), noticeably less than \(\sqrt{2gh}\) (the frictionless sliding result).

Work done by a torque

Just as a force does work when it moves through a displacement (\(W = F\,d\cos\theta\)), a torque does work when it rotates through an angular displacement:

$$W_\tau = \tau \, \Delta\theta$$

And just as the work-energy theorem says \(W_{\text{net}} = \Delta K\) for translation, the rotational version gives \(W_{\tau,\text{net}} = \Delta K_{\text{rot}}\). A torque applied over an angular displacement changes the object's rotational kinetic energy.

Angular momentum

Every translational quantity has a rotational twin. Momentum \(p = mv\) becomes angular momentum:

$$L = I\omega$$

Units: kg·m²/s. You'll also see angular momentum for a point mass traveling in a circle (or, more generally, a point with some linear velocity measured from a pivot): \(L = mvr\), where \(r\) is the perpendicular distance from the pivot to the line of motion. These two forms are equivalent — a point mass at distance \(r\) has \(I = mr^2\) and \(\omega = v/r\), so \(I\omega = mr^2(v/r) = mvr\). Same thing, different costume.

Angular impulse-momentum theorem

Exactly analogous to linear impulse \(F\Delta t = \Delta p\), we have:

$$\tau \Delta t = \Delta L$$

A torque applied for a time produces a change in angular momentum. This is sometimes called the angular impulse-momentum theorem. On a graph of \(\tau\) vs \(t\), the area under the curve equals \(\Delta L\).

Conservation of angular momentum

Here's the big idea. If no external torque acts on a system, angular momentum is conserved:

$$\text{If } \tau_{\text{ext}} = 0, \text{ then } L_i = L_f$$

This is the rotational analog of conservation of linear momentum (which requires zero external force). It's ruthlessly general. It applies to figure skaters, planets, galaxies, and gymnasts in mid-air.

When a skater pulls her arms in, she reduces her rotational inertia \(I\). No external torque acts on her (the ice provides no torque about the vertical axis because friction on the blade is very small). So \(L = I\omega\) is constant. If \(I\) goes down, \(\omega\) must go up. Fast. This is the entire explanation — she doesn't push off anything, she just rearranges her mass.

Worked example 3 · angular momentum conservation

The figure skater

A figure skater spins at 2.0 rad/s with her arms extended. Her rotational inertia in this configuration is 4.0 kg·m². She pulls her arms in, reducing her rotational inertia to 1.5 kg·m². Find her new angular velocity.

Apply conservation of angular momentum

\(I_i \omega_i = I_f \omega_f\), so \((4.0)(2.0) = (1.5)\omega_f\), giving \(\omega_f = 8.0/1.5 \approx 5.3\) rad/s.

New angular velocity: 5.3 rad/s, about 2.7× her initial speed.

Bonus: what happened to kinetic energy?

\(K_i = \tfrac{1}{2}(4.0)(2.0)^2 = 8\) J. \(K_f = \tfrac{1}{2}(1.5)(5.3)^2 \approx 21\) J. Kinetic energy increased by about 13 J. Where did that energy come from? The skater did work pulling her arms in against the "centrifugal tendency" of her extended body. Her muscles supplied it.

Careful: angular momentum conserved ≠ energy conserved

In the figure skater example, \(L\) is conserved but \(K\) is not. This is different from ordinary collision problems where you often say "momentum conserved, energy conserved (if elastic)." In rotation, angular momentum conservation by itself doesn't imply energy conservation — the system can do internal work on itself, or internal work can be dissipated.

Rotational collisions

Two students grab onto a spinning merry-go-round. A clay blob lands on a rotating disk and sticks. A child jumps off a rotating platform. These are rotational collisions: two objects interact through internal forces only, and the combined angular momentum of the system is conserved.

$$L_{\text{total, before}} = L_{\text{total, after}}$$

This is the workhorse equation for every "blob lands on rotating disk" problem. Compute total angular momentum before (often \(I_{\text{disk}}\omega + 0\) for a blob that's initially at rest). Compute rotational inertia after (the disk + blob system has different \(I\)). Solve for the new \(\omega\).

Where this unit goes next

Unit 7 will introduce oscillations — periodic motion where angular frequency \(\omega\) appears again, but in a different context. Many oscillation problems (the physical pendulum, the torsion pendulum) rely on Unit 5–6 concepts. The rotational energy and angular momentum ideas you learn here also appear in Kepler's laws of planetary motion (CED topic 6.6), which explain why planets speed up near the sun: as a planet approaches the sun, \(r\) decreases, and to conserve angular momentum \(L = mvr\), \(v\) must increase. The universe runs on conservation of angular momentum.

Conceptual summary

Unit 6 introduces the energy and momentum concepts for rotating systems. Rotational kinetic energy \(\tfrac{1}{2}I\omega^2\) is the rotational analog of \(\tfrac{1}{2}mv^2\) and can be used in energy-conservation problems. Rolling without slipping requires both translational and rotational KE, and the constraint \(v = R\omega\). Torque does work when rotation occurs: \(W = \tau\Delta\theta\). Angular momentum \(L = I\omega = mvr\) is conserved in any system with no external torque, which explains phenomena from figure skaters to orbiting planets. Angular momentum conservation does not imply energy conservation — when an object changes shape, internal work can add or remove kinetic energy while \(L\) stays fixed.

· · ·
02
Reference sheet

Every equation in Unit 6

★ Rotational kinetic energy

$$K_{\text{rot}} = \tfrac{1}{2}I\omega^2$$
\(I\)
rotational inertia (kg·m²)
\(\omega\)
angular velocity (rad/s)

Use when you need the kinetic energy stored in rotation. Scalar — no direction.

★ Total KE for rolling/moving objects

$$K_{\text{tot}} = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2$$
\(v\)
linear speed of center of mass
\(\omega\)
angular velocity about CM

Use when an object is both translating and rotating (rolling ball, thrown frisbee, rolling cylinder).

