Motion of mass constrained to rail

[Dale]

I have been assuming the track is a monorail and the rod attached fore and aft.

That is a different scenario than the one in the OP, which explicitly assumes a thin rod meaning no difference between fore and aft.

The frictionless track, the thin rod, and the rigid rod are simplifying assumptions. Are you familiar with the term "simplifying assumption"?

 

[sophiecentaur]

That is a different scenario than the one in the OP, which explicitly assumes a thin rod meaning no difference between fore and aft.

The frictionless track, the thin rod, and the rigid rod are simplifying assumptions. Are you familiar with the term "simplifying assumption"?

That's just being cheeky!

Yes, I get it now.

 

[Fantasist]

No, but the force which provides the torque can. I.e. a single force can provide both torque and acceleration.

Not if the force attaches at a right angle to the rod, and especially not if you have opposite force vectors on both sides of the rod.

So what? What has any of that got to do with the expression for the KE? Do you believe that the expression for KE is automatically $\frac{1}{2}m \dot q^2$ for any generalized coordinate, q?

No, not for any generalized coordinate. It depends obviously on the scenario i.e. on the specific constraints. And in this case, if I am not mistaken, the constraint is such that it does (as I tried to explain through the graphical illustration above).

But if you think that your derivation wraps up all possible scenarios, why don't you answer on this basis my above question what happens if the moving rod collides with a stationary point mass (same mass as the rod) in the curved section of the rail? What are the velocities of the rod and the point mass after the collision? An answer to this question should help to clarify the issue.

 

[A.T.]

Consider a rod, with the entire mass concentrated in two equal point masses, at the same distance from the rod center. Make that distance equal to the curve radius, so that the inner mass stops when the rod goes around the curve, and transfers all KE to the outer mass. The KE of the outer mass will double, so its speed will be √2*v₀. The speed of the rod center in the cruve is half of that v₀/√2, which is less than v₀.

You can generalize this, because due to the squared velocity, you always need more KE to accelerate by Δv, than you gain by slowing down the same mass from the same v by the same Δv. You can consider the uniform rod as a collection of such symmetrical point mass pairs.

 

[Dale]

Not if the force attaches at a right angle to the rod, and especially not if you have opposite force vectors on both sides of the rod.

Who said the force attaches at a right angle to the rod and is on both sides of the rod? Those weren't any of your constraints. Your constraints were that the COM followed the track and that the thin rod was perpendicular to the track at each point.

Under the constraints in the OP you get the above results. The forces from the track are whatever are needed to achieve those constraints on the motion. You cannot constrain both the motion and also the forces. Once you fix one then the other is determined.

No, not for any generalized coordinate. It depends obviously on the scenario i.e. on the specific constraints. And in this case, if I am not mistaken, the constraint is such that it does (as I tried to explain through the graphical illustration above).

Yes, you are mistaken. Your graphical illustration is irrelevant. For any rigid object you can always easily find generalized coordinates where the object is stationary in those coordinates.

So the fact that the plot of the generalized coordinates matches your illustration is completely uninformative. Any rigid body in any scenario undergoing any possible motion can match your illustration for some suitable choice of generalized coordinates.

Therefore, you have to derive the expression for the KE in the generalized coordinates some other way. For this problem, I showed how in 5 and again in 33.

But if you think that your derivation wraps up all possible scenarios

I never made that claim. My derivation wraps up THIS scenario without doubt, but certainly not all possible scenarios.

why don't you answer on this basis my above question what happens if the moving rod collides with a stationary point mass (same mass as the rod) in the curved section of the rail? What are the velocities of the rod and the point mass after the collision? An answer to this question should help to clarify the issue.

I fail to see the relevance of this new scenario to the scenario of the OP. You keep on bringing in irrelevant side issues. I don't know why.

If you want to solve this new system (I don't) then all you need to do is introduce another generalized coordinate representing the position of the point mass along the track. I assume that you still want KE conserved (rigid, frictionless, etc.). The KE of the system is the KE for the rod (given above) plus the KE for the point mass (easy to derive). The only hard part is handling the inequality constraint that the point mass must always be in front of the rod, but there are standard methods for doing that in Lagrangian mechanics.

 

[Fantasist]

Who said the force attaches at a right angle to the rod and is on both sides of the rod? Those weren't any of your constraints. Your constraints were that the COM followed the track and that the thin rod was perpendicular to the track at each point.

Under the constraints in the OP you get the above results. The forces from the track are whatever are needed to achieve those constraints on the motion. You cannot constrain both the motion and also the forces. Once you fix one then the other is determined.

