Set up a 1.0 m ramp at about 15° in front of the class. Hold up three objects of roughly the same radius (≈ 3 cm) — a solid steel sphere (a large ball bearing), a solid disk/cylinder (a hockey puck on edge or a solid wooden dowel section), and a thin metal ring or PVC pipe cap. Ask: 'If I release all three from the same height at the same instant, in what order do they reach the bottom?' Take a show-of-hands vote for four options: (a) they tie, (b) heaviest first, (c) sphere → disk → ring, (d) ring → disk → sphere. Release them (a starting gate is a ruler you pull away). The sphere wins, the disk is second, the ring is last, regardless of mass. Write the finish order on the board and tell them: 'By the end of class you will be able to predict this order from first principles and prove it with an energy argument.' Do NOT explain why yet — leave the puzzle open.

1. 7m

   The rolling constraint: v = rω and the velocity pattern on a wheel

   Content

   Rolling without slipping means the point on the wheel currently touching the ground is momentarily at rest relative to the ground — it is not sliding. Because that contact point is instantaneously at rest, the whole wheel can be treated as rotating about the contact point (the instantaneous axis) at angular speed ω. That makes the center of mass, one radius above the contact, move forward at vcm = rω, and the top of the wheel, two radii above the contact, move forward at 2rω = 2v. So on any rolling wheel: bottom = 0, center = v, top = 2v. The rolling constraint links translation and rotation: vcm = rω, and taking a time derivative, acm = rα. Worked check: a bicycle tire of radius r = 0.35 m rolls at v = 7.0 m/s. Then ω = v/r = 7.0/0.35 = 20 rad/s, and the top of the tire moves forward through the air at 2v = 14 m/s while the tread patch on the ground has speed 0 at the instant it touches.

   Delivery

   Emphasize the word 'instantaneously' — the contact point isn't stuck forever, it just has zero velocity in the frame of the ground at the moment it touches. Ask a student to explain why a stationary tire tread is what leaves a clean footprint in mud (no smearing) while a skidding tire leaves a streak. Pre-empt the misconception that the bottom of the wheel is moving fastest because 'it's where the action is' — it is the SLOWEST point on the wheel. Reinforce vcm = rω by asking: 'If I double ω without changing r, what happens to v?' and 'If two wheels roll at the same v but one has half the radius, which spins faster?'

2. 7m

   Energy split: translational + rotational, and why shape matters

   Content

   A rolling body carries kinetic energy in two accounts: translation of the center of mass, KEtrans = ½mv², and rotation about the center of mass, KErot = ½Iω². For rolling without slipping, ω = v/r, so KErot = ½I(v/r)² = ½(I/r²)v². Total KE = ½mv²(1 + I/mr²). The dimensionless number β = I/mr² tells you what fraction of the KE is 'stuck' in rotation: solid sphere β = ⅖, solid disk β = ½, thin ring β = 1. Now apply energy conservation from rest down a ramp of vertical drop h: mgh = ½mv²(1 + β), giving v² = 2gh/(1 + β). Notice mass cancels — only shape matters. Plug in: sphere v² = 2gh/1.4 = (10/7)gh; disk v² = 2gh/1.5 = (4/3)gh; ring v² = 2gh/2 = gh. So for h = 1.0 m: vsphere ≈ 3.74 m/s, vdisk ≈ 3.62 m/s, vring ≈ 3.13 m/s. The sphere wins because the smallest fraction of its energy goes into spin, leaving more for forward motion.

   Delivery

   Walk through the algebra deliberately — the substitution ω = v/r is where students fumble. Ask, 'Where did mass go?' after mass cancels to drive home that a bowling ball and a marble made of the same material tie. Then push on the shape dependence: 'Why does the ring lose?' The ring has all its mass at radius r, so I = mr² is as large as it can be for a given m and r, and half of its KE is trapped in rotation. Head off the misconception that 'all round things roll the same' by pointing at the demo they just watched — the ranking is a direct consequence of β.