Rolling-without-slipping constraint

$$v_{cm} = R\omega, \quad a_{cm} = R\alpha$$
\(R\)
radius of rolling object

Use when an object rolls without slipping. Turns rolling problems into two linked equations sharing one \(v\) (or \(a\)).

Work done by torque

$$W_\tau = \tau \Delta\theta$$
\(\Delta\theta\)
angular displacement (rad)

Use when a torque acts through a rotation. Rotational analog of \(W = Fd\).

★ Angular momentum — rigid body

$$L = I\omega$$
\(L\)
angular momentum (kg·m²/s)

Use when a rigid body rotates about a fixed axis (wheel, disk, skater, rod).

★ Angular momentum — point mass

$$L = mvr_\perp$$
\(r_\perp\)
perpendicular distance from axis to line of motion

Use when a point mass (or small object) moves with some velocity measured relative to an axis. Equivalent to \(I\omega\) for a point mass.

★ Angular impulse-momentum theorem

$$\tau_{\text{ext}} \Delta t = \Delta L$$
\(\tau_{\text{ext}}\)
average external torque

Use when a torque acts for a known time and you need the change in \(L\). Rotational analog of \(F\Delta t = \Delta p\).

★ Conservation of angular momentum

$$L_i = L_f \; \text{if} \; \tau_{\text{ext}} = 0$$
Condition
zero net external torque on system

Use when an isolated rotating system changes shape (skater pulling arms in) or has a rotational collision (blob landing on a disk).

· · ·
· · ·
04
Worksheet A · Foundation

Ten problems building from basic to moderate

Use \(g = 10\) m/s² throughout. Rolling means rolling without slipping unless stated otherwise.
Foundation

A disk of rotational inertia 0.5 kg·m² spins at 10 rad/s. Find its rotational kinetic energy.

Foundation

A wheel of rotational inertia 2 kg·m² rotates at 5 rad/s. Find its angular momentum.

Foundation

A 0.3 kg ball moves in a circle of radius 0.5 m at 4 m/s. Find its angular momentum about the center of the circle.

Moderate

A torque of 6 N·m is applied to a wheel for 5 seconds. If the wheel starts at rest, what is its final angular momentum?

Moderate

A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping at 3 m/s. Find (a) its translational KE, (b) its rotational KE, (c) its total KE.

Moderate

A figure skater has rotational inertia 5.0 kg·m² with arms extended, spinning at 3.0 rad/s. She pulls her arms in, reducing her rotational inertia to 2.0 kg·m². Find her new angular velocity.

Moderate

A solid cylinder rolls without slipping from rest down a ramp of height 2.0 m. Find its linear speed at the bottom.

Moderate

A hoop and a solid cylinder, both of equal mass and radius, race down the same ramp from the same height starting from rest. Which reaches the bottom with greater speed? Justify numerically.

Moderate

A 2.0 kg clay ball moving at 5.0 m/s strikes and sticks to the end of a uniform thin rod of mass 4.0 kg and length 2.0 m that is initially at rest and free to rotate about its far end. Find the angular velocity of the rod + ball system immediately after the collision. (Use \(I_{\text{rod, end}} = \tfrac{1}{3}ML^2\).)

Moderate

A merry-go-round (rotational inertia 400 kg·m²) is spinning at 1.0 rad/s. A 50 kg child jumps on at a point 2.0 m from the axis. Find the new angular velocity.

· · ·
05
Worksheet B · AP exam style

Ten problems in AP exam format

Part I — Multiple Choice

AP MC

A solid sphere and a hoop have the same mass, same radius, and same angular velocity. Which statement is true?

  1. They have the same rotational kinetic energy and the same angular momentum.
  2. The hoop has more rotational KE and more angular momentum.
  3. The hoop has more angular momentum but the same KE.
  4. The sphere has more angular momentum because it rotates faster internally.
AP MC

A figure skater spinning at 2 rad/s with \(I = 3\) kg·m² pulls her arms in to reduce \(I\) to 1 kg·m². What happens to her kinetic energy?

  1. It stays the same because angular momentum is conserved.
  2. It decreases by a factor of 3.
  3. It increases by a factor of 3.
  4. It decreases by a factor of 9.
AP MC

A solid sphere and a solid cylinder, both of mass \(M\) and radius \(R\), roll without slipping down identical ramps starting from rest. Which wins the race?

  1. The sphere wins because it has less rotational inertia.
  2. The cylinder wins because it has more rotational inertia.
  3. They tie because they have the same mass and radius.
  4. Impossible to tell without knowing the ramp angle.
AP MC

A torque is applied to a wheel according to the graph below. What is the change in the wheel's angular momentum from \(t = 0\) to \(t = 6\) s?

0 4 2 4 6 t (s) τ (N·m)

  1. 8 kg·m²/s
  2. 12 kg·m²/s
  3. 16 kg·m²/s
  4. 24 kg·m²/s
AP MC

A 1 kg puck moves in a straight line at 5 m/s across a frictionless table. Pick an axis perpendicular to the table at a point 0.3 m from the puck's path of motion. What is the puck's angular momentum about that axis?

  1. Zero, because the puck is not rotating.
  2. 1.5 kg·m²/s
  3. 5 kg·m²/s
  4. Impossible to determine without knowing the puck's moment of inertia.

Part II — Short Free Response

AP SFR

(~7 min) A uniform rod of mass \(M\) and length \(L\) hangs vertically from a pivot at its top end and is initially at rest. A bullet of mass \(m\) moving horizontally with speed \(v_0\) strikes and embeds in the bottom of the rod. The rod's rotational inertia about its pivot is \(\tfrac{1}{3}ML^2\).

