Conical Pendulums: Net Force, Changing Omega & More

[PFuser1232]

In the book "Introduction to Mechanics" by K&K, in the section on conical pendulums, the net force in the ##\hat{k}## direction is set to zero, since the ##z##-coordinate of the particle doesn't change. However, later on the effect of changing ##\omega## on ##\alpha## (the angle the rod makes with the vertical) is discussed. If we change ##\alpha## according to the equation we obtained by applying Newton's second law in the ##\hat{r}## and ##\hat{k}## directions, then ##\ddot{z}## is most probably no longer zero, which contradicts the equation we used earlier. How are we allowed to change ##\omega## if it gets in the way of some of the assumptions we made earlier (such as ##\ddot{z} = 0##)? Or does the author mean comparing two different systems with different values of ##\omega##, but not actually changing the value of ##\omega## for a particular system? What am I missing here?

 

[Simon Bridge]

Is the text assuming steady state at a fixed angle - which you may change to a new angle?
Then the formula may not apply while the angle is changing, only while the angle is not changing.

 

[PFuser1232]

Is the text assuming steady state at a fixed angle - which you may change to a new angle?
Then the formula may not apply while the angle is changing, only while the angle is not changing.

 

[Simon Bridge]

Nothing in the attached files to contradict what I wrote in post #2.
The problem statement says that the motion is at a constant radius.

 

[PFuser1232]

Nothing in the attached files to contradict what I wrote in post #2.
The problem statement says that the motion is at a constant radius.

So the equation holds for constant ##\omega## and ##\alpha##.
But what exactly are we doing when we "observe the effect of changing ##\omega## on the angle ##\alpha##"? Are we comparing two pendulums with different angular frequencies? Or are we actually changing the angular frequency of the pendulum?
If it's the latter, then it contradicts the equation.
Is that what you meant?

 

[Simon Bridge]

You start with a pendulum with some frequency ##\omega## and elevation ##\alpha##; you do something to it, and you end up with a pendulum with a different frequency and elevation. You are doing a calculation on the initial and final states, not on the transition ... which you probably do not know anyway.

Whether this is the same pendulum or a different one is irrelevant to the maths.
You seem to think there is a contradiction involved - please explain.

You've done this sort of thing before - when you did conservation of momentum in collisions: you did not know the details of the collision, just some information about momentum and energy before and after the collision. Of course, during the collision all kinds of things can happen.

Note: I could not read the second pic in detail.

 

[PFuser1232]

You start with a pendulum with some frequency ##\omega## and elevation ##\alpha##; you do something to it, and you end up with a pendulum with a different frequency and elevation. You are doing a calculation on the initial and final states, not on the transition ... which you probably do not know anyway.

Whether this is the same pendulum or a different one is irrelevant to the maths.
You seem to think there is a contradiction involved - please explain.

You've done this sort of thing before - when you did conservation of momentum in collisions: you did not know the details of the collision, just some information about momentum and energy before and after the collision. Of course, during the collision all kinds of things can happen.

Note: I could not read the second pic in detail.

So the equation, namely ##\cos{\alpha} = \frac{g}{l \omega^2}## does not tell us what happens in between the change (probably something complicated), instead it relates the values of ##\omega## and ##\alpha## before the change and after the change, right?
Thanks Simon!

 

[Simon Bridge]

That's the one - well done.
To change the state you could, for example, stop the motion completely, then start it again with it's new motion.