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Consider the question from this post where I had asked about a scenario where a pendulum was attached to the roof of a car and I wished to understand how people in different frames would understand the same behavior of the pendulum.

The person outside understands the pendulum as initially staying in place but then is dragged along due to the constraint would understand it using the concept of pseudo forces.

With all that in mind, I tried to find the angle which the pendulum rope makes with the vertical from the external frame and it seems impossible to do. Below I present firstly the derivation of finding the angle from the accelerating frame and then in an inertial frame.


Diagram:

enter image description here


In the accelerating frame of reference:

The pendulum has a force of $-ma_0$ as a pseudo force and this is balanced by the horizontal component of tension. Considering this and balancing the vertical component of tension with gravity, for equilibrium:

$$ T \cos \theta = mg$$

$$ T \sin \theta = ma_0$$

This leads to:

$$ \tan \theta = \frac{a_0}{g}$$

Hence,

$$ \theta = \tan^{-1} \frac{a_0}{g}$$


In the inertial frame, the net horizontal force is equal to $ma_x$ and since the body is in equilibrium in the vertical direction (it doesn't move up and down) is equal to $ ma_y$ hence:

$$ T \sin \theta = ma_x$$

And,

$$ T \cos \theta =mg$$

Or,

$$ \tan \theta = \frac{a_x}{g}$$


The problem: Inside the moving frame of the car, it is pretty easy to understand the equilibrium condition where the pseudo force balances the tension's component and motion halts.

However, how does the observer outside realize the same equilibrium? From the pseudo force case, we can say that it is when the tension equals $ma_0$ but how would a person outside conclude this?

I considered arguing that once the bob gets fixed, the pendulum just acts like an extended part of the car and hence all accelerations must be the same but I don't think I could have concluded this from the equations without seeing the answer from the pseudo force method. Or is it impossible to do in this method and you have to consider that the body would move with the same acceleration as a given?

Tl;dr: How do I prove that $ a_x = a_0$ by viewing the body in an inertial frame alone without using any external physical assumptions?

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  • $\begingroup$ which equilibrium are you talking about for the observer in inertial frame ? $\endgroup$
    – Ankit
    Oct 24 '20 at 9:38
  • $\begingroup$ Vertical , it stops moving up and down after some point $\endgroup$
    – Buraian
    Oct 24 '20 at 9:40
  • $\begingroup$ Could you rephrase what you have said? I don't quite understand $\endgroup$
    – Buraian
    Oct 24 '20 at 9:43
  • $\begingroup$ @Buraianit could be due to my misunderstanding of your question . Are you asking how does the Bob comes in equilibrium ? $\endgroup$
    – Ankit
    Oct 24 '20 at 9:48
  • $\begingroup$ The bob comes to equilibrium because at some point the angle adjusts in a way such that the tension can balance of the pseudo force. Using this we conclude that in equilibrium , it accelerates at same rate as car in horizontal $\endgroup$
    – Buraian
    Oct 24 '20 at 9:57
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The condition for “equilibrium” in both frames is that $ \theta = const. = \theta_0$ (so $\dot \theta=0$ and $\ddot \theta =0$). In the inertial frame for $\theta = \theta_0$ to hold we must have $\ddot x=a_x=a_0$ and $\ddot y=a_y=0$. Any other values will produce a non-constant $\theta$.

To see this, consider the motion of the pivot point, $(X,Y)$ where $(x,y)$ is the position of the pendulum mass and $(X,Y)$ is the position of the pivot point and $\theta$ is the angle from vertical as drawn in your figure.