Your suggestion was that the force of constraint changes the dynamics of the mass moving on the rail. This is impossible according to d'Alembert's principle. The force of constraint can't do any work. It just constrains. It would only be noticeable in the form of internal stress forces in the material.

Your graphical illustration is irrelevant. For any rigid object you can always easily find generalized coordinates where the object is stationary in those coordinates.

So the fact that the plot of the generalized coordinates matches your illustration is completely uninformative. Any rigid body in any scenario undergoing any possible motion can match your illustration for some suitable choice of generalized coordinates.

You make it sound as if I had plucked the general coordinate out of thin air. It strictly relates to the physical constraint here. The rail and rigidly connected rod nail down the path of any mass element of the rod to be a curve parallel to the rail. So taking the general coordinate as the path along the rail is not only the natural thing to do here but even what you should do unless you want to get into difficulties.

Therefore, you have to derive the expression for the KE in the generalized coordinates some other way. For this problem, I showed how in 5 and again in 33.

You have assumed a rotational degree of freedom that simply isn't there. 'Degree of freedom' means just that: the object is free to move in this dimension. But in this case it isn't because of the constraint. You can't rotate the rod without translating it. The system has only 1 degree of freedom. The rod may get rotated but it is not rotating. If you assume it is rotating you count something twice here.

I fail to see the relevance of this new scenario to the scenario of the OP. You keep on bringing in irrelevant side issues. I don't know why.

It is not irrelevant. An elastic collision can conveniently separate the linear momentum from the rotational momentum. If you assume the rod with linear momentum p hits elastically a point mass with the same mass at rest on the rail, the conservation laws for the linear momentum and kinetic energy require that after the collision

1) the point mass has linear momentum p
2) the rod has linear momentum 0

But because the orientation of the rod is constrained to its linear motion, 2) implies also

3) the rod has intrinsic angular momentum 0

So since the total energy is p^2/2m after the collision, energy conservation requires that it is also p^2/2m before the collision, i.e. there was no intrinsic rotational energy of the rod present in the first place.

 

[Dale]

Your suggestion was that the force of constraint changes the dynamics of the mass moving on the rail. This is impossible according to d'Alembert's principle. The force of constraint can't do any work. It just constrains.

I agree fully with d'Alembert's principle. The KE is conserved, therefore the constraint force does no work.

Your suggestion is the one which violates d'Alembert's principle since you would have the constraint do work to increase the KE around the bend so that the COM velocity is unchanged.

You make it sound as if I had plucked the general coordinate out of thin air. It strictly relates to the physical constraint here. The rail and rigidly connected rod nail down the path of any mass element of the rod to be a curve parallel to the rail. So taking the general coordinate as the path along the rail is not only the natural thing to do here but even what you should do unless you want to get into difficulties.

I agree that your s is a very natural coordinate, but that is not relevant to the argument. The naturalness of the coordinate is not in discussion, only the expression for the KE in terms of the coordinate.

You have assumed a rotational degree of freedom that simply isn't there. 'Degree of freedom' means just that: the object is free to move in this dimension. But in this case it isn't because of the constraint. You can't rotate the rod without translating it. The system has only 1 degree of freedom. The rod may get rotated but it is not rotating. If you assume it is rotating you count something twice here.

Again, completely irrelevant. It doesn't matter if it is free to rotate or constrained to rotate. Either way it undergoes rotational motion wrt any inertial coordinate system and therefore has rotational KE. Remember, s is a generalized coordinate not an inertial coordinate. So you cannot simply assume that pure translation wrt s implies an absence of rotation wrt inertial coordinates.

My derivation in 33 starts with the known expression for KE in an inertial coordinate system and then calculates the expression in the generalized coordinates. If you disagree with my conclusion then please point out the exact step where I made my mistake.

It is not irrelevant. An elastic collision can conveniently separate the linear momentum from the rotational momentum.

Momentum isn't conserved in this problem, just energy. Again, I fail to see the relevance, but if you wish to work the problem, please be my guest. I gave instructions for doing so previously.

 

[Fantasist]

f you disagree with my conclusion then please point out the exact step where I made my mistake.

The mistake is in your last step

KE = \frac{1}{2} m R^2 \omega^2 + \frac{1}{24} L^2 m\omega^2
KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2

where effectively you dropped the constraint.
Note that in the first equation the same angular frequency \omega appears in both terms, which means that if one term gets larger (due to an increase of \omega), the other would have to get larger as well, so it would violate energy conservation. Your final equation just hides this fact and pretends that you can compensate the increase/decrease of one term by correspondingly adjusting the other. This would be justified if you had actually two degrees of freedom (e.g. if the rod would only accidentally have a synchronous rotation) but not if the orientation is rigidly fixed relatively to the path.