  1. Derive an expression for the angular velocity of the rod+bullet system immediately after the collision.
  2. Is kinetic energy conserved during this collision? Justify your answer with a one-sentence reason.
  3. Calculate the fraction of the bullet's original kinetic energy that is lost in the collision if \(m = M/4\).
AP SFR

(~7 min) A solid cylinder rolls without slipping on a horizontal surface at speed \(v\). It reaches the foot of a ramp and rolls up.

  1. Derive an expression for the maximum height \(h\) the cylinder reaches on the ramp, in terms of \(v\) and \(g\).
  2. How does this compare with the height a frictionless sliding block of the same mass and speed would reach? Justify your comparison.
  3. The ramp is now made slick so the cylinder slides instead of rolling. Does the cylinder go higher, lower, or the same distance up? Explain.
AP SFR

(~7 min) Two identical disks of rotational inertia \(I\) are mounted on a common axle. Disk 1 spins at \(\omega_0\); disk 2 is at rest. The disks are released so that they couple together (friction between them causes them to eventually spin at the same rate).

  1. Find the final angular velocity in terms of \(\omega_0\).
  2. Find the ratio of the final kinetic energy to the initial kinetic energy. Where did the missing energy go?

Part III — Extended Free Response

AP EFR

(~18 min) A merry-go-round of rotational inertia \(I_M\) rotates freely about a vertical axis at angular velocity \(\omega_0\). A child of mass \(m\) stands on the edge of the merry-go-round at radius \(R\). The child walks radially inward from the edge to the center.

  1. Treating the child as a point mass, write an expression for the total angular momentum of the system when the child is at radius \(r\) from the center, in terms of \(I_M\), \(m\), \(r\), and the angular velocity \(\omega\) at that instant.
  2. Using angular momentum conservation, derive an expression for \(\omega\) as a function of \(r\) as the child walks inward.
  3. Evaluate \(\omega\) when the child reaches the center (\(r = 0\)).
  4. Calculate the change in the system's kinetic energy from when the child is at the edge (\(r = R\)) to when the child reaches the center. Explain where this energy came from or went.
  5. If the child walks back out to \(r = R\), does the system return to its original angular velocity? Justify your answer.
AP EFR

(~18 min — rolling on a ramp with energy) A hollow sphere of mass \(M\) and radius \(R\) (rotational inertia \(\tfrac{2}{3}MR^2\)) rolls without slipping down a ramp that makes angle \(\theta\) with the horizontal.

  1. Draw a free-body diagram showing all forces on the hollow sphere. Include friction.
  2. Derive an expression for the acceleration of the center of mass down the ramp, using Newton's second law in both translational and rotational form.
  3. Derive an expression for the friction force. Confirm it is less than the maximum static friction \(\mu_s Mg\cos\theta\) for the rolling-without-slipping assumption to hold. What does this condition tell you about \(\mu_s\) and \(\theta\)?
  4. A block of the same mass \(M\) slides down the same ramp with zero friction. Compare the time it takes the block to reach the bottom with the time it takes the rolling sphere. Justify with calculation.
· · ·
06
Worksheet C · Challenge

Five problems at olympiad level

Challenge

Kepler's law via angular momentum. A satellite orbits a planet in an elliptical orbit. At perihelion (closest approach, distance \(r_p\)) its speed is \(v_p\). At aphelion (farthest, distance \(r_a\)) its speed is \(v_a\). Show \(v_p r_p = v_a r_a\). Then, using energy conservation, derive an expression for \(v_p\) in terms of \(r_p\), \(r_a\), \(G\), and the planet's mass \(M\).

Challenge

Sliding-to-rolling transition. A bowling ball (solid sphere, mass \(M\), radius \(R\)) is thrown with initial linear velocity \(v_0\) but with zero angular velocity on a surface with kinetic friction coefficient \(\mu_k\). Friction both slows the ball's translation and spins it up. Eventually it rolls without slipping. (a) Find the time at which rolling begins. (b) Find the linear velocity when rolling begins. (c) Compute the energy dissipated as a fraction of initial KE, and verify using energy accounting.

Challenge

The physical pendulum. A uniform rod of mass \(M\) and length \(L\) is pivoted at one end and released from a horizontal position. (a) Using energy conservation, find the angular velocity at the moment the rod passes through the vertical. (b) Find the linear speed of the rod's center of mass at this instant, and compare with a point mass dropped freely from the same height. Why are they different?

Challenge

The ballerina with dumbbells. A ballerina has rotational inertia \(I_0\) with arms close to her body, spinning at \(\omega_0\). She holds two dumbbells of mass \(m\) each, initially at distance \(r_0\) from her axis. She extends them to a new distance \(r_1 > r_0\). (a) Find her new angular velocity. (b) Compute the change in kinetic energy. (c) Explain physically where the energy came from or went.

Challenge

Rolling into a loop-de-loop. A solid sphere of mass \(M\) and radius \(R\) starts from rest at height \(h\) on a track and rolls without slipping into a vertical loop of radius \(r\) (where \(r \gg R\), so treat the sphere as a point for motion but still include its rotational KE). Find the minimum height \(h\) such that the sphere maintains contact with the top of the loop.

· · ·
07
Complete worked solutions

Every problem, every step

Worksheet A — Foundation

· rotational KE

\(K = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(0.5)(10)^2 = 25\) J.

\(K = 25\) J
· angular momentum of wheel

\(L = I\omega = (2)(5) = 10\) kg·m²/s.

\(L = 10\) kg·m²/s
· angular momentum of ball in circle

Point mass traveling tangentially: \(L = mvr = (0.3)(4)(0.5) = 0.6\) kg·m²/s.

\(L = 0.6\) kg·m²/s
For circular motion, the velocity is always perpendicular to the position vector from the center, so \(r_\perp = r\) and no sine factor is needed.
· angular impulse

\(\Delta L = \tau \Delta t = (6)(5) = 30\) kg·m²/s. Starting from rest, final \(L = 30\) kg·m²/s.

