# Rotation of a Bicycle Wheel

Circular motion at a constant velocity requires a net force toward the center of rotation.

If I stand a bicycle on its seat, wheels upwards and spin the wheels with my hands, they start rotating. Where is the net force toward the center of the wheel coming from. Is there one in this situation?

• The wheel's structure (spokes, rim, etc.) provide the necessary force – user1936752 Nov 24 '18 at 23:07

The spokes are under tension. That's the force toward the center. Provided by molecular bonds, to counteract the "centrifugal force".

Not just the spokes of course. The metal rim and the rubber tire also have molecular forces that oppose the pull to expand outward.

Think of it as a mass attached to a string and you apply a force to make it do circular motion. The force that you apply makes the wheel turn, the centripetal force and the centrifugal force cancel each other out. I think you are talking about the centripetal force when you say'net force towards the center'.

In rigid body rotation the applied force acts (you're pushing on the tire) a certain distance away from the axis of rotation (radius). The cross product of the applied force and radius give torque.

In this situation you are giving the wheel a torque. There is a centripetal force but it's effects are not immediately noticeable. The object is structured, and we use torque to describe its rotation. A centripetal force would occur if you tied a ball to a string and spun it in circles. The only obvious forces in rigid body rotation are the tangential force and radial force.

The radial force acts outward. The tangential force acts at a $$90^o$$ angle to the line of action. The line of action is the imaginary line through which your force acts. It is always connected to the axis of rotation. The vector sum of these forces is your net force.

So in summary there is a net inward force, it's just not as applicable in rigid-body rotation. Your net inward force (centripetal force) simply acts in the direction of change in velocity. This $$\Delta v$$ is inward along the circular path.