# Pendulum hung from the ceiling of a train

• brotherbobby
In summary, two observers confined to a train compartment cannot determine the motion of the train by looking at a plumb line or a marble rolling on a frictionless floor. However, they can feel different accelerations and use a pendulum to determine the horizontal component of the net force to compare the two situations.
brotherbobby
Homework Statement
A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration ##a_0## along a straight horizontal track, the string supporting the bob makes an angle ##\tan^{-1}(a_0/g)## with the normal to the ceiling. Suppose the train moves on an inclined but straight track with uniform velocity. If the angle of incline is ##\tan^{-1} (a_0/g)##, the string again makes the same angle with the normal to the ceiling.
(a) Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerated on a horizontal straight track or going uniformly up along an incline ?
(b) If yes, how ? If no, suggest a method to do so.
Relevant Equations
In an inertial frame, Newton's second law states that the net (external) force on a particle ##\Sigma F = ma##. This force is real - meaning it is due to physical interaction of the particle and its surroundings.
In a non-inertial frame that accelerates with an acceleration ##a_0## relative to an inertial frame, the net force acting on a particle of mass ##m## : ##\Sigma F = ma+ma_0##. Here the force is ##F_0 = ma_0## is not due to interactions - it is an apparent force observed in the non-inertial frame acting on the object.
(a) No, a person seated inside the train compartment will not be able to tell whether the train is accelerating on a horizontal track or moving uniformly up an inclined track by observing the plumb line.

(b) I am assuming that both observers are not allowed to look "out" of the boundaries of the apartment. If they are, the former will notice surrounding objects accelerate with an acceleration ##a_0## for which no cause (real force) can be found. However, the latter observer will not notice any such accelerating bodies.
Assuming they are both confined to the boundaries of the compartment, I wonder if they can find out the motion of their trains by letting a marble roll "down" the (frictionless) floor. For the first observer, the marble will roll to the "back" of the train with an acceleration ##a_0##, equal to the acceleration of the train relative to ground.
For the second observer, the marble will roll to the "back" of the compartment with an acceleration ##a = g \sin \theta = \frac{g}{ \operatorname{cosec} \theta} = \frac{g}{\sqrt{1+\cot^2 \theta}} = \frac{g}{\sqrt{1+g^2/a_0^2}}\; (\text{since we are given}\;\tan \theta = a_0/g) = \frac{a_0}{\sqrt{a_0^2+g^2}}g\;\boxed{\neq a_0}##.

The different accelerations of the marble "down" the compartment floor will inform the observers as to status of their motion.

Am I correct?

Delta2
Your answers seem reasonable to me. You have reasoned that we cannot determine the difference by looking at the angle of the apparent acceleration of gravity but that we can determine the difference by looking at the magnitude of that apparent acceleration.

One might measure the length of the cord on the plumb bob, set it in motion and time the oscillations.

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Part (a) is correct.
Part (b) is half correct, the second observer half. The first observer sees an effective acceleration of gravity ##g_{\text{eff.}}=\sqrt{a_0^2+g^2}## that forms an angle relative to the floor. You first need to find the component of the net force parallel to the floor, set it equal to ##m a##, then compare.

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kuruman said:
Part (a) is correct.
Part (b) is half correct, the second observer half. The first observer sees an effective acceleration of gravity ##g_{\text{eff.}}=\sqrt{a_0^2+g^2}## that forms an angle relative to the floor. You first need to find the component of the net force parallel to the floor, set it equal to ##m a##, then compare.
Since we are in non-inertial FoR, the pendulum bob will be in equilibrium when we take pseudo force
at the bob.
in parallel to the floor direction, the equation is given by: T##sin\theta##=##ma_0##
, where T is the tension in the string. The net force in the horizontal direction is thus zero. Is it not?

NTesla said:
Since we are in non-inertial FoR, the pendulum bob will be in equilibrium when we take pseudo force
at the bob.
in parallel to the floor direction, the equation is given by: T##sin\theta##=##ma_0##
, where T is the tension in the string. The net force in the horizontal direction is thus zero. Is it not?
I had a difficult time parsing what you were saying there at first.

You are correct that there is a horizontal pseudo-force arising from our decision to adopt the accelerating frame. And there is also a real horizontal force arising from the tension in the string supporting the bob. Yes, these are equal and opposite, so their sum in the horizontal direction is zero.

This fits with the fact that the plumb bob hangs stationary in the accelerating frame.

However, there is some concern about what "horizontal" means in the accelerating frame. Does it mean parallel to the surface of the earth? Or does it mean perpendicular to the apparent force of gravity?

The effective force of gravity acts diagonal to the surface of the Earth in this case. It has contributions from both the force of the Earth's gravity and the parallel-to-the-earth's-surface pseudo-force from the choice of reference frame. Those are perpendicular and add as vectors -- using the Pythagorean theorem.

@jbriggs444 , yes I already understand what you've written. I was referring to the point made by @kuruman , wherein he had said
You first need to find the component of the net force parallel to the floor, set it equal to ##ma##, then compare.
In the reference frame of the train, the net force parallel to the floor is zero. So how can it be equated to ma.?

