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Homework Statement
You have a rectangular frame with side lengths l and l', and a rigid rod attached to this frame at two points, r1 and -r1. The rod passes through the center of mass, and makes an angle theta with the horizontal. Then you begin to rotate the the frame around the rod with angular velocity [tex]\omega[/tex]. What is the force exerted on the two points r1 and r2?
Homework Equations
[tex]\tau = r \times F[/tex]
[tex]\ dL/dt = \omega \times L[/tex]
[tex]\L = I\omega[/tex]
Here omega denotes the angular velocity vector,
[tex]\omega = \omega (-cos \theta, sin \theta, 0)[/tex]
I along the three principle axis = ml'^2, ml^2, ml'^2 + ml^2
The Attempt at a Solution
Using [tex]\L=I\omega[/tex] we obtain
[tex]\L=m\omega(l'^{2}cos \theta, l^{2} sin \theta, 0)[/tex]
then, using
[tex]\ dL/dt = \omega \times L[/tex], we obtain
[tex]\tau = m \omega^{2} sin \theta cos \theta (l^{2}-l'^{2})[/tex]
Which is equal to the torque. Torque is then equal to [tex]\tau = r \times F[/tex]
the position of r1 as a function of theta is
[tex]\ r_{1} = l/2 (-1, tan \theta, 0)[/tex].
Now I have the solutions, but I don't really understand the rest of the solution; and I was hoping someone could help clarify.
According to the solution guide:
Since the rod is frictionless, we want a force perpendicular to the rod (why?) of the form [tex]\ F_{1} = f (sin \theta, cos \theta, 0)[/tex] (Why!?)
Which satisfies [tex]\2r_{1} \times F_{1} = (fl/cos \theta)z = \tau[/tex](WHY!?)
z denotes the unit vector.
Then we just set it equal to the expression I understand for torque, solve for f (the magnitude of the force), and we have the solution.
The steps with a "Why" are not very clear to me, and I'd really love to change that.