Forces acting when turning a circle on a unicycle

[Fibo112]

Hello.
Lets say I am riding a unicycle rolling along the wheels direction with some velocity v0. I know that for the unicycle to turn a curve the centripedal force must be provided by the static friction with the rode that acts against any slipping of the wheel.

I am however having trouble picturing how this tales place. When the unicycle begins to turn the curve the part of it in contact with the ground is moving at some velocity. The way I picture it I keep thinking the friction should act in the oposite direction of the velocity, but for the turn to work is has to be normal to the velocity...

 

[lekh2003]

The way I picture it I keep thinking the friction should act in the oposite direction of the velocity,

Friction has to be opposite to the force acting on the unicycle. There is the force of you moving the unicycle forward and then friction pushing back. When you turn, the centripetal force moves inward and then the force of friction acts in the opposite direction.

Hope that helps.

 

[A.T.]

...friction should act in the oposite direction of the velocity...

For a non-slipping wheel there is no relative velocity between the contact patches, so it's static friction, which can have any direction needed to prevent slippage.

 

[Fibo112]

Lets say the unicycle is rolling without slipping straight ahead with some velocity. Then the direction of the wheel changes instantaneously by some angle. What will happen next? The wheel cannot keep rolling without sliping because that would mean an instantaneous change in velocty.

 

[jbriggs444]

Lets say the unicycle is rolling without slipping straight ahead with some velocity. Then the direction of the wheel changes instantaneously by some angle. What will happen next? The wheel cannot keep rolling without sliping because that would mean an instantaneous change in velocty.

First, understand that a unicycle wheel is powered. The rider has their feet on the pedals. It is entirely possible for the static friction at the contact patch to have a forward or rearward component in addition to any sideways component. This means that if you twist the unicycle wheel in place nothing changes immediately. The same forces that were there before the twist can still be there after the twist.

What does change is the ability of the wheel to roll forward or backward [primarily] under control of the pedals. The unicyclist uses such twists together with forward and backward force on the pedals to maintain balance.

If one imagines a unicycle wheel where the rider has imprudently removed their feet from the pedals along with an inward twist to the wheel, there would be an effect that would manifest rapidly. With nothing to restrain it, the wheel would no longer be able to apply or resist any fore-and-aft net force at the contact patch. The wheel would roll rapidly backward and the unicyclist would fall down.

[The net force on the wheel being the vector sum of the mostly lengthwise force from the unicycle shaft, the normal contact force from the road and any sideways frictional force from the road. The normal force is neutral with respect to any fore and aft movement. The sideways frictional force is neutral with respect to any fore and aft movement. The lengthwise force along the unicycle axis then results in a rearward roll]

If we imagine an outward twist, the same applies, but the wheel would roll rapidly forward and the unicyclist would again fall down.

 

[rcgldr]

After initiating a lean by countersteering, the wheel is turned inwards, and the wheel attempts to roll inwards of the current velocity. This results in an outward force applied by the tire to the pavement, and an inwards force applied by the pavement to the tire, which is the source of the centripetal force (the force perpendicular to current velocity).

As for the static friction issue, the tire pushes backwards on the pavement, and the pavement pushes forward on the bike. The forwards force times distance equals the work done in overcoming drag and/or accelerating the unicycle and rider. Since the point of contact moves forward with the unicycle, static friction can do work (from the pavement frame of reference, the work is done on the unicycle, rider, and air, from the unicycle frame of reference, the work is done on the Earth (it's moved / rotated backwards at a very tiny velocity)), and negative work is done on the air (it's slowed down wrt unicycle).

 

[jbriggs444]

Since the point of contact moves forward with the unicycle

This is erroneous. The motion of the point of contact is irrelevant to work done on the wheel. The motion of the wheel material at the point of contact is relevant. That velocity is zero as long as slipping does not occur.

If one were contemplating center-of-mass work then the motion of the rider, the unicycle and the bulk of the wheel would be relevant to work done. But that is still not the same as the motion of the point of contact.

 

[A.T.]

Lets say the unicycle is rolling without slipping straight ahead with some velocity. Then the direction of the wheel changes instantaneously by some angle.

...which never happens instantaneously in real life.

What will happen next? The wheel cannot keep rolling without sliping because that would mean an instantaneous change in velocty.

If the wheels slips, then the kinetic friction will usually be in the direction opposite of the relative motion of the contact patches, which for a spinning and slipping wheel is different than the motion of the vehicle. So the kinetic friction can have a centripetal component.

 

[rcgldr]

This is erroneous. The motion of the point of contact is irrelevant to work done on the wheel. The motion of the wheel material at the point of contact is relevant. That velocity is zero as long as slipping does not occur.

If "point of contact" is defined as the current position where the tire touches the pavement with respect to the pavement, then the "point of contact" has a velocity with respect to the pavement. Note that "contact patch" as used by the vehicle industry has this same dynamic definition.

If a specific point on the tire surface is observed from the pavement frame of reference, then the path of that point is a cycloid, and the relative velocity is zero only for an instant in time, once per revolution, while the average velocity of that point wrt pavement will be the same as the average velocity of the tire axle wrt pavement.

 

[jbriggs444]

If "point of contact" is defined as the current position where the tire touches the pavement with respect to the pavement, then the "point of contact" has a velocity with respect to the pavement. Note that "contact patch" as used by the vehicle industry has this same dynamic definition.

None of which has the slightest relationship to the definition of work.

 

[rcgldr]

As for the static friction issue, the tire pushes backwards on the pavement, and the pavement pushes forward on the bike. The forwards force times distance equals the work done in overcoming drag and/or accelerating the unicycle and rider.

If "point of contact" is defined as the current position where the tire touches the pavement with respect to the pavement, then the "point of contact" has a velocity with respect to the pavement.

None of which has the slightest relationship to the definition of work.

I posted that the force times distance traveled by the point of contact equals the work done. I never claimed that the pavement was actually performing this work, as the power source is the rider, not the pavement.

 

[A.T.]

I posted that the force times distance traveled by the point of contact equals the work done.

It happens to equal "centre of mass work" in this case, because the geometrical point of contact happens to move at the same velocity as the centre of mass. But as jbriggs444 said, that geometrical point has no relevance for work done.