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Finding the period of a pendulum in motion along a curve

Homework Statement
A wagon is moving down a curve with friction. There's a pendulum of length ##l## inside it. Find its period
Homework Equations
Period of a pendulum
I was solving problems about the period of a pendulum inside an elevator. They're all the same. If the elevator accelerates upwards you have that the period is shorter and it's longer if the direction is downwards.

But I tried to solve something more difficult and I thought about a pendulum inside a wagon which is going down a curve (not an inclined plane)

I wanted to find its period, but I don't know how to start. I mean, I wanted to use polar coordinates so that the radial axis is aligned with the string and you have the tension. But what's ##\theta##? Because it changes. How should I decompose the weight?
Then, are there any pseudo-forces? Because I think about centrifugal force since it is moving along a circle. And what about Coriolis? Because the pendulum has a relative velocity

And how does the friction change the situation?

And then which acceleration should I plug in the equation of period??

If you can, please attach an image with your explanation to make it easier to understand what angles and forces I should take into account
 

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Hi.
Let ##a## be acceleration along the slope with gravity and friction corporated, people in the wagon observe acceleration applying to the pendulum be instead of -g for pendulum on the Earth,
[tex]g'_x=-a\ cos\theta, g'_y=a\ sin\theta - g[/tex]
where [tex]a= g\ sin\theta - \mu[/tex] and ##\mu## is dynamic friction coefficient. The magnitude of acceleration is
[tex]g'=\sqrt{ g_x' ^2 + g_y' ^2 }[/tex]
and direction from the vertical line be angle ##\alpha## whose cotangent is
[tex]cot\ \alpha=\frac{g'_y}{g'_x}[/tex]. And the period T' be
[tex]T'=\sqrt{\frac{g}{g'}}T[/tex] where T is period of the pendulum on the Earth.

Please check whether the case of ##\theta=\pi/2## and ##\mu=0## meets your previous cases of elevator.
 

kuruman

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The key to answering this question is figuring out the acceleration ##\vec a##. That's because in the non-inertial frame, it appears, as you know, as a fictitious force ##-m\vec a## which then gets added vectorially to the fixed force of gravity ##m\vec g##. The magnitude of the resultant divided by the mass is the effective acceleration of gravity, ##g'=|\vec g-\vec a|## that can be used instead of ##g## in the usual expression for the period. So the problem reduces to finding the acceleration ##\vec a##.

Finding the acceleration if the incine is flat is no problem. If the incline is curved, the problem becomes much more complicated. Not only you need to know the shape of the curve, but the magnitude and direction of the acceleration are not constant and depend continuously on position. Furthermore, the standard definition of period is the time required for the pendulum to perform one oscillation and return to the starting position. Even if you could find an analytical expression for the acceleration, you need to ask yourself the question, what is the meaning of "period" in this case where ##\sqrt{\frac{l}{ |\vec g-\vec a|} }## and the equilibrium position for small oscillations change continuously with time? Friction and Coriolis acceleration should be the least of your worries.
 
Let ##a## be acceleration along the slope with gravity and friction corporated, people in the wagon observe acceleration applying to the pendulum be instead of -g for pendulum on the Earth,
[tex]g'_x=-a\ cos\theta, g'_y=a\ sin\theta - g[/tex]
where [tex]a= g\ sin\theta - \mu[/tex] and ##\mu## is dynamic friction coefficient.
Excuse me, why do you say that ##a## along the slope is ##g.cos(\alpha)-\mu##? Because I get
##x) mg.sin(\alpha)-mg.cos(\alpha).\mu=m.a##
##g.sin(\alpha)-g.cos(\alpha).\mu=a##

And for:
##y) N-mg.cos(\alpha)=0##
##N=mg.cos(\alpha)##

So the acceleration in ##y## is 0 and in ##x## is ##g.sin(\alpha)-g.cos(\alpha).\mu##


And then when doing the free body diagram of the pendulum, tell me if I'm wrong please, I get:
##e_r)-T+mg.sin(\theta)=-r\dot \theta^2##
##e_{\theta})mg.cos(\alpha)=r.(\ddot \theta)##
But where should I plug the pseudo-force?
 
59
31
Hi.
I have to correct ##a## as you pointed out
[tex]a=g\ sin\theta - \mu\ cos\theta[/tex]
where ##\theta## is angle of slope. Thanks.
 

kuruman

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Hi.
I have to correct ##a## as you pointed out
[tex]a=g\ sin\theta - \mu\ cos\theta[/tex]
where ##\theta## is angle of slope. Thanks.
This equation is dimensionally incorrect.
 
59
31
Hi.
Once you know magnitude of ##g'## and its direction from vertical line by angle ##\alpha##, I do not think you have to write down equation of motion to know period. The pendulum goes to and fro around ##\alpha## line with period T' I wrote.
 
59
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So correction again
[tex]a=g(sin\theta-\mu' \ cos\theta)[/tex]
where ##\mu'## is driving friction coefficient I put dash to make it clear.
Thanks.
 

kuruman

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So correction again
[tex]a=g(sin\theta-\mu' \ cos\theta)[/tex]
where ##\mu'## is driving friction coefficient I put dash to make it clear.
Thanks.
Yes, you do need the equation of motion. More correctly the acceleration is a function of time: ##a(t)=g\{\sin[\theta(t)]-\mu \cos[\theta(t)]\}##; the angle with respect to the horizontal changes as the mass slides down the curved incline. Without knowing ##\theta(t)##, you don't really know the acceleration.

Let's say the track is circular. Then the equation of motion of the sliding mass in the absence of friction is exactly the same as that of a simple pendulum. To solve that for arbitrary angle ##\theta## you need to do an elliptic integral, hence the small angle approximation. Think about it.
 

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