Rolling: Rotation Plus Translation

A bicycle wheel neither simply slides nor simply spins. It does both at once, and the combination has some surprises: the bit touching the road is standing still.

Translation plus rotation#

When a body both moves and turns, describe it in two parts: the motion of its centre of mass, and its rotation about an axis through the centre of mass. The first obeys $\vec F_{\text{ext}} = M\vec a_{cm}$ as always. For the second there is a pleasant fact: although the centre-of-mass frame may be accelerating, the pseudo-forces −mᵢa all act like a single force through the centre of mass, so they have no torque about it. Hence τ = Iα still holds there.

Pure rolling#

Paint one spoke of a bicycle wheel. Each time the spoke returns to pointing straight down, the wheel has turned once and the bike has moved one circumference, 2πR. Over a short time the centre moves Δx while the wheel turns Δθ, with Δx = RΔθ:

Now add velocities. Relative to the centre, every rim point moves at Rω along the rim. At the bottom that is backward, at the top forward. Adding the centre’s forward velocity v:

If the condition fails, the wheel slips:

• $v_{cm} > R\omega$: forward slipping. The contact point rubs forward, as when you brake hard on a wet road and the wheel barely turns.
• $v_{cm} < R\omega$: backward slipping. The contact point rubs backward, as when a wheel spins in mud.

Kinetic energy of a rolling body#

Write each particle’s velocity as the centre’s velocity plus its velocity relative to the centre, and expand $\sum\tfrac12 m_i v_i^2$. The cross term is proportional to the total momentum in the centre-of-mass frame, which is zero, leaving

For pure rolling, put $v_{cm} = R\omega$: $K = \tfrac12(I_{cm} + MR^2)\omega^2 = \tfrac12 I_P\omega^2$, where $I_P$ is the moment of inertia about the contact line (parallel axis theorem). At each instant a rolling body is in pure rotation about its contact point.

Down a slope#

Release a body from rest on a slope of angle θ. If it rolls without slipping, static friction does no work (the contact point doesn’t move), so after a drop h, with $I_{cm} = kMR^2$:

Angular momentum of a rolling body#

About a fixed point O, splitting each position and velocity into centre-of-mass and relative parts gives two terms (the cross terms vanish just as for energy):

the spin about the centre of mass, plus the angular momentum of the whole mass imagined concentrated at the centre of mass.