3. 6m

   Static friction on the incline: the force that does no work

   Content

   For a sphere accelerating down a ramp of angle θ, three forces act at the center of mass description: gravity mg (straight down), normal force N (perpendicular to the ramp), and static friction fₛ acting along the surface. Which way does fₛ point? If the ramp were frictionless the sphere would slide without spinning; friction is what makes it spin. To produce angular acceleration in the direction of rolling (front of the sphere rotating downhill), friction at the contact point must push UP the incline, providing a torque about the center of mass. Newton's second law along the incline: mg sinθ − fₛ = ma. Torque about the center: fₛ r = Iα, and rolling gives α = a/r, so fₛ = Ia/r² = βma. Combining: a = g sinθ / (1 + β). Same shape ranking as before. Now the subtle point: fₛ is static friction, not kinetic. The contact patch does not slide, so the point where fₛ acts has zero velocity. Work is force times displacement of the point of application, and that displacement is zero at every instant — so fₛ does no work on the sphere and dissipates no energy. That is exactly why energy conservation was valid in the previous beat.

   Delivery

   Ask the class to predict the direction of the friction arrow BEFORE you reveal it — many will guess 'down the incline' by analogy with sliding friction opposing motion. Then reason it out: without friction, no spin; friction must be the torque-provider; therefore it points up the incline. Then hit the misconception head-on: 'Doesn't friction always dissipate energy?' No — kinetic friction does, because the surfaces slide. Static friction on a rolling object acts at a point that is momentarily at rest, so its instantaneous power (F·v) is zero. Connect back to the demo: the sphere rolled without heating up the ramp because no energy leaked to friction.

1. 25m

   Rolling Race Lab: Ranking Shapes and Testing v = rωLab

   Groups of 3–4. Each group gets the ramp, the three shapes, and a photogate/motion sensor at the bottom. Targets SP 1 (Creating Representations — FBD and energy-bar reasoning), SP 2 (Mathematical Routines — apply v² = 2gh/(1+β)), and SP 3 (Argumentation — justify the ranking from data). Before lab: hand each student the handout below (or project it). Walk around and check that each group has measured h correctly (vertical drop, not ramp length) and identifies β for each shape from the given moments of inertia. Enforce release from rest at the same starting line. Expect the sphere to win by a comfortable margin, the disk a close second, and the ring clearly last. Groups will compute predicted v at the bottom from energy conservation and compare to measured v from the photogate. Reserve the last 4 minutes for whiteboarding — ask each group to defend their ranking with ONE sentence citing β and ONE sentence citing static friction. Student handout: Part 1 — Setup and measurements Measure and record: - Ramp angle θ = ______ ° - Vertical drop from release line to photogate h = ______ m - Radius of sphere rₛ = ______ m - Radius of disk r_d = ______ m - Radius of ring r_r = ______ m Part 2 — Predict before you roll (SP 2) Use v² = 2gh/(1 + β) with g = 9.8 m/s². - Sphere (β = ⅖): predicted v = ______ m/s - Disk (β = ½): predicted v = ______ m/s - Ring (β = 1): predicted v = ______ m/s Predicted finish order (fastest first): ______, ______, ______ Part 3 — Measure (SP 3) Release each shape from rest at the release line. Record the speed at the photogate. Do three trials each and average. - Sphere: v₁ = ___ v₂ = ___ v₃ = ___ average = ___ m/s - Disk: v₁ = ___ v₂ = ___ v₃ = ___ average = ___ m/s - Ring: v₁ = ___ v₂ = ___ v₃ = ___ average = ___ m/s Part 4 — Test the rolling constraint (SP 2) For the sphere trial, use vcm = rω to compute the angular speed at the bottom: - ω = vcm / rₛ = ______ rad/s - How fast is the TOP of the sphere moving at that instant? ______ m/s - How fast is the CONTACT POINT moving? ______ m/s Part 5 — Argue from evidence (SP 3) Answer in complete sentences: 1. Did your measured ranking match your prediction? Cite specific speeds. 2. Use β to explain WHY the ring finished last even though it may have been heavier than the sphere. 3. Static friction acts on each object as it rolls. Explain, using the definition of work, why this friction does not appear as a loss term in your energy calculation. The contact point's velocity is the key. 4. If you replaced the ring with a heavier ring of the same radius, what would happen to its finish time? Justify.

