Q.A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

In the full wheel, internal (interatomic) forces provide each particle's centripetal acceleration; by Newton's third law these cancel in pairs, so they contribute zero net external force or torque — the wheel needs no external help to keep spinning uniformly. …

Reconciling equilibrium with centripetal acceleration

Mechanical equilibrium of a rigid body means: zero net external force (so the centre of mass doesn't accelerate) and zero net external torque (so the angular velocity doesn't change). A uniformly spinning wheel satisfies both — its centre of mass sits still at the axis, and $\omega$ is constant.

But every particle of the wheel does have centripetal acceleration, directed toward the centre. This is not a contradiction: that acceleration is supplied by internal forces — the elastic (interatomic) forces holding the rigid body together. By Newton's third law, these internal forces occur in equal-and-opposite pairs between neighbouring particles. Summed over the whole wheel, they cancel completely and contribute zero net force and zero net torque on the wheel as a whole. Internal forces can accelerate individual particles; they cannot move or spin up the body as a whole. That is why no external force or torque is required to sustain the wheel's uniform rotation, even though each particle is constantly accelerating.

Does the half-wheel need an external torque to keep spinning?

The same internal-force argument applies to any rigid body, symmetric or not, as long as it rotates about an axis through its own centre of mass. For a flat (planar) body like a half-wheel, an axis perpendicular to its plane through its centre of mass is automatically a principal axis of the body — a mathematical fact true for every lamina, regardless of how irregular its shape is. Rotating about a principal axis means the angular momentum vector $\vec{L}=I\vec{\omega}$ stays fixed along that same axis as the body spins, with no tendency to wander.

So once the half-wheel is spinning at angular velocity $\omega$ about an axis through its own centre of mass:

• its centre of mass is on the axis and stationary — no net external force needed;
• $\vec{L}$ stays constant along the axis — no net external torque needed. …

Concept: Mechanical Equilibrium Means Zero Net External Force/Torque — Not Zero Acceleration of Every Particle

Step 1: State the equilibrium conditions precisely

A rigid body is in mechanical equilibrium when the net external force is zero (centre of mass doesn't accelerate) and the net external torque is zero (angular velocity doesn't change) — this says nothing about individual particles.

Step 2: Reconcile this with centripetal acceleration

Every particle of a spinning wheel does have centripetal acceleration toward the centre. This acceleration is supplied entirely by internal (interatomic/elastic) forces holding the rigid body together. By Newton's third law these internal forces occur in equal-and-opposite pairs; summed over the whole body they cancel completely, contributing zero net external force and zero net external torque. Internal forces can accelerate individual particles but cannot move or spin up the body as a whole — hence no external force/torque is needed to sustain uniform rotation.

Step 3: Apply the same argument to a half-wheel

The internal-force cancellation argument above works for any rigid body — symmetric or not — as long as it rotates about an axis through its own centre of mass. For a flat (planar) body like a half-wheel, an axis perpendicular to its plane through its centre of mass is automatically a principal axis (a general fact for any lamina, regardless of shape): the angular momentum $\vec L = I\vec\omega$ then stays fixed along that axis, with no tendency to wander.

Step 4: Conclude

Once spinning at constant $\omega$ about an axis through its centre of mass:

• the centre of mass sits on the axis and stays stationary → no net external force needed; …