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

[Legit]

Homework Statement

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.)

Homework Equations

ΔK + ΔU = -f_kd + ƩW
F = ma
Ʃτ = Iα
MI_wheel = MR^2 + MR^2 = 2MR^2
MI_rod1 = ML^2/3
MI_rod2 = ML^2/12 + m(L/2)^2 = ML^2/3 (parallel-axis theorem?)

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 = (I_wheel + I_rod1 + I_rod2)α, where I_rod1 is the inertia of the rod with the axle at its end (MI = ML²/3) and I_rod2 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 F_axle = ma + mgcosθ. As you can see I am completely lost . Thanks for any assistance.

 

[anathema]

Hello,

Thank you for your question. It seems like you are on the right track with using the torque equation and the moment of inertia of the rods. However, I think there are a few things that need to be clarified before proceeding with the solution.

First, the equation you wrote for the moment of inertia of the wheel and the two rods is not correct. The moment of inertia of a thin ring is given by I = MR^2, so the moment of inertia of the wheel would be MIwheel = 2MR^2, not MR^2 + MR^2 = 2MR^2. Similarly, the moment of inertia of the rods would be MIrod1 = ML^2/12 (using the parallel-axis theorem) and MIrod2 = ML^2/12 + M(L/2)^2 = ML^2/8 (again using the parallel-axis theorem). So the total moment of inertia in the torque equation would be Ʃτ = (2MR^2 + 2ML^2/12 + ML^2/8)α.

Second, the free-body diagram you have drawn is not quite correct. The force exerted by the axle should be perpendicular to the axle, not at an angle. This means that the only forces acting on the wheel are the normal force from the axle and the weight of the wheel (assuming no other external forces). Therefore, the equation should be Faxle = N - mgcosθ = ma.

With these corrections, you can proceed with solving the problem. You are correct in using the torque equation to find α, and then using Rα = a to find the acceleration. Once you have the acceleration, you can plug it into the force equation to find the force exerted by the axle. Remember to include the negative sign in the torque equation since the torque is acting in the opposite direction of the angular acceleration.

I hope this helps. Good luck with your solution!