Without loss of generality we can set $X=\frac{1}{2} a_0 t^2$ and $Y=0$. Then $x=X-r \ \sin (\theta)$ and $y= Y - r \ \cos (\theta)$ so $$\ddot x = r \ \sin (\theta) \ \dot \theta^2+\ddot X - r \ \cos (\theta) \ \ddot \theta=a_0$$ $$\ddot y= r \cos(\theta) \ \dot \theta^2 + \ddot Y + r \ \sin(\theta) \ \ddot \theta =0$$

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  • $\begingroup$ Makes sense but could you explain your coordinates a bit more $\endgroup$
    – Buraian
    Oct 24 '20 at 14:34
  • $\begingroup$ @Buraian certainly, I added clarification in the answer itself $\endgroup$
    – Dale
    Oct 24 '20 at 14:47
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The person outside understands the pendulum as initially staying in place but then is dragged along due to the constraint would understand it using the concept of pseudo forces.

The person outside does not see anything as a pseudo force. The person the bob initially staying in place due to its inertia (Newton's 1st Law). The person subsequently sees the bob being dragged along due to a net horizontal force equal to the horizontal component of the tension in the string (Newton' 2nd Law).

The problem: Inside the moving frame of the car, it is pretty easy to understand the equilibrium condition where the pseudo force balances the tension's component and motion halts.

Correct.

However, how does the observer outside realize the same equilibrium?

The observer outside sees the same vertical equilibrium and the same constant angle. But the observer outside does not see the bob in horizontal equilibrium in the observers inertial frame. It sees the bob accelerating along with the car. The observer inside sees the bob in both vertical and horizontal equilibrium.

From the pseudo force case, we can say that it is when the tension equals $ma_0$ but how would a person outside conclude this?

The pseudo force case only applies to the person inside. The inside person only knows that some (invisible) force must be acting on the bob to oppose the horizontal component of the tension in the string.

The person outside sees the bob accelerating horizontally in the person's frame which requires, per Newton's 2nd law, a net horizontal force acting on the bob. The only thing contacting the bob is the string. So the person concludes that the net horizontal force is the horizontal component of the tension in the string.

I considered arguing that once the bob gets fixed, the pendulum just acts like an extended part of the car

Correct. The car is accelerating in the frame of the outside observer and the string is attached to the car, so the pendulum is in effect an extension of the car. Both observers would conclude that.

and hence all accelerations must be the same but I don't think I could have concluded this from the equations without seeing the answer from the pseudo force method.

Once the angle is a maximum, only the person outside sees the car and bob accelerating. In the reference frame neither the bob nor the car is accelerating. The pseudo force acting on the bob is an inertial force and is due solely to the fact that the frame of the car, and the observer in it, is accelerating.

The only difference in your equations for the inside and outside observer is how you labeled the acceleration. For the outside observer you labeled it $a_x$, or the acceleration of the bob and car on the x-axis of the outside observers frame of reference. The label $a_0$ as it pertains to the inside observer is the acceleration of the frame of the car.

How do I prove that $ a_x = a_0$ by viewing the body in an inertial frame alone without using any external physical assumptions?

You already have proven it. You have already stated that the pendulum is an extension of the car. This applies to both the inside and outside persons. Therefore the horizontal acceleration of the bob in the inertial frame, $a_x$ must equal the horizontal acceleration of the car in the same inertial frame, $a_0$.

You've also indirectly proven it with respect to the maximum pendulum angle. That maximum angle $θ$, and any changes in the angle that occurred initially, should appear the same to both the inside and outside observers. The only way that can be true is if $a_x$ = $a_0$

Hope this helps

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  • $\begingroup$ The reason I didn't want to use the extended body argument to state the acceleration was equal was because in the accelerating frame, you prove it using only mathematics. Saying that it acts like extended body is like adding one more physical fact into it. Other than that, I did find it interesting equilibrium in one frame may not equilibrium in all frames. $\endgroup$
    – Buraian
    Oct 26 '20 at 17:05
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    $\begingroup$ Forget the extended body argument. If you agree the status of the angle looks the same to both observers then $a_x$ must equal $a_0$, as you’ve already proved $\endgroup$
    – Bob D
    Oct 26 '20 at 17:49

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