How Do Two Rotating Rods Interact?

[eitan77]

Homework Statement

Two long rods rotate with the speed of omega and 2*omega around the parallel axes that pass through the points A and B respectively (and orthogonal to the plane of the drawing). At the moment in time t=0, both rods are turned to the right. What is the trajectory of the intersection point of the rods as a function of time?

I was able to find a mathematical expression for the trajectory of the intersection point in time except for the times when the two rods are horizontal, I understand that in this case there are infinite intersection points but I cannot see it mathematically.

Relevant Equations

R_ca = L + R _cb

 

[scottdave]

Check if there are other times when the rods are parallel? If the time is allowed to continue long enough, they could be parallel at any angle. My mistake.

 

[eitan77]

What about other times when the rods are parallel? If the time is allowed to continue long enough, they could be parallel at any angle.

Thanks for the comment. So where do you think my mistake is? Did I miss something during the solution?

 

[TSny]

So where do you think my mistake is?

I don't see anything wrong with your solution.

Did I miss something during the solution?

What do you get for the final expression for $\vec{R}_{cb}$? Interpret.

 

[scottdave]

How did you get the left hand side of step 2 in the e₁ calculations?

You change it from r_ca to r_cb and use the double angle formula.

I circled what I'm talking about.

 

[haruspex]

How did you get the left hand side of step 2 in the e1 calculations?

It is a substitution using the last equation under the preceding "e2:" analysis.

What about other times when the rods are parallel? If the time is allowed to continue long enough, they could be parallel at any angle.

Not so.

 

[haruspex]

Thanks for the comment. So where do you think my mistake is? Did I miss something during the solution?

For what values of $\omega t$ are the rods parallel/antiparallel?

 

[Steve4Physics]

@eitan77, not an answer to your question but possibly worth noting…

The trajectory of the intersection point (C) is not continuous over time. There are time-intervals when there is no intersection, e.g. consider the positions of the rods when when $\omega t = 135^o$ so $2\omega t = 270^o$.

When intersection occurs, using simple geometry you can show triangle ABC is isosceles (AB=BC). This provides a simple way to find C's trajectory.

 

[haruspex]

When intersection occurs, using simple geometry you can show triangle ABC is isosceles (AB=BC). This provides a simple way to find C's trajectory.

That doesn’t work for $\omega t>\pi$.

 

[eitan77]

@eitan77, not an answer to your question but possibly worth noting…

The trajectory of the intersection point (C) is not continuous over time. There are time-intervals when there is no intersection, e.g. consider the positions of the rods when when $\omega t = 135^o$ so $2\omega t = 270^o$.

When intersection occurs, using simple geometry you can show triangle ABC is isosceles (AB=BC). This provides a simple way to find C's trajectory.

Now that you mention that I noticed that my answer restricts C to having only positive e1 components

 

[eitan77]

Now that you mention that I noticed that my answer restricts C to having only positive e1 components
View attachment 335718

@eitan77, not an answer to your question but possibly worth noting…

The trajectory of the intersection point (C) is not continuous over time. There are time-intervals when there is no intersection, e.g. consider the positions of the rods when when $\omega t = 135^o$ so $2\omega t = 270^o$.

When intersection occurs, using simple geometry you can show triangle ABC is isosceles (AB=BC). This provides a simple way to find C's trajectory.

So where do you think my mistake is?

 

[haruspex]

Now that you mention that I noticed that my answer restricts C to having only positive e1 components
View attachment 335718

So? Seems ok to me.

 

[TSny]

So where do you think my mistake is?

Does this equation restrict the allowed values of $t$?

 

[haruspex]

So where do you think my mistake is?

No mistake. Try answering the questions in posts #4, #7.

 

[eitan77]

View attachment 335719

Does this equation restrict the allowed values of $t$?

I didn't notice it until now, since r_cb represents length it cannot be negative so this limits it to values of t to be 0<t<pi/2.

 

[eitan77]

I didn't notice it until now, since r_cb represents length it cannot be negative so this limits it to values of t to be 0<t<pi/2.

0<omega*t<pi/2

 

[TSny]

I didn't notice it until now, since r_cb represents length it cannot be negative so this limits it to values of t to be 0<t<pi/2.

ok, that's the right idea. Your inequality should include $\omega$ in it. Also, there will be other allowed ranges.

 

[eitan77]

I don't see anything wrong with your solution.What do you get for the final expression for $\vec{R}_{cb}$? Interpret.