   Materials

   • Adjustable ramp or board (≈ 1.0–1.5 m long) set at 10–15°
   • Meter stick and protractor
   • Solid steel ball or marble (radius ~1–2 cm)
   • Solid disk or short solid cylinder (a hockey puck rolled on edge, or a solid wooden cylinder)
   • Thin-walled ring or hoop (metal ring or short section of PVC pipe)
   • Two photogates OR one motion sensor at the bottom of the ramp
   • Stopwatch as backup
   • Digital calipers or ruler to measure radii
   • Whiteboard for group results

   Example outputs

   • Sample data (θ = 12°, h = 0.25 m): sphere measured 1.85 m/s (predicted 1.89 m/s), disk 1.78 m/s (predicted 1.81 m/s), ring 1.55 m/s (predicted 1.57 m/s). Ranking matches prediction; small deficits attributed to axle/bearing effects and slight slipping.
   • Sample Part 5 answer: 'The ring finished last because β = 1 for a thin ring, so half of its KE was tied up in rotation, leaving less for translation. A heavier ring of the same radius would finish at the same time because mass cancels in v² = 2gh/(1+β). Static friction did no work because the contact point had zero velocity at each instant, so no energy was lost to it — that is why our measured speeds landed within a few percent of the energy-conservation prediction.'

1. A uniform solid sphere (I = ⅖mr²) and a thin hollow ring (I = mr²) have equal masses and equal radii. Both are released from rest at the same height h on the same incline and roll without slipping to the bottom. Determine the ratio v(sphere) / v(ring) at the bottom, and state which object arrives first. Targets SP 2.

   calculationEnergy conservation: mgh = ½mv²(1 + β). Sphere: v² = 2gh/(1 + ⅖) = (10/7)gh Ring: v² = 2gh/(1 + 1) = gh Ratio: v(sphere)/v(ring) = √((10/7)/1) = √(10/7) ≈ 1.20 The sphere arrives first because a larger fraction of its energy goes into translation (β is smaller).

2. A bowling ball rolls without slipping across a level floor at 4.0 m/s. Its radius is 0.11 m. (a) What is its angular speed ω? (b) What is the instantaneous velocity of the point on the ball currently touching the floor, and of the point on the top of the ball? (c) Explain in one sentence why the point of contact has that velocity even though the ball is clearly moving. Targets SP 1 and SP 2.

   short answer(a) ω = v/r = 4.0 / 0.11 ≈ 36 rad/s. (b) Contact point: 0 m/s. Top of ball: 2v = 8.0 m/s (forward). (c) For rolling without slipping the contact point is instantaneously at rest relative to the floor — the translational velocity of the center (+v forward) exactly cancels the rotational velocity of that point (−rω = −v).

3. A solid disk rolls without slipping down a ramp inclined at angle θ. A student claims: 'Static friction acts up the incline on the disk, and because this force has a component along the disk's motion, it does negative work on the disk and drains its kinetic energy.' Identify what is correct and what is wrong with this claim. Justify your answer using a specific physical quantity. Targets SP 3.

   short answerCorrect: static friction does point UP the incline; it provides the torque that produces the angular acceleration needed for rolling. Wrong: it does NOT do work on the disk. Work depends on the displacement of the point of application of the force, not the displacement of the center of mass. Because the disk rolls without slipping, the contact point (where fₛ acts) has zero velocity at every instant, so the instantaneous power P = fₛ · v(contact) = 0 and the work done by fₛ is zero. Energy is therefore conserved, and mgh converts entirely into ½mv² + ½Iω².

4. Two solid spheres roll down the same incline from the same height without slipping. Sphere 1 has mass m and radius r. Sphere 2 has mass 2m and radius 2r. Which sphere reaches the bottom with the greater translational speed? (A) Sphere 1 (B) Sphere 2 (C) They tie (D) Cannot be determined without knowing θ. Briefly justify. Targets SP 2 and SP 3.

   multiple choice(C) They tie. From v² = 2gh/(1 + β) and β = ⅖ for any solid sphere regardless of mass or radius, both spheres have the same v at the bottom. Mass and radius cancel — only the shape (through β) matters.

• ☐Adjustable ramp (≈ 1 m) with protractor
• ☐Solid sphere, solid disk/cylinder, and thin ring of comparable radius
• ☐Two photogates OR a motion sensor with LabQuest/computer interface
• ☐Meter stick, calipers, stopwatch as backup
• ☐Handout (Parts 1–5, one per student)
• ☐Whiteboards and markers for group ranking arguments
• ☐Calculator per student