\(L = 30\) kg·m²/s
· rolling sphere KE breakdown

(a) \(K_{\text{trans}} = \tfrac{1}{2}Mv^2 = \tfrac{1}{2}(2)(9) = 9\) J.

(b) For a solid sphere, \(I = \tfrac{2}{5}MR^2\) and rolling gives \(\omega = v/R\). So \(K_{\text{rot}} = \tfrac{1}{2}(\tfrac{2}{5}MR^2)(v/R)^2 = \tfrac{1}{5}Mv^2 = \tfrac{1}{5}(2)(9) = 3.6\) J.

(c) Total: 12.6 J.

Trans: 9 J; Rot: 3.6 J; Total: 12.6 J.
For a rolling sphere, rotational KE is always exactly \(\tfrac{2}{7}\) of the total. Memorize this ratio — it shows up constantly. Solid cylinder: \(\tfrac{1}{3}\). Hoop: \(\tfrac{1}{2}\).
· figure skater

Conservation of \(L\): \(I_i \omega_i = I_f \omega_f\), so \((5.0)(3.0) = (2.0)\omega_f\), giving \(\omega_f = 7.5\) rad/s.

\(\omega_f = 7.5\) rad/s
· cylinder rolling down ramp

Energy conservation: \(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2\). For solid cylinder with \(I = \tfrac{1}{2}MR^2\) and \(\omega = v/R\): \(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{4}Mv^2 = \tfrac{3}{4}Mv^2\). Solving: \(v = \sqrt{4gh/3} = \sqrt{4(10)(2)/3} = \sqrt{80/3} \approx 5.16\) m/s.

\(v \approx 5.16\) m/s
· hoop vs cylinder race

Same setup. Hoop (\(I = MR^2\)): \(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}Mv^2 = Mv^2\), so \(v = \sqrt{gh}\). Cylinder (\(I = \tfrac{1}{2}MR^2\)): \(v = \sqrt{4gh/3}\). Since \(4/3 > 1\), cylinder wins.

Cylinder reaches bottom with greater speed; ratio \(v_{cyl}/v_{hoop} = \sqrt{4/3} \approx 1.155\).
Smaller shape factor \(\beta\) (cylinder 1/2 vs hoop 1) means less energy diverted to rotation, more left for translation. Neither mass nor radius appears — only \(\beta\) matters for rolling races.
· ballistic pendulum with rod

Angular momentum about pivot is conserved (the pivot exerts no torque about itself). Before: only the ball moves. \(L_i = m v r_\perp = (2.0)(5.0)(2.0) = 20\) kg·m²/s (the ball strikes at the end, distance 2.0 m from pivot).

After: rod + embedded ball rotate together. \(I_{\text{after}} = \tfrac{1}{3}ML^2 + mL^2 = \tfrac{1}{3}(4.0)(4) + (2.0)(4) = 5.33 + 8.0 = 13.33\) kg·m².

\(L_f = I\omega\): \(20 = 13.33 \omega\) → \(\omega = 1.5\) rad/s.

\(\omega = 1.5\) rad/s
Even though the ball travels in a straight line before the collision, it has angular momentum about the rod's pivot. The formula \(L = mvr_\perp\) gives the right answer whether the motion is straight-line or curved.
· child on merry-go-round

Treat child as a point mass. Initially only the merry-go-round rotates: \(L_i = I_M \omega_0 = (400)(1.0) = 400\) kg·m²/s.

After: child at radius 2.0 m adds \(I_c = mr^2 = (50)(4) = 200\) kg·m² to the total. \(I_{\text{total}} = 600\) kg·m².

\(L\) conserved: \(400 = 600 \omega_f\) → \(\omega_f \approx 0.67\) rad/s.

\(\omega_f \approx 0.67\) rad/s
Adding mass at a nonzero radius increases the system's \(I\). To keep \(L\) constant, \(\omega\) must drop. This is why playground equipment slows down when kids jump on.

Worksheet B — AP Exam Style

· sphere vs hoop, same \(\omega\)

Hoop \(I = MR^2\); sphere \(I = \tfrac{2}{5}MR^2\). Same \(M\), \(R\), \(\omega\). Hoop has larger \(I\), so both \(L = I\omega\) and \(K = \tfrac{1}{2}I\omega^2\) are larger for the hoop.

(B) Hoop has more of both.
· skater KE change

\(L\) is conserved: \(I_i \omega_i = I_f \omega_f\), so \(\omega_f = (I_i/I_f)\omega_i = 3\omega_i\). KE ratio: \(K_f/K_i = (I_f \omega_f^2)/(I_i \omega_i^2) = (1/3)(9) = 3\). KE triples.

(C) Increases by a factor of 3.
Quick shortcut: for fixed \(L\), \(K = L^2/(2I)\), so \(K\) is inversely proportional to \(I\). Halve \(I\) → double \(K\). Reduce \(I\) by a third → triple \(K\).
· sphere vs cylinder race

Smaller \(\beta\) wins. Sphere: \(\beta = 2/5 = 0.4\). Cylinder: \(\beta = 1/2 = 0.5\). Sphere wins.

(A) Sphere wins.
· area under torque graph

\(\Delta L\) = area under \(\tau\)-\(t\) graph. Shape is a triangle rising from 0 to 4 over \(t = 0\) to 2, then a rectangle (4 N·m) from \(t = 2\) to 4, then a triangle falling from 4 to 0 from \(t = 4\) to 6.

Area = \(\tfrac{1}{2}(2)(4) + (2)(4) + \tfrac{1}{2}(2)(4) = 4 + 8 + 4 = 16\) kg·m²/s.

