Rotational Mechanics · Rolling Without Slipping

Two disks roll around a fixed disk — when do they meet again?

Two small disks sit on the rim of a large stationary disk, touching each other. They roll without slipping in opposite directions, one spinning at ω, the other at 2ω. They drift apart, travel all the way round, and the question is the time τ until they touch again. Watch the two of them go.

Separating · relative angle
Drag to rotate · scroll to zoom
Animation speed 1.0×
Large disk (radius R)held stationary
Small disk · spin ωrolls one way around
Small disk · spin 2ωrolls the opposite way
Contact flashwhen the disks touch
For this problem (r = R/50)
Orbit speed Ω = ω·r/(R+r)ω / 51
Relative speed ω_rel3ω / 51
Start gap Δθ = 2r/(R+r)2 / 51
Angle to close2π − 4/51
The 3D disks are drawn much larger than R/50 so you can see them spin — the physics (rolling, opposite directions, meeting point) is the same.

What is happening

01 / ROLL

Spin drives the orbit

Because the big disk is fixed, the point of contact is momentarily still. A small disk spinning at ω therefore creeps around the rim at a much slower orbital rate Ω.

Ω = ω · r / (R + r)
02 / CHASE

Opposite ways add up

One disk goes clockwise, the other anticlockwise, so their orbital speeds add. The gap between them closes at the sum of the two rates.

ω_rel = (ω + 2ω) · r/(R+r) = 3ω r/(R+r)
03 / CLOSE

Almost a full lap

They begin only a sliver Δθ apart, so to touch again the centres must sweep through everything except that sliver on each side: 2π − 2Δθ.

τ = (2π − 2Δθ) / ω_rel
Putting in r = R/50
r/(R+r) = (R/50)/(51R/50) = 1/51  →  Δθ = 2/51,   ω_rel = 3ω/51
τ = (2π − 2·(2/51)) / (3ω/51) = 51 × (2π − 4/51) / 3ω
A51 (2π − 4/51) / ω
B51 (2π − 2/51) / 3ω
C51 (2π − 4/51) / 3ω  ✓
D51 (2π − 2/51) / ω