Bike Wheel Angular momentum problem

[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?

 

[Nenad]

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

Regards,

Nenad

 

[Punchlinegirl]

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...

 

[Nenad]

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,

Nenad

 

[blz5k]

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.

 

[Felix83]

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

 

[blz5k]

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.

 

[Punchlinegirl]

Thanks. Yes I do go to PSB

 

[Felix83]

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