(C) 16 kg·m²/s
· angular momentum of straight-line puck

\(L = mvr_\perp = (1)(5)(0.3) = 1.5\) kg·m²/s. The puck has nonzero \(L\) even though it moves in a straight line — \(L\) depends on the choice of axis.

(B) 1.5 kg·m²/s
Angular momentum is not a property of the object alone — it depends on the axis you choose. A straight-line-moving puck has \(L = 0\) about any axis lying along its path, but nonzero \(L\) about any axis off to the side.
· bullet-rod collision

(a) Angular momentum conservation about pivot. Before: bullet at distance \(L\) from pivot, \(L_i = m v_0 L\). After: rod + bullet rotate together with \(I_{\text{total}} = \tfrac{1}{3}ML^2 + mL^2\). So:

$$\omega_f = \dfrac{m v_0 L}{\tfrac{1}{3}ML^2 + mL^2} = \dfrac{m v_0}{(M/3 + m)L} = \dfrac{3m v_0}{(M + 3m)L}$$

(b) No — this is a perfectly inelastic collision. The bullet embeds in the rod, so they have a common velocity afterward; kinetic energy is always lost in perfectly inelastic collisions.

(c) With \(m = M/4\): \(\omega_f = 3(M/4)v_0/[(M + 3M/4)L] = (3M/4)v_0/[(7M/4)L] = 3v_0/(7L)\).

\(K_i = \tfrac{1}{2}mv_0^2 = \tfrac{M v_0^2}{8}\). \(K_f = \tfrac{1}{2}I_{\text{total}}\omega_f^2 = \tfrac{1}{2}(\tfrac{7M L^2}{12})(3v_0/(7L))^2 = \tfrac{1}{2}(\tfrac{7ML^2}{12})(\tfrac{9v_0^2}{49L^2}) = \tfrac{3Mv_0^2}{56}\).

Fraction lost: \(1 - K_f/K_i = 1 - (3Mv_0^2/56)/(Mv_0^2/8) = 1 - 24/56 = 1 - 3/7 = 4/7\).

\(\omega_f = 3mv_0/[(M+3m)L]\); not conserved; 4/7 of KE lost.
· cylinder rolls up ramp

(a) Energy conservation: \(\tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = Mgh\). For solid cylinder: \(\tfrac{3}{4}Mv^2 = Mgh\) → \(h = 3v^2/(4g)\).

(b) A frictionless sliding block: \(\tfrac{1}{2}Mv^2 = Mgh\) → \(h = v^2/(2g)\). The rolling cylinder reaches 1.5× as high because it carries additional rotational KE that also converts to PE.

(c) On a slick ramp, the cylinder's rotation can't change (no friction to exert torque). Translational KE converts to PE: \(h = v^2/(2g)\). Cylinder reaches the same height as the block (lower than rolling). Note: rolling cylinder actually goes higher on rough ramp than slick one because rotational KE also converts to height.

Rolling: \(h = 3v^2/(4g)\); sliding/slick: \(h = v^2/(2g)\). Rolling goes higher.
· coupled disks

(a) \(L\) conserved: \(I\omega_0 + 0 = (2I)\omega_f\) → \(\omega_f = \omega_0/2\).

(b) \(K_i = \tfrac{1}{2}I\omega_0^2\). \(K_f = \tfrac{1}{2}(2I)(\omega_0/2)^2 = \tfrac{1}{4}I\omega_0^2\). So \(K_f/K_i = 1/2\). Half the KE was lost to friction between the coupling surfaces as they brought each other to a common angular velocity.

\(\omega_f = \omega_0/2\); half the energy lost to friction.
This is the rotational analog of two blocks colliding perfectly inelastically — momentum conserved, half the energy lost. Same math, different variables.
· child walking inward on merry-go-round

(a) \(L(r) = (I_M + mr^2)\omega\).

(b) Initially \(L_0 = (I_M + mR^2)\omega_0\). Conservation: \((I_M + mr^2)\omega = (I_M + mR^2)\omega_0\), so

$$\omega(r) = \omega_0 \cdot \dfrac{I_M + mR^2}{I_M + mr^2}$$

(c) At \(r = 0\): \(\omega = \omega_0 (I_M + mR^2)/I_M = \omega_0 (1 + mR^2/I_M)\).

(d) \(K_i = \tfrac{1}{2}(I_M + mR^2)\omega_0^2\). \(K_f = \tfrac{1}{2}I_M\omega_f^2 = \tfrac{1}{2}I_M \omega_0^2 (I_M + mR^2)^2/I_M^2 = \tfrac{(I_M + mR^2)^2 \omega_0^2}{2I_M}\).

Ratio \(K_f/K_i = (I_M + mR^2)/I_M > 1\). KE increases. The child does work by pulling himself inward — muscular effort supplies the extra energy.

(e) Yes. Angular momentum is conserved throughout; if the child walks back out to \(r = R\), the system's \(I\) returns to its original value, so \(\omega\) must too. KE also returns to original — the child does negative work on the way back out (the "centrifugal tendency" pushes him outward, he resists it, doing negative work).

\(\omega(0) = \omega_0(I_M + mR^2)/I_M\); KE increases; reversible.
· hollow sphere on ramp

(a) Forces: weight \(Mg\) at center pointing down; normal \(N\) perpendicular to ramp surface; static friction \(f_s\) up the ramp (along the surface).

(b) Translational: \(Mg\sin\theta - f_s = Ma\). Rotational about center: \(f_s R = I\alpha = \tfrac{2}{3}MR^2 \alpha\). Rolling: \(a = R\alpha\). Substitute: \(f_s = \tfrac{2}{3}Ma\). Into first equation: \(Mg\sin\theta - \tfrac{2}{3}Ma = Ma\), so

$$a = \dfrac{g\sin\theta}{1 + 2/3} = \dfrac{3g\sin\theta}{5}$$

(c) \(f_s = \tfrac{2}{3}Ma = \tfrac{2}{3}M(3g\sin\theta/5) = \tfrac{2Mg\sin\theta}{5}\). For rolling without slipping: \(f_s \leq \mu_s Mg\cos\theta\) → \(\tan\theta \leq \tfrac{5\mu_s}{2}\). Below this critical angle, the sphere rolls; above, it slips.