From my solution it is obtained if there is a point of intersection

 

[haruspex]

From my solution it is obtained if there is a point of intersection
View attachment 335720

Which looks like... ?

 

[kuruman]

If the time is allowed to continue long enough, they could be parallel at any angle.

Is that obvious? You can take the cross product between the vectors and demand that it be zero. That will give you the times at which the vectors are parallel or antiparallel. In this case "long enough" is the period of the slower moving vector $T=\frac{2\pi}{\omega}$ at the end of which the vectors are back where they started.

 

[eitan77]

From my solution it is obtained if there is a point of intersection
View attachment 335720

Which looks like... ?

Sorry I don't think I understand what you mean.

 

[eitan77]

For what values of $\omega t$ are the rods parallel/antiparallel?

they are parallel for $\pi +2\pi k =\omega t$ for k =0,1,2...

 

[eitan77]

So it is correct to say that there is an intersection point that holds :

when $2\pi k <\omega t <= \pi /2 +2\pi k$ for k = 0,1,2... ?

 

[haruspex]

they are parallel for $\pi +2\pi k =\omega t$ for k =0,1,2...

That's for parallel in the strict sense, but allowing antiparallel it is $\omega t=n\pi$.
In your algebra you wrote "While $\sin(\omega t)\neq 0$", thereby sidelining these cases.

Sorry I don't think I understand what you mean.

Can you recognise View attachment 335720
as a simple geometric shape?

 

[eitan77]

That's for parallel in the strict sense, but allowing antiparallel it is $\omega t=n\pi$.
In your algebra you wrote "While $\sin(\omega t)\neq 0$", thereby sidelining these cases.

Can you recognise View attachment 335720
as a simple geometric shape?

It's a circle

 

[haruspex]

It's a circle

Yes. But the algebra that is based on treats the rods as doubly infinite. What portion of the circle is ruled out by taking them as semi infinite, as illustrated?

 

[eitan77]

Yes. But the algebra that is based on treats the rods as doubly infinite. What portion of the circle is ruled out by taking them as semi infinite, as illustrated?

My intuition says that the bottom half of the circle should be ruled out, But I'm not sure how I'm supposed to see this mathematically. (if I am right)

And thank you very much for your patience.

 

[haruspex]

My intuition says that the bottom half of the circle should be ruled out, But I'm not sure how I'm supposed to see this mathematically. (if I am right)

And thank you very much for your patience.

You have shown that the intersection lies in the half plane to the right of A. For what range(s) of $\omega t$ is that consistent with the position of the rod hinged at A?

 

[scottdave]

It is a substitution using the last equation under the preceding "e2:" analysis.

Not so.

Ok thanks.... My mistake.

 

[Steve4Physics]

That doesn’t work for $\omega t>\pi$.

I see it like this. A full cycle of the system occurs when rod A performs 1 full rotation and rod B performs 2 full rotations – we’re then back to the originaL start position.

Take point A as the origin so AB lies on the x-axis.

$\theta_A$ is rod A’s direction ($\theta_A = \omega t$) and similarly for rod B, $\theta_B = 2\omega t = 2\theta_A$.

There rods don' intersect for $\pi/2 <\theta_A<3\pi/2$.

But, for example, consider $\theta_A =7\pi/4$. Rod A bisects the 4th quadrant. Rod B points at $7\pi/2$ which is the -y direction. The intersection (point C) is ($l, -l$) so that ABC is a right-angled isosceles triangle.

Is there a mistake in my logic?

 

[haruspex]

I see it like this. A full cycle of the system occurs when rod A performs 1 full rotation and rod B performs 2 full rotations – we’re then back to the originaL start position.

Take point A as the origin so AB lies on the x-axis.

$\theta_A$ is rod A’s direction ($\theta_A = \omega t$) and similarly for rod B, $\theta_B = 2\omega t = 2\theta_A$.

There rods don' intersect for $\pi/2 <\theta_A<3\pi/2$.

But, for example, consider $\theta_A =7\pi/4$. Rod A bisects the 4th quadrant. Rod B points at $7\pi/2$ which is the -y direction. The intersection (point C) is ($l, -l$) so that ABC is a right-angled isosceles triangle.

Is there a mistake in my logic?

No, you are right. It wasn’t as obvious as I thought you were implying, but I didn't follow it through far enough.

 

[eitan77]

Thank you very much to all of you, you helped me a lot!