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Rotational Motion · Chapter 10

The Rolling Race

Predict the finishing order of a solid sphere, solid cylinder, and hollow cylinder rolling down an inclined plane using conservation of energy and rotational kinematics — then test your predictions.

Theory — Conservation of Energy in Rolling Motion

The Rolling Constraint

When an object rolls without slipping, its linear velocity (v) and angular velocity (ω) are linked by the radius R:

Rolling Without Slipping v = ω · R     therefore     ω = v / R

Energy Conservation — The Master Equation

As each object rolls from height h to the bottom of the incline, all gravitational potential energy converts into translational kinetic energy (motion of the center of mass) and rotational kinetic energy (spinning). Setting PE = KE_total:

Conservation of Energy Mgh = ½Mv² + ½Iω²

Substitute ω = v/R:
Mgh = ½Mv² + ½I(v/R)²
Mgh = ½v²(M + I/R²)
Mgh = ½Mv²(1 + I/MR²)

Solve for v:
v = √[ 2gh / (1 + I/MR²) ]

The critical insight: the factor (1 + I/MR²) controls the final speed. A smaller I/MR² gives a higher final velocity. Since M and R cancel completely, the finish order depends only on the shape — not the mass or size.

Moment of Inertia for Each Object

Solid Sphere
I = (2/5)MR²
I/MR² = 0.400
1 + I/MR² = 1.400
v = √(10gh/7)
coefficient: 1.4286
Solid Cylinder
I = (1/2)MR²
I/MR² = 0.500
1 + I/MR² = 1.500
v = √(4gh/3)
coefficient: 1.3333
Hollow Cylinder
I = ½M(R²+R₁²)
R₁ = 2R/3 → I/MR² = 13/18
1 + I/MR² = 1.722
v = √(36gh/31)
coefficient: 1.1613

Step-by-Step Derivations

Solid Sphere — I = 2MR²/5

Mgh = ½Mv² + ½(2MR²/5)(v/R)²
gh = ½v² + (1/5)v²
gh = v²(5/10 + 2/10) = 7v²/10
v = √(10gh/7) ≈ √(1.4286·gh)

Solid Cylinder — I = MR²/2

Mgh = ½Mv² + ½(MR²/2)(v/R)²
gh = ½v² + ¼v²
gh = 3v²/4
v = √(4gh/3) ≈ √(1.3333·gh)

Hollow Cylinder — I = ½M(R² + R₁²), where R₁ = 2R/3

R₁ = 2R/3 → R₁² = 4R²/9
I = ½M(R² + 4R²/9) = ½M(13R²/9) = 13MR²/18
I/MR² = 13/18 ≈ 0.7222

Mgh = ½Mv² + ½(13MR²/18)(v/R)²
gh = ½v² + (13/36)v² = (18/36 + 13/36)v² = 31v²/36
v = √(36gh/31) ≈ √(1.1613·gh)

Predicted Finish Order

Higher coefficient → higher final speed → wins the race. Sphere has the largest coefficient:

1
Solid Sphere
coeff = 1.4286 · gh
2
Solid Cylinder
coeff = 1.3333 · gh
3
Hollow Cylinder
coeff = 1.1613 · gh

Key result: Mass and radius cancel out entirely. A heavier or larger sphere still beats a lighter cylinder — shape is everything. The hollow cylinder loses because more of its mass is far from the center, storing more energy as rotation and leaving less as translational speed.

Instructions — Running the Virtual Experiment

1
Record your predictions first. Before running the simulation, use the theory equations to calculate the expected final velocity for each object at your chosen height h. Write these in your data table.
2
Set the height (h) using the slider in the simulation. This is the vertical drop. Start with h = 0.50 m.
3
Set the incline angle using its slider. Note that changing the angle does NOT change the final speed — final velocity depends only on h, not the slope. This is an important result of energy conservation.
4
Click Start Race to release all three objects simultaneously from the top of the ramp. Watch which reaches the bottom first. The objects roll realistically — notice the hollow cylinder spins faster for the same forward speed.
5
Record your calculated final velocities using the theory equations in your lab notebook, then enter them in the table. Compare to the simulation values and compute the percent difference.
6
Test the mass independence: Use the Mass Multiplier slider to double the solid cylinder's mass. Run the race again. The cylinder should finish in the same position — confirming that mass does not affect the outcome.
7
Answer the practice questions at the end of the lab page to prepare for exams.
8
Complete the results table and write your lab report following the example in the Report section. Your grade depends on the accuracy of your calculations and the clarity of your work.

Simulation — The Rolling Race

Rolling Race Simulation | 🔴 Sphere   🔵 Solid Cylinder   🟢 Hollow Cylinder Set parameters and click Start Race

Ramp Parameters

0.50 m
30°

Object Properties

1× (normal)
Hollow cylinder inner radius R₁ = 2R/3 (fixed as per lab specification)
Object Formula Calculated v (m/s) Simulated v (m/s) % Difference Your Calculated v
Solid Sphere √(10gh/7)
Solid Cylinder √(4gh/3)
Hollow Cylinder √(36gh/31)

Example Lab Report

The following is a sample lab report demonstrating the expected format, level of detail, and equation work required. Use this as a guide when writing your own report.

Rotational Motion: The Rolling Race

Physics 171 | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To calculate the final velocity of three rolling objects — a solid sphere, solid cylinder, and hollow cylinder — as they roll without slipping down an inclined plane, and to verify these predictions experimentally. This lab demonstrates conservation of energy when gravitational potential energy converts to both translational and rotational kinetic energy.

Theory

When an object rolls without slipping down an incline of height h, energy conservation gives:

Mgh = ½Mv² + ½Iω²

Using the rolling constraint ω = v/R and solving for v:

v = √[ 2gh / (1 + I/MR²) ]

Note that mass M and radius R cancel, so the result depends only on the shape factor I/MR².

Calculations — h = 0.50 m, g = 9.81 m/s²

Solid Sphere (I = 2MR²/5 → I/MR² = 0.400):

v = √[2(9.81)(0.50) / (1 + 0.400)] = √[9.81 / 1.400] = √7.007 = 2.647 m/s

Solid Cylinder (I = MR²/2 → I/MR² = 0.500):

v = √[2(9.81)(0.50) / (1 + 0.500)] = √[9.81 / 1.500] = √6.540 = 2.557 m/s

Hollow Cylinder (R₁ = 2R/3 → I = 13MR²/18 → I/MR² = 13/18 = 0.722):

v = √[2(9.81)(0.50) / (1 + 0.722)] = √[9.81 / 1.722] = √5.697 = 2.387 m/s

Results Table

ObjectSimulation v (m/s)Your Calculated v (m/s)% DifferenceFinish Position
Solid Sphere2.6472.630.6%1st
Solid Cylinder2.5572.540.7%2nd
Hollow Cylinder2.3872.370.7%3rd

Discussion

The student-calculated velocities (using energy conservation equations) agreed with the simulation values to within 1%, confirming that the theory accurately predicts rolling motion. Any small differences are due to rounding in the calculations.

The most significant finding is that mass and radius have no effect on the finish order — only the distribution of mass (captured by I/MR²) determines which object wins. This is a powerful demonstration of how the moment of inertia governs rotational dynamics.

The hollow cylinder stored the greatest proportion of its energy as rotation (41.9%) because its mass is concentrated far from the axis, giving it a large moment of inertia relative to MR². This explains why it loses the race despite having the same mass as the solid cylinder.

Conclusion

The rolling race experiment successfully demonstrated conservation of energy in rotational motion. The predicted finish order (sphere first, solid cylinder second, hollow cylinder third) was confirmed by the simulation. Student-calculated velocities matched simulation values within 1%. The key conclusion is that for rolling objects, the shape — not the mass or size — determines the outcome.

Practice Questions

Use these questions to prepare for exams. Show all work and include units.

Question 1
A solid sphere rolls down a ramp of height h = 1.20 m. What is its final speed at the bottom? How does this compare to a solid cylinder released from the same height?
Hint: Use v = √[2gh/(1 + I/MR²)] for each object separately.
Question 2
A hollow cylinder has an inner radius R₁ = R/2 (not 2R/3). Calculate the new I/MR² ratio and determine whether this hollow cylinder would beat a solid cylinder in a rolling race.
Hint: I = ½M(R² + R₁²). Substitute R₁ = R/2 and simplify.
Question 3
Two solid cylinders are released from rest at the top of the same incline. Cylinder A has mass 0.5 kg and radius 3 cm. Cylinder B has mass 2.0 kg and radius 8 cm. Which reaches the bottom first? Justify your answer mathematically.
Hint: For a solid cylinder, I/MR² = 1/2 regardless of mass or radius. What does that tell you?
Question 4
A solid sphere rolls down a ramp of height h = 0.80 m. What percentage of its total kinetic energy at the bottom is stored as rotational kinetic energy?
Hint: KE_rot / KE_total = (½Iω²) / (Mgh). For sphere: this equals (I/MR²)/(1 + I/MR²).
Question 5
An object slides (no rolling) down a frictionless ramp of height h = 0.60 m. How does its final speed compare to a solid sphere rolling (without slipping) down a ramp of the same height? Which is faster and by what factor?
Hint: For sliding: v = √(2gh). For rolling sphere: v = √(10gh/7). Take the ratio.
Question 6 — Challenge
A solid cylinder rolls without slipping along a horizontal surface at v = 3.0 m/s and then rolls up an incline. How high does it rise before stopping? Does a sphere released with the same initial speed rise to the same height?
Hint: Use energy conservation in reverse — all KE converts back to PE. Does mass appear in the equation?