Question on Rotational Dynamics (Just understanding something, have solutions)

[Quantumpencil]

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.

 

[tiny-tim]

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!?)

Hi Quantumpencil!

A frictionless body can only exert a force normal to its surface.

And cos + sin*tan = 1/cos

 

[Quantumpencil]

Right, ok. I feel a little ridiculous for not thinking of that... It's just like a flat surface or whatever, the force being applied is a normal force just like in all the Mechanics problems from HS. Ugh -_-

So I'm trying to construct some good follow-up questions to the set up posted here.

I thought of one involving the motion of the object if the rod is quickly yanked when l=/=l' and l=l', but I'm afraid it's too elementary. So here's my next stab.

9 - 3: The diagram shows our favorite light rigid rectangular frame with masses at the corners (that we have been talking about in lecture). But instead of letting it float in space, we drill holes through the frame at r1 and -r1 and put a frictionless rod through the holes as shown. Now we set the frame in rotation about the fixed rod with angular velocity ω with its center of mass fixed and the masses above the rod (in the diagram) going in the negative z direction, into the plane of the paper.

Follow up - Assume that immediately after the impulse, the particle at r1 making up the rod magically passes through the frame, and the frame is attached only at - r1. Afterwards, the frame:

1: will continue to rotate with angle velocity ω, because the body is rigid

2: will precess wildly around the axis of the rod, because there is now net torque on r1.

3: will precess until settles at a new angular velocity, ω'.


Answer - 3



After the body connection at r1, the motion begins to change in complicated ways, as I varies. The initial angular velocity, ω = ω (-cos θ, sin θ, 0); lies along no principle axes; The torque which was applied by our force at r1 has effectively vanished, but the torque being applied at r2 has not. Since r1 is now free to vary, but not wildy (the distance with respect to the rod can change, but the distance between r1 and r2 can't), and there is still a force exerted on r1, because there is a force exerted on -r1 and our frame is rigid; the effective torque acting on the frame at r1 must be perpendicular to it in the other direction (since the two forces canceled when it was attached at both points); meaning that the frame will be pushed against perpendicular to the rod, and since the distance from the point r1 to r2 cannot change (Due to the bodies rigidity.) The motion will settle to rotation about the rod (Think of circular motion, with center -r1, and radius (r2-r1)).

Does this make sense? Anyways to make the problem more creative or interesting?

Thanks in advance my fellow PFers.