 

[mfb]

where effectively you dropped the constraint.

This does not change the validity of the equations.

2x=2x remains true even if I define y=x and replace it to get 2x=x+y as modified equation.

Note that in the first equation the same angular frequency ω appears in both terms, which means that if one term gets larger (due to an increase of ω), the other would have to get larger as well, so it would violate energy conservation.

This just shows ω is constant for constant curvature. A reduced curvature radius corresponds to an increased angular velocity - but not in a linear relation due to the second term, which is exactly the effect we are looking at.

 

[Dale]

The mistake is in your last step
where effectively you dropped the constraint.

Where is the mistake?
$v=R\omega$
and
$I=\frac{1}{12}L^2 m$

So the last equation follows from the previous by simple algebraic substitution. This is very basic and valid algebra. The math doesn't care about whether two variables are constrained by some separate equation. The substitution remains valid regardless.

Note that in the first equation the same angular frequency \omega appears in both terms, which means that if one term gets larger (due to an increase of \omega), the other would have to get larger as well, so it would violate energy conservation.

Obviously, if one gets larger then the other would indeed have to get larger as well. I don't know why you think this is a mistake.

This does not in any way imply a violation of energy conservation. The only way that one could get larger is by some external force doing work on the system. In such a circumstance the external force would have to increase both terms, as shown by the equation.

The only way that you would get a violation of conservation of energy here is if ω were to change without any external work. If that were to happen then the conservation of energy would still be violated even if we used only the translational KE term. So the expression does not indicate anything amiss, and I don't know why you think it does.

 

[Fantasist]

Where is the mistake?
$v=R\omega$
and
$I=\frac{1}{12}L^2 m$

So the last equation follows from the previous by simple algebraic substitution. This is very basic and valid algebra. The math doesn't care about whether two variables are constrained by some separate equation. The substitution remains valid regardless.

The substitution is valid, but not your conclusions from it. Your argument was that if in

KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2

the second (rotational) term on the right hand side gets larger due to an increase in \omega, the translational velocity v has to get smaller to keep the total energy KE constant. But from the original equation

KE = \frac{1}{2} m R^2 \omega^2 + \frac{1}{24} L^2 m\omega^2

you can see that this is not possible as the same angular frequency \omega appears in both terms (and R does not change at the moment when the rod hits the curved section).

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

 

[Dale]

The substitution is valid

If the substitution is valid then the derivation is correct and that is the correct equation for the KE during the curve. You cannot have it both ways.

Either the math is valid and the derivation is correct or there is a mistake in the math. You cannot admit that the math is valid and yet logically claim that the result is wrong.

and R does not change at the moment when the rod hits the curved section

Yes it does. It is infinite before the curve and finite after.

 

[A.T.]

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

See my post #54. It uses only linear KE, and arrives at the same conclusion (center must slow down in the curve).

 

[sophiecentaur]

Just lost a half made-up post. Try again.

Can one of you chaps help me with your disagreement, please?

Is it, basically, that momentum conservation and energy conservation are not both followed in your model? Can you not just equate KE before and after and see how the speed along the rail is changed due to the change of linear KE into some rotational KE? The point of transition onto the curve should not be a fundamental problem because any change of curvature would introduce the same problem - however great the radius.

Take time off to give me a heads up on the story so far. (I'll hand round orange segments to suck)

 

[AlephZero]

The substitution is valid, but not your conclusions from it...

... you can see that this is not possible as the same angular frequency \omega appears in both terms (and R does not change at the moment when the rod hits the curved section).

From this you can conclude that it was a mistake in the first place to assume a separate rotational energy term when there is no rotational degree of freedom (because of the constraint).

No, your mistake is trying to describe the KE using variables that don't make sense. On the straight part of the track, $\omega = 0$ and $R$ is "infinite". You are trying to multiply "0 by infinity" to get the finite value of the KE.

The problem goes away if you write the KE in terms of $R$ and $v$, not $R$ and $\omega$.

$$\begin{aligned}KE &= \frac{1}{2} m v^2 + \frac{1}{24} mL^2 \frac{v^2}{R^2}\\
&= \frac{1}{2}mv^2\left(1 + \frac{L^2}{12R^2}\right)\end{aligned}$$

Now it is clear that on the straight track, the "infinite" value $R$ gives the correct KE of $\frac{1}{2}mv^2$, and that if the KE is constant (because there are no forces acting that do work), then if $R$ decreases, so does $v$.

 

[Dale]

Can one of you chaps help me with your disagreement, please?

Fantastist believes that the correct formula for KE is KE = \frac{1}{2} m v^2 even though multiple references and a derivation have been provