§ 18. Moment of Inertia & Rolling Kinetic Energy (Grade 10–11 · Uni)
Race between a solid cylinder and a hollow hoop rolling without slipping. In rolling without slipping, potential energy splits into translational and rotational kinetic energy: the body with smaller dimensionless inertia β wins.
Theory Summary & Scientific Mechanism
When an axisymmetric rigid body of mass m and radius R rolls without slipping (v = ω R) down an incline, König’s theorem states that its total kinetic energy is the sum of translational kinetic energy Eₜrans = (1)/(2)mv² and rotational kinetic energy Eᵣot = (1)/(2)Iω².
Expressing the moment of inertia via the dimensionless shape factor β = (I)/(mR²) (β = 0.4 for a solid sphere, β = 0.5 for a solid cylinder/disk, β = 1.0 for a thin hoop) gives Eₖ = (1)/(2)mv²(1 + β). Conservation of energy mgh = (1)/(2)mv²(1 + β) yields the linear downhill acceleration a = (gsin α)/(1 + β).
Remarkably, the rolling acceleration a is completely independent of total mass m and radius R, depending only on the mass distribution factor β! Because a thin hoop (β = 1.0) diverts 50% of its energy into rotation whereas a solid disk (β = 0.5) diverts only 33.3%, the solid disk or sphere always beats the hoop to the bottom.
Everyday Analogy
Race a solid marble against a hollow ring (like a roll of tape) down a ramp! The solid sphere always wins regardless of weight because the ring’s mass sits far out on the rim, taking more energy just to spin up.
Interactive Question & Derivation
A solid disk (β = 0.5) and a thin-walled hoop (β = 1.0) of equal mass and radius are released simultaneously from rest at the top of an incline and roll without slipping. Which body reaches the bottom first?
- A: Both arrive simultaneously because they have identical mass m and radius R
- B: The solid disk wins because its inertia factor β = 0.5 is smaller, so a smaller fraction of energy goes into spinning it up
- C: The hoop finishes first because all of its mass is concentrated at the rim
Study Hint: Compare the rotational inertia factor β = I/(mR²): is it smaller for a solid sphere (0.4) or a hoop (1.0), and which one gets the higher linear acceleration a = (gsin α)/(1+β)?
Proof & Explanation: By conservation of energy mgh = (1)/(2)mv² + (1)/(2)Iω² = (1)/(2)mv²(1 + β), where β = I/(mR²). The linear acceleration down the slope is a = (gsin α)/(1 + β). Since a solid cylinder has β = 0.5 (a = (2)/(3)gsin α) while a thin hoop has β = 1.0 (a = (1)/(2)gsin α), the solid disk always beats the hoop regardless of mass or radius!
Key Equations & 3D Simulation Parameters
a = (gsin α)/(1 + β), vₑnd = √((2gh)/(1 + β)), (Eᵣot)/(Eₜrans) = β
- Inertia factor β = I/(mR²): 0.2 … 1 (default: 0.5)
- Incline angle α: 14 … 40 ° (default: 26 °)
- Rolling body radius R: 0.45 … 0.9 m (default: 0.65 m)
- Body mass m: 1 … 4 kg (default: 2 kg)