(d) Frictionless block: \(a = g\sin\theta\). Rolling hollow sphere: \(a = 3g\sin\theta/5 = 0.6 g\sin\theta\). Block accelerates faster by factor 5/3. Time to cover distance \(L\) down the ramp: \(L = \tfrac{1}{2}at^2\), so \(t \propto 1/\sqrt{a}\). Ratio \(t_{\text{roll}}/t_{\text{block}} = \sqrt{5/3} \approx 1.29\). Block arrives 29% faster.

\(a = 3g\sin\theta/5\); block beats sphere by factor \(\sqrt{5/3}\) in time.

Worksheet C — Challenge

· Kepler via angular momentum

Gravity from a central body acts along the line between the satellite and that body — always toward the center. About the central body, this force produces zero torque (force is radial, lever arm zero). So the satellite's angular momentum about the central body is conserved. At perihelion and aphelion, velocity is perpendicular to the position vector (these are the extremes), so \(L = mvr\). Equating: \(m v_p r_p = m v_a r_a\), giving \(v_p r_p = v_a r_a\).

Energy conservation: \(\tfrac{1}{2}mv_p^2 - GMm/r_p = \tfrac{1}{2}mv_a^2 - GMm/r_a\). Combine with \(v_a = v_p r_p/r_a\): \(v_p^2 - v_p^2 r_p^2/r_a^2 = 2GM(1/r_p - 1/r_a)\). Factor: \(v_p^2 (r_a^2 - r_p^2)/r_a^2 = 2GM(r_a - r_p)/(r_p r_a)\). Simplify using \(r_a^2 - r_p^2 = (r_a - r_p)(r_a + r_p)\):

$$v_p = \sqrt{\dfrac{2GM r_a}{r_p (r_p + r_a)}}$$

\(v_p r_p = v_a r_a\); \(v_p = \sqrt{2GM r_a/[r_p(r_p + r_a)]}\).
Kepler's second law (equal areas in equal times) is exactly angular momentum conservation. The "sweep rate" \(\tfrac{1}{2}r^2 \omega\) is proportional to \(L\), which is constant.
· bowling ball slide-to-roll

Friction decelerates translation: \(\mu_k Mg = Ma\) → linear deceleration \(\mu_k g\). After time \(t\): \(v(t) = v_0 - \mu_k g t\).

Friction torques up rotation: \(\mu_k Mg R = I\alpha = \tfrac{2}{5}MR^2 \alpha\) → \(\alpha = 5\mu_k g/(2R)\). Starting from rest: \(\omega(t) = \alpha t = 5\mu_k g t /(2R)\).

(a) Rolling begins when \(v = R\omega\): \(v_0 - \mu_k g t = R \cdot 5\mu_k g t/(2R) = \tfrac{5}{2}\mu_k g t\). So \(v_0 = \tfrac{7}{2}\mu_k g t\), giving

$$t^* = \dfrac{2v_0}{7\mu_k g}$$

(b) At this time: \(v^* = v_0 - \mu_k g t^* = v_0 - 2v_0/7 = 5v_0/7\).

(c) \(K_i = \tfrac{1}{2}Mv_0^2\). \(K_f = \tfrac{1}{2}Mv^{*2} + \tfrac{1}{2}I\omega^{*2} = \tfrac{1}{2}M(5v_0/7)^2 + \tfrac{1}{5}M(5v_0/7)^2 = \tfrac{7}{10}M(5v_0/7)^2 = \tfrac{7}{10}M \cdot \tfrac{25v_0^2}{49} = \tfrac{5Mv_0^2}{14}\).

\(K_f/K_i = (5/14)/(1/2) = 10/14 = 5/7\). So 2/7 of KE is lost to kinetic friction.

\(t^* = 2v_0/(7\mu_k g)\); \(v^* = 5v_0/7\); 2/7 of KE dissipated.
· physical pendulum (rod)

(a) Rod released horizontal. CM drops by \(L/2\) when rod reaches vertical. Energy conservation: \(Mg(L/2) = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(\tfrac{1}{3}ML^2)\omega^2 = \tfrac{1}{6}ML^2\omega^2\). So \(\omega^2 = 3g/L\) → \(\omega = \sqrt{3g/L}\).

(b) CM is at \(L/2\) from pivot, so \(v_{cm} = (L/2)\omega = (L/2)\sqrt{3g/L} = \tfrac{1}{2}\sqrt{3gL}\).

Point mass freely dropped from height \(L/2\): \(v = \sqrt{2g(L/2)} = \sqrt{gL}\). Ratio: \(v_{cm}/v_{\text{free}} = \tfrac{1}{2}\sqrt{3}/1 \approx 0.866\). The rod's CM is slower because some of the PE goes into rotational KE of the rod about its CM, not just translational motion of the CM.

\(\omega = \sqrt{3g/L}\); \(v_{cm} = \tfrac{1}{2}\sqrt{3gL}\) — slower than free-fall.
The tip of the rod, at distance \(L\) from pivot, travels at \(v_{\text{tip}} = L\omega = \sqrt{3gL}\), faster than free-fall (\(\sqrt{2gL}\)) from the same height. Extended rotating objects have tip speeds that can exceed free-fall values.
· ballerina extends dumbbells

(a) \(L\) conserved: \((I_0 + 2mr_0^2)\omega_0 = (I_0 + 2mr_1^2)\omega_f\). Solve:

$$\omega_f = \omega_0 \cdot \dfrac{I_0 + 2mr_0^2}{I_0 + 2mr_1^2}$$

(b) Since \(r_1 > r_0\), \(\omega_f < \omega_0\). \(K_f/K_i = \omega_f/\omega_0 < 1\) (using \(K = L^2/(2I)\) with \(L\) fixed). KE decreases.