NTesla said:
@jbriggs444 , yes I already understand what you've written. I was referring to the point made by @kuruman , wherein he had said In the reference frame of the train, the net force parallel to the floor is zero. So how can it be equated to ma.?
Ahh, that makes sense. The net that it seems that @kuruman has in mind is the net effective gravity times mass of the rolling ball. In both cases, the component parallel to the floor will be ##m g_\text{eff}\sin \theta##. It's just that in the one case, ##g_\text{eff}## is plain old g and in the other case, ##g_\text{eff}## is ##\sqrt{g^2+a^2}##.

kuruman
jbriggs444 said:
Ahh, that makes sense. The net that it seems that @kuruman has in mind is the net effective gravity times mass of the rolling ball. In both cases, the component parallel to the floor will be ##m g_\text{eff}\sin \theta##. It's just that in the one case, ##g_\text{eff}## is plain old g and in the other case, ##g_\text{eff}## is ##\sqrt{g^2+a^2}##.
Yes, that is my understanding too. The marble will experience different accelerations in the two different situations. You've correctly written the magnitude of the accelerations in the two cases.

The man in the bogie of the train will not be able to tell only by looking at the pendulum whether the train is accelerating on the horizontal track, or is going up the incline with uniform velocity. Having said that, the man will definitely FEEL (in his body) different amounts of acceleration in the two different cases, but will not be able to tell when he is on horizontal track or going up the incline.
@kuruman, is this right ?

NTesla said:
The man in the bogie of the train will not be able to tell only by looking at the pendulum whether the train is accelerating on the horizontal track, or is going up the incline with uniform velocity. Having said that, the man will definitely FEEL (in his body) different amounts of acceleration in the two different cases, but will not be able to tell when he is on horizontal track or going up the incline.
Agreed except, perhaps, for the man being able to feel the difference. At an acceleration of 1/10 of a gee (quite high for a train), the result would be an effective g about 0.5% higher than normal. One can get a similar magnitude effect by simply standing at different places on the surface of the Earth. Or by eating a big breakfast.

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NTesla
In the non-inertial frame, the passenger will not feel any different from just standing up on a inclined non-accelerating floor. He will look at the pendulum at rest and will adjust his stance instinctively so that his body is parallel to the pendulum string that defines his vertical. You can find more details on that plus more here

https://www.physicsforums.com/insights/frames-reference-linear-acceleration-view/

NTesla and jbriggs444
@kuruman,
The article in the link was quite nice.

In post#3, you've written: "You first need to find the component of the net force parallel to the floor, set it equal to ##ma##, then compare."
However, in the direction parallel to the floor, net force is zero. So could you please let us know, what did you mean to convey when you wrote that, as my understanding is that since the net force in direction parallel to the surface is zero, it can't be compared to anything of purpose.

NTesla said:
@kuruman,
The article in the link was quite nice.

In post#3, you've written: "You first need to find the component of the net force parallel to the floor, set it equal to ##ma##, then compare."
However, in the direction parallel to the floor, net force is zero. So could you please let us know, what did you mean to convey when you wrote that, as my understanding is that since the net force in direction parallel to the surface is zero, it can't be compared to anything of purpose.
If you accept that the passenger in the non-inertial frame believes that he is standing on an inclined plane, then he is experiencing an effective acceleration of gravity ##g_{\text{eff.}}=\sqrt{a_0^2+g^2}.## The angle of the incline ##\alpha## is such that $$\sin\alpha=\frac{a_0}{\sqrt{a_0^2+g^2}}~;~~\cos\alpha=\frac{g}{\sqrt{a_0^2+g^2}}.$$ The net force is zero which means that the component of his weight "in the parallel to the floor" direction must be matched by the force of static friction. Then $$f_s=mg_{\text{eff.}}\sin\alpha=m\sqrt{a_0^2+g^2}\frac{a_0}{\sqrt{a_0^2+g^2}}=ma_0.$$The needed force of static friction is the same as the fictitious force ##ma_0.## That's what I meant to convey. I apologize if I ended up confusing you.

@kuruman, in the article in the link and also in your post # 12 above, you've assumed that alice and bob both are looking at the same situation. The bogie in the article and in your post above, is not assumed to be traveling up the incline with uniform velocity, but is assumed to be moving horizontally right with uniform acceleration.
Thus your calculation is correct in that respect.

But in the question asked by the OP, the bogie of the train is actually moving up the incline in the 2nd situation. In that case, ##g_{\text{eff.}}##= g= 9.8##m/s^2##. Isn't it ?

I think the original author of the question was fishing for the following answer (if already provided previously my apologies):

The static pendulum will hang at the same angle in either case. If the pendulum is allowed to oscillate slightly about the new equilibrium the period will be shorter for the flat accelerated train because ##g_{eff}## is larger . The effective g has been calculated above.

NTesla said:
But in the question asked by the OP, the bogie of the train is actually moving up the incline in the 2nd situation. In that case, = g= 9.8. Isn't it ?
Yes, if the passenger (why do you call him "bogie"?) traveling up the incline with constant speed were to measure the acceleration of gravity, he would get 9.8 m/s2. The other passenger would measure something greater than that but not if he were on Mars and his acceleration were juuust right.

Because Lauren Bacall was in the other seat. Or maybe he was talking about the train wheels. I like the first one.

NTesla and Delta2
kuruman said:
Yes, if the passenger (why do you call him "bogie"?) traveling up the incline with constant speed were to measure the acceleration of gravity, he would get 9.8 m/s2. The other passenger would measure something greater than that but not if he were on Mars and his acceleration were juuust right.
When I'm writing bogie, I'm referring to the train's car, in which the passenger is sitting.

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kuruman and Delta2
kuruman said:
Part (a) is correct.
Part (b) is half correct, the second observer half. The first observer sees an effective acceleration of gravity ##g_{\text{eff.}}=\sqrt{a_0^2+g^2}## that forms an angle relative to the floor. You first need to find the component of the net force parallel to the floor, set it equal to ##m a##, then compare.

Sorry for coming into the discussion late, not least because the thread was mine. For some reason, PF alerts me only when a reply is sent to something that I have written - not to replies which people send to others in a thread that is (originally) my creation.

There are two observers in the problem. One is an observer in an accelerating train on horizontal ground with acceleration ##a_0##. The other is an observer in a uniformly moving train up an incline. If the first observer releases a marble towards the front of his carriage (compartment/boogie), he will observe the marble to accelerate relative to the carriage with an acceleration ##-a_0##. I don't see why you need to compute the effective gravity here, as the acceleration due to gravity ##g## will have no component parallel to the floor of the carriage.

It would be better to do as you have said. I draw a small diagram alongside. The value of ##g_{eff} = \sqrt{a_0^2+g^2}##. The component of this parallel to the floor of the carriage is ##\left(g_{\text{eff}}\right)_x = g_{\text{eff}} \cos \alpha = \sqrt{a_0^2+g^2} / \sec \alpha = \sqrt{a_0^2+g^2} / \sqrt{1+\tan^2 \alpha} = \frac{\sqrt{a_0^2+g^2}}{\sqrt{1+g^2/{a_0}^2}} = \mathbf {a_0}##, as claimed above.

brotherbobby said:
Sorry for coming into the discussion late, not least because the thread was mine. For some reason, PF alerts me only when a reply is sent to something that I have written - not to replies which people send to others in a thread that is (originally) my creation.

There are two observers in the problem. One is an observer in an accelerating train on horizontal ground with acceleration ##a_0##. The other is an observer in a uniformly moving train up an incline. If the first observer releases a marble towards the front of his carriage (compartment/boogie), he will observe the marble to accelerate relative to the carriage with an acceleration ##-a_0##. I don't see why you need to compute the effective gravity here, as the acceleration due to gravity ##g## will have no component parallel to the floor of the carriage.

View attachment 275008It would be better to do as you have said. I draw a small diagram alongside. The value of ##g_{eff} = \sqrt{a_0^2+g^2}##. The component of this parallel to the floor of the carriage is ##\left(g_{\text{eff}}\right)_x = g_{\text{eff}} \cos \alpha = \sqrt{a_0^2+g^2} / \sec \alpha = \sqrt{a_0^2+g^2} / \sqrt{1+\tan^2 \alpha} = \frac{\sqrt{a_0^2+g^2}}{\sqrt{1+g^2/{a_0}^2}} = \mathbf {a_0}##, as claimed above.
See post #12.

## 1. How does the pendulum behave when the train is moving?

The pendulum will appear to swing back and forth in a straight line, just as it would if it were stationary. This is because the train's movement does not affect the pendulum's natural frequency of oscillation.

## 2. Does the length of the pendulum affect its behavior in a moving train?

Yes, the length of the pendulum does affect its behavior. A longer pendulum will have a slower frequency of oscillation, while a shorter pendulum will have a faster frequency. This means that a longer pendulum will appear to swing more slowly in a moving train, while a shorter pendulum will appear to swing faster.

## 3. Can the pendulum swing in a circular motion in a moving train?

No, the pendulum will not swing in a circular motion in a moving train. Its motion will always be back and forth in a straight line, regardless of the train's movement. This is because the pendulum's motion is determined by gravity, not the train's movement.

## 4. Will the pendulum's behavior be affected if the train is accelerating or decelerating?

Yes, the pendulum's behavior will be affected if the train is accelerating or decelerating. This is because the train's movement will cause a change in the pendulum's speed and direction, resulting in a change in its frequency of oscillation.

## 5. How does the pendulum's behavior in a moving train differ from its behavior in a stationary train?

In a stationary train, the pendulum will appear to swing back and forth in a straight line at a constant frequency. In a moving train, the pendulum will also swing back and forth in a straight line, but its frequency may be affected by the train's movement. Additionally, the pendulum may experience slight changes in its speed and direction in a moving train due to the train's acceleration or deceleration.

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