# Bike Wheel Angular momentum problem

• Punchlinegirl
In summary, the bike wheel, with a mass of 1.36 kg and angular momentum of 10.8 kgm^2/s, is placed at the end of a rod 0.530 m in length which can pivot freely. After using the equation L=Iw with I=MR^2+MG^2 and the parallel axis theorem, the correct precessional angular velocity is determined to be 22.6 rad/s. However, there is some confusion over the moment of inertia for a hoop and the radius of the bicycle wheel.
Punchlinegirl
A bike wheel, of mass 1.36 kg is placed at the end of a rod 0.530 m in length, which can pivot freely about the other end.
The rod is of negligble mass. The wheel is turning rapidly such that it has an angular momentum of 10.8 kgm^2/s. At what angular speed does the wheel revolve horizontally about the pivot?

I used the equation
L= Iw, substituting .5MR^2 for I
L= .5MR^2 w
10.8=.5(1.36)(.530)^2 w
Solving for w gives 56.5 rad/s which isn't right

Can anyone tell me what I did wrong?

Your moment of inertia is incorrect. You need to use the parallel axis theorem.

Regards,

Ok I tried using
L=Iw, with I= MR^2 +MG^2
10.8= (1.36)(.265)^2+ (1.36)(.530)^2
solving for w gives 22.6 rad/s... which isn't right
I think I'm using the Parallel Axis Theorem wrong...

what is the moment of inertia or a hoop pivoted in the centre? use this and add it to the length between the centre to the pivot point squared, times mass. The moment of inertia for a hoop is not MR^2. Just look it up on google.

Regards,

Actually, a google search turns up I=MR^2 to be the moment of inertia of a hoop. But where does that play into this problem, seeing as you don't know the radius of the bicycle wheel? I'm having difficulty with a similar problem.

they are asking you for the precessional angular velocity which is mgr/L. where r is the length of the rod that the wheel is on. I got this straight out of a physics book, I haven't worked through it to understand how they derived it but it works. punchlinegirl, where do u go to school at, I had a very similar problem in my physics class

Thanks Felix. It looks like punchlinegirl actually goes to the same school as me (form of the question and date of posting), which is Penn State Behrend.

Thanks. Yes I do go to PSB

ha, me too. i guess a lot of people have discovered this site

## What is angular momentum and how does it apply to a bike wheel?

Angular momentum is a measure of an object's rotational motion. In the case of a bike wheel, it is the product of the wheel's moment of inertia and its angular velocity. This means that the faster a bike wheel spins, the greater its angular momentum will be.

## Why is it important to consider angular momentum when riding a bike?

Angular momentum plays a crucial role in maintaining the stability and steering of a bike. When a cyclist leans into a turn, the angular momentum of the bike wheel helps to keep the bike upright and aids in steering. Without it, the bike would be much more difficult to control.

## What factors affect the angular momentum of a bike wheel?

The angular momentum of a bike wheel is affected by its moment of inertia, which is determined by the mass distribution of the wheel and its radius. The angular velocity of the wheel is also a factor, as well as any external forces acting on the wheel, such as friction or air resistance.

## How can one change the angular momentum of a bike wheel?

The angular momentum of a bike wheel can be changed by altering its moment of inertia or its angular velocity. This can be done by changing the mass distribution of the wheel, adjusting the tire pressure, or changing the speed at which the wheel is spinning.

## What are some real-life applications of the bike wheel angular momentum problem?

The principles of angular momentum in a bike wheel have many practical applications, such as in the design of gyroscopes, spinning tops, and other rotating devices. It is also important in sports like figure skating and gymnastics, where rotational motion is a key component of many moves and routines.

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