What exactly does the damping parameter beta control?

Beta only acts at the two turning points, top and bottom, multiplying the rebounding speed by a factor each time the wheel reaches an extreme. It represents losses from string flexing and axle friction; with beta equal to zero the wheel bounces forever at constant amplitude, and larger beta makes each successive oscillation visibly smaller, like a damped pendulum losing energy per swing.

What does the velocity versus position phase-space plot show?

It plots every simulated position and velocity pair as a fading dot, tracing the wheel's trajectory in phase space. With no damping this trace would retrace the same closed loop forever; as damping increases, successive loops shrink inward toward the origin, visually showing the oscillation's amplitude decaying the same way the energy bars do.

⚙️ Maxwell's Wheel

Maxwell's wheel (a yo-yo on a string) converts gravitational potential energy into translational and rotational kinetic energy as it descends, then back again as it rises. Adjust the wheel's mass distribution and observe how the period and maximum speed change.

y = 0.000 m v = 0.000 m/s

Wheel Parameters

Mass m 0.5 kg Axle radius r 0.02 m Wheel radius R 0.10 m Damping β 0.02

Energy

KE trans—

KE rot—

PE grav—

Total E—

α = r²/(R²+r²)—

I = ½mR² (solid disk)
a = g/(1 + I/mr²)
= g·r²/(R²+r²) [disk]
T = 2√(2h/a)
ω_max = v_max/r
v_max = √(2gh·r²/(R²+r²))

Rotational Dynamics

As Maxwell's wheel descends with string unrolling from its axle (radius r), both translational velocity v and angular velocity ω increase with the constraint v = ω·r. For a solid disk, moment of inertia I = ½mR², so the equation of motion gives acceleration a = g·r²/(R² + r²). The effective "gyration radius" k² = I/m = R²/2 determines how much of the gravitational work appears as rotation vs. translation. When r≪R, almost all energy goes into rotation and the descent is very slow. The wheel bounces back perfectly (no damping) or loses amplitude due to string imperfection (modeled by damping β).

About Maxwell's Wheel

This simulation numerically integrates a simplified model of Maxwell's wheel — a solid disk of mass m and radius R suspended on two strings wound around a thin axle of radius r. At every animation frame an explicit-Euler step advances the fall using the constant descent acceleration a = g·r²/(R²+r²), obtained from Newton's second law together with the rolling constraint v = ω·r and the solid-disk moment of inertia I = ½mR². When the wheel reaches the top or bottom of its travel, the velocity reverses and is scaled by (1−10β) to approximate string and friction losses, while gravitational, translational, and rotational kinetic energy are recomputed and charted every step.

🔬 What It Shows

Real-time position, spin, and a three-way energy split — gravitational (PE), translational kinetic (KE trans), and rotational kinetic (KE rot) — as the wheel descends and rebounds, charted over time and traced as a v-vs-y phase-space plot that reveals how each cycle's amplitude evolves.

🎮 How To Use It

Drag Mass m, Axle radius r, and Wheel radius R to change how the descent acceleration a = g·r²/(R²+r²) splits released energy between falling and spinning; raise Damping β to watch the bounces shrink; use Reset to restart from rest and Pause to freeze the motion for inspection.

💡 Did You Know?

Because acceleration scales with r²/(R²+r²), a wheel whose axle is ten times thinner than its rim falls at roughly 1% of free-fall acceleration — turning a drop that would take a fraction of a second in free fall into a slow-motion descent lasting many seconds, entirely from stored rotational inertia.

Frequently Asked Questions

What determines how fast Maxwell's wheel falls in this simulation?

Each frame the simulator applies a constant descent acceleration a = g·r²/(R²+r²), derived from Newton's second law for the falling axle combined with the rolling constraint v = ω·r and the solid-disk moment of inertia I = ½mR². Because r is much smaller than R for a realistic wheel, this acceleration is only a small fraction of g — most of the released energy has to go into spinning up the disk rather than speeding its fall.

Why does the wheel bounce back up instead of falling straight through the floor?

When the simulated height y reaches the bottom (y = −H0) or the string is fully wound back at the top (y = 0), the code flips the sign of the velocity and shrinks its magnitude by the factor (1−10β), then continues integrating. This mimics the string reversing direction as it rewinds around the axle at the end of each real descent or ascent.

What exactly does the damping parameter β control?

β only acts at the two turning points (top and bottom), multiplying the rebounding speed by (1−10β) each time the wheel reaches an extreme. It represents losses from string flexing and axle friction; with β = 0 the wheel bounces forever at constant amplitude, and larger β makes each successive oscillation visibly smaller, like a damped pendulum losing energy per swing.

How is the rotational kinetic energy calculated?

The simulator assumes a solid uniform disk, so moment of inertia I = ½mR² using the outer Wheel radius R (not the thin axle radius r). Angular velocity is recovered from the rolling constraint as ω = v/r, and rotational kinetic energy is KE_rot = ½Iω², plotted alongside translational KE_trans = ½mv² and gravitational PE = mg(y+H0) so their sum (Total E) stays visibly constant.

What does the v-versus-y phase-space plot show?

It plots every simulated (position, velocity) pair as a fading dot, tracing out the wheel's trajectory in phase space. With no damping this trace would retrace the same closed loop forever; as β increases, successive loops shrink inward toward the origin, visually showing the oscillation's amplitude decaying exactly the way the energy bars do.