How Does the No-Slipping Condition Affect Wheel Rotation Dynamics?

[schuksj]

I have a question about this problem. A rotationally symmetric axle wheel system is smoothly joined to a fixed point O.

l=distance from O to CM
m=mass of axle wheel system
R=Radius of wheel
Is and I are the principal moments of inertia relative to CM.
The axle is horizonal as the wheel rolls over the horizonal ground without slipping. The radius of the circle desrcibed by the wheel is L. Show that he no-slipping condition leads to:

d(phi)/dt=-(R/L)*S

I started with Euler's equation for the x component and set theta double dot to zero.

so d(phi)dt*sin(theta)*Is*S-(d(phi)/dt)^2*sin(theta)*cost(theta)=-mLsing(theta)

I said that I=mR^2/2 and Is-mR^2

plugging those into the equation I got:

d(phi)/dt*sin(theta)*mR^2*S-(d(phi)/dt)^2*cos(theta)*sin(theta)*mR^2/2=-mLsin(theta). I then took away the 2nd part of the equation because it is a higher order term and got that d(phi)/dt=-L/(R^2*S). This isn't exactly the right answer and I was wondering what I was doing wrong! Thanks.

 

[schuksj]

I sitll have no idea how to do this problem. If someone could give me a clue I would really appreciate it. Thanks