Bike tire momentum problem

• Rockazella
In summary, if you let a bike tire with X mass and Y moment of inertia drop to the ground, it will have an acceleration and final velocity that depends on the coefficient of friction between the tire and the ground.

Rockazella

You have a bike tire with X mass and Y moment of inertia.
You hold the tire on a vertical plane just slightly above the ground. You then spin it to give it Z amount of angular momentum.

If you were to let the tire drop to the ground, what would its acceleration and final velocity be? (not taking into account drag)

anyone?

Originally posted by Rockazella
You have a bike tire with X mass and Y moment of inertia.
You hold the tire on a vertical plane just slightly above the ground. You then spin it to give it Z amount of angular momentum.

If you were to let the tire drop to the ground, what would its acceleration and final velocity be? (not taking into account drag)

It would depend on the coefficient of friction (both static and kinetic) between the tire and the ground. If the wheel bounces, you'd need the coefficient of restitution as well.

It's not a simple problem (unless you make a bunch of simplifying assumptions)

The answer depends on the friction coefficients between the tire and ground. For example if there was no friction at all, the tire would keep on rotating around its axis with constant angular velocity (conservation of angular momentum relative to the center of the wheel)and at the most could bounce up and down. With "low" friction the tire would start rolling and sliding, energy would be dissipated continuously until pure rolling sets in. With "high" friction the tire would start rolling without sliding and no energy would be dissipated after landing.
In the general case (with friction and considering an almost zero vertical landing speed just to disregard bouncing) what is conserved (immediately before and after touching the ground) is the total angular momentum with respect to landing point since all the interaction forces are bounded there and hence produce a null external momentum with respect to this pole.
The system is also pretty unstable (a small angle with the vertical direction will create a totally different evolution).

Hope this helps!

What is the "Bike tire momentum problem"?

The "Bike tire momentum problem" refers to the phenomenon where a bicycle tire, once set in motion, will continue to rotate even when the bicycle is not being actively pedaled.

What causes the "Bike tire momentum problem"?

The "Bike tire momentum problem" is caused by the conservation of angular momentum. This means that once the bicycle tire is set in motion, it will continue to spin unless acted upon by an external force.

How does the "Bike tire momentum problem" affect a cyclist?

The "Bike tire momentum problem" can affect a cyclist in a number of ways. It can make it easier for a rider to maintain their speed, but it can also make it more challenging to change direction or come to a stop.

How can a cyclist use the "Bike tire momentum problem" to their advantage?

A cyclist can use the "Bike tire momentum problem" to their advantage by learning how to control and manipulate their bicycle's momentum. This can be done through techniques such as leaning, counter-steering, and proper braking.

Are there any downsides to the "Bike tire momentum problem"?

In some cases, the "Bike tire momentum problem" can lead to accidents or falls if a cyclist is not skilled in handling their bicycle's momentum. It can also make it more difficult to navigate certain terrain, such as sharp turns or steep hills.