# Wheel pivoted about a frictionless axle - moment of inertia, force of axle

1. Apr 22, 2012

### Legit

1. The problem statement, all variables and given/known data

A wheel has 4 spokes, see sketch ( http://i44.tinypic.com/213hp52.png ), and is pivoted about the indicated frictionless axle which is perpendicular to the page. The wheel is lifted such that θ = 90° (above the negative y axis) and released from rest. What is the force (magnitude and direction) the axle exerts on the wheel when θ = 30° (above the negative y axis)? (Treat the wheel as a uniform thin ring of radius R and mass M and the 4 spokes as uniform thin rods of mass M.)

2. Relevant equations

ΔK + ΔU = -fkd + ƩW
F = ma
Ʃτ = Iα
MIwheel = MR^2 + MR^2 = 2MR^2
MIrod1 = ML^2/3
MIrod2 = ML^2/12 + m(L/2)^2 = ML^2/3 (parallel-axis theorem?)

3. The attempt at a solution

What I think I'm supposed to do is use the torque equation to find α, then use Rα = a to get acceleration, and plug that into F = ma? But I am unsure as to how to implement the moment of inertia of the rods. I was thinking that instead of looking at the rods as 4 small rods, I could look at them as 2 larger rods of length 2R. So could the equation look something like Ʃτ = (Iwheel + Irod1 + Irod2)α, where Irod1 is the inertia of the rod with the axle at its end (MI = ML2/3) and Irod2 is the inertia of the rod rotating around the axle but not touching it (also MI = ML2/3)? I'm not sure if my equation there is even legal, and I am unsure if i have the moment of inertia of the second rod correct. I also have a free-body diagram where Faxle = ma + mgcosθ. As you can see I am completely lost . Thanks for any assistance.

Last edited: Apr 22, 2012