(c) The ballerina's muscles do negative work on the dumbbells as they extend outward — the "centrifugal tendency" pushes the dumbbells outward, but her arms resist and control the motion. Alternatively: the dumbbells, as they extend, slow down the whole system. The KE lost went into muscular effort and heat — if the extension is controlled (as in a real ballerina), the energy is dissipated as muscle heat.

\(\omega_f = \omega_0(I_0 + 2mr_0^2)/(I_0 + 2mr_1^2)\); KE decreases; energy dissipated to muscular effort.
· sphere in loop-de-loop

At the top of the loop, minimum condition: gravity provides all the centripetal force, so \(mg = mv_{\text{top}}^2/r\) → \(v_{\text{top}}^2 = gr\).

Energy conservation from release height \(h\) to top of loop (height \(2r\)): \(Mgh = Mg(2r) + \tfrac{7}{10}Mv_{\text{top}}^2\) (using total KE = \(\tfrac{7}{10}Mv^2\) for rolling sphere).

\(gh = 2gr + \tfrac{7}{10}v_{\text{top}}^2 = 2gr + \tfrac{7}{10}gr = \tfrac{27}{10}gr\). So \(h_{\min} = 2.7r\).

\(h_{\min} = 2.7r\) (vs 2.5r for a frictionless sliding block).
A rolling ball needs a higher start than a sliding block because some of the PE it converts goes into rotational KE, not just translational KE needed to maintain contact at the top of the loop.
· · ·
08
Common mistakes

Seven errors that cost students points on Unit 6

Forgetting rotational KE when setting up energy conservation for a rolling object

A sphere rolls from the top of a 2 m ramp. Find its speed at the bottom.

\(mgh = \tfrac{1}{2}mv^2\) → \(v = \sqrt{2gh} = \sqrt{40} \approx 6.3\) m/s.
\(mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2\) with \(\omega = v/R\). For a solid sphere: \(mgh = \tfrac{7}{10}mv^2\) → \(v = \sqrt{10gh/7} \approx 5.3\) m/s.

Students apply the "slide down a frictionless ramp" formula out of habit. But a rolling object is also spinning — some PE is converted to rotational KE.

A solid cylinder starts from rest at the top of a 3 m ramp and rolls without slipping. Find its speed at the bottom.

Assuming energy is conserved whenever angular momentum is conserved

A skater pulls her arms in. Her \(\omega\) doubles. Is her kinetic energy the same?

Yes, because no external torque acted, so both \(L\) and \(K\) are conserved.
No. \(L\) is conserved (no external torque), but \(K = L^2/(2I)\). If \(I\) halves, \(K\) doubles. The skater's muscles did work.

Conservation laws are independent. Conservation of angular momentum (no external torque) and conservation of kinetic energy (no dissipative work) are separate conditions. A system changing its own shape can keep \(L\) constant while \(K\) changes.

A ballerina extends her arms and dumbbells, tripling her rotational inertia. By what factor does her kinetic energy change?

Confusing \(L = I\omega\) with \(L = mvr_\perp\) for a point mass

A puck of mass 2 kg moves at 4 m/s in a straight line 1.5 m from an axis. Find its angular momentum.

The puck isn't rotating, so \(\omega = 0\) and \(L = I\omega = 0\).
Treat it as a point mass. \(L = mvr_\perp = (2)(4)(1.5) = 12\) kg·m²/s.

Even an object moving in a straight line has angular momentum about any axis not on its path. The point-mass formula \(L = mvr_\perp\) applies whether the motion is straight or curved.

A 0.2 kg ball is thrown with a horizontal velocity of 10 m/s. At one instant, the ball is at coordinates (3 m, 4 m) from the origin. Find the magnitude of the ball's angular momentum about the origin. (Hint: find \(r_\perp\) from the position and velocity direction.)

Forgetting the rolling constraint \(v = R\omega\)

A cylinder of mass \(M\) and radius \(R\) rolls without slipping down a ramp. Set up energy conservation.

\(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2\). Two unknowns (\(v\) and \(\omega\)), one equation — can't solve.
Rolling without slipping gives \(v = R\omega\), so \(\omega = v/R\). Substituting: \(Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}(\tfrac{1}{2}MR^2)(v/R)^2 = \tfrac{3}{4}Mv^2\). Now solvable.

Rolling without slipping is a constraint, not an equation you derive — it's given by the problem setup. Without it, every rolling problem is underdetermined.

A hoop of mass 0.5 kg and radius 0.2 m rolls without slipping at 2 m/s. Find (a) its angular velocity, (b) its total kinetic energy.

Thinking friction does work on a rolling object

A ball rolls without slipping down a ramp. Does friction do work on the ball?

Yes, friction acts over the distance the ball travels, so it does work.
No. For rolling without slipping, the contact point between the ball and the surface is instantaneously at rest. Static friction at that point acts on a point with zero velocity, so the power \(P = \vec F \cdot \vec v = 0\), and no work is done. Mechanical energy is conserved for rolling without slipping.

Kinetic friction on a sliding object dissipates energy because the contact point has a nonzero velocity relative to the surface. Static friction on a rolling object has zero such velocity, so it cannot dissipate energy. This is why you can use energy conservation for rolling-without-slipping problems.

A solid sphere rolls down one ramp without slipping; an identical sphere slides (without rotating) down an identical ramp of identical height. Compare their speeds at the bottom.

Thinking the rolling race depends on mass or radius

Which wins the rolling race: a heavy hoop, or a light hoop of the same radius?

The heavy hoop wins because it has more rotational kinetic energy at the bottom.
They tie. The acceleration for a rolling object on a ramp is \(a = g\sin\theta/(1 + \beta)\) where \(\beta = I/(MR^2)\) is the shape factor. \(M\) and \(R\) drop out completely. Only the shape matters.

This is a classic conceptual trap. The intuition "more mass = more force" is correct, but more mass also means more inertia — they cancel exactly. This is why Galileo's observation "all objects fall at the same rate" extends to rolling.

A bowling ball (solid sphere, 7 kg, 0.1 m radius) and a small marble (solid sphere, 0.01 kg, 0.005 m radius) are released at the top of the same ramp. Which reaches the bottom first?

Forgetting that a straight-line-moving point mass still has angular momentum

A bullet flies past a post. About the post, does the bullet have angular momentum?

No, because the bullet is flying straight, not going around the post.
Yes. \(L = mvr_\perp\), where \(r_\perp\) is the perpendicular distance from the post to the bullet's line of motion. The bullet has a well-defined, nonzero angular momentum about the post (unless the bullet's path passes through the post exactly).

Angular momentum depends on the axis, not on whether the motion is circular. This becomes critical in "bullet embeds in rod" collision problems — you need the bullet's pre-collision \(L\) about the rod's pivot even though the bullet was moving in a straight line.

A 0.05 kg bullet moving at 200 m/s strikes and embeds in the end of a 2 m rod (pivoted at the other end, mass 5 kg, at rest). Find the rod+bullet angular velocity right after the collision.

· · ·
09
Tutor's corner

What a decade of teaching this unit has taught me

What every student struggles with (in order)

The #1 sticking point in Unit 6 is subtle: students who aced Units 3 and 4 (translational energy and momentum) try to apply the exact same templates here, and end up conflating angular momentum conservation with energy conservation. These are separate laws. Translational linear momentum and kinetic energy are both conserved in elastic collisions and neither in perfectly inelastic ones. But in rotational problems, especially ones where the system reshapes itself (a skater pulling in arms, a child walking inward on a merry-go-round), angular momentum is conserved while kinetic energy changes. The skater's muscles supply energy. This possibility doesn't exist for a simple point-mass collision in Units 3–4, which is why students have never had to separate the two concepts until now.

The #2 struggle is recognizing that a straight-line-moving point mass has angular momentum about a chosen axis. This shows up in every "bullet embeds in rod" problem. The equation \(L = mvr_\perp\) is the cleanest way to handle it.

What separates a 4 from a 5

Instinct for when angular momentum is conserved. Students who score 5 read a problem and immediately classify: "This is a rotational collision → \(L\) conservation. This is a ramp rolling problem → energy conservation. This is combined translation+rotation → two Newton's laws linked by rolling constraint." Students who score 4 have to derive the correct approach for each problem, which costs time on the AP exam. Build the classification reflex; it's what turns a 45-minute FRQ problem into a 15-minute one.

The question that reveals true understanding

"A figure skater pulls her arms in and speeds up. Where did the extra kinetic energy come from?" Students who say "from conservation of angular momentum" have missed the point — conservation of \(L\) predicts the new angular velocity, but doesn't supply any energy. The correct answer: the skater's muscles did positive work pulling her arms in against the outward "centrifugal tendency" felt by her extended limbs. Energy came from her body, specifically from ATP in her muscles. This question separates students who've memorized formulas from those who've integrated Units 3–4 with Unit 6.

Exam-day strategy

Classify the problem first. Every Unit 6 problem falls into one of three buckets:

(1) Rolling energy problems — object rolls down/up a ramp or surface → energy conservation, \(K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2\), use \(v = R\omega\).

(2) Angular momentum conservation — isolated rotating system, either changes shape or has internal collisions → \(L_i = L_f\). If you need to find \(\omega\) after, use this. If you need energy after, compute \(K = \tfrac{1}{2}I\omega^2\) with the new \(I\) and \(\omega\).

(3) Torque applied over time or angle → use \(\tau \Delta t = \Delta L\) or \(W = \tau \Delta\theta = \Delta K_{\text{rot}}\). These show up as graph-reading problems on the AP exam.

Before starting any calculation, write at the top of your page which bucket the problem is in. This 10-second check saves minutes of wasted work.

A surprising real-world connection

Neutron stars are spinning remnants of collapsed massive stars. Before collapse, the star is roughly the size of the sun and rotates once every few days. After collapse, the same mass is compressed into a sphere about 10 km across — rotational inertia drops by a factor of \(10^{10}\). By angular momentum conservation, \(\omega\) must go up by the same factor. A typical pulsar rotates hundreds of times per second, with some rotating over a thousand times per second. This is the exact same physics as the figure skater pulling in her arms, scaled up by a factor of \(10^{10}\). When astronomers first discovered pulsars in 1967, conservation of angular momentum was what let them immediately identify the rotating object as something very small. Scale independence of physical law made stellar-physics predictions from ice-rink physics.

One piece of advice before tomorrow's test

If your test is combined Units 5 and 6: the most likely free-response is either (a) a rolling-object-on-a-ramp problem or (b) an angular momentum conservation problem with a collision or shape change. Both of those are drilled in Worksheets B and C here. If you only have time for two problems tonight, do Problem 16 (bullet+rod collision) and Problem 20 (hollow sphere rolling on ramp). These two templates cover 60% of what Unit 6 tests.

Also: for any problem involving both translation and rotation (rolling, yo-yos, pulleys with mass), always use \(v = R\omega\) explicitly as your third equation. The most common way students lose points on these problems is by setting up two Newton's laws correctly and then forgetting the no-slip constraint — that's the single step that makes the algebra work.

Finally: don't panic if a problem looks unfamiliar. Every Unit 6 problem reduces to one of three templates. Spend the first 30 seconds identifying which. Then the rest is just algebra.

End of Unit 6 · Energy & Momentum of Rotating Systems   ·  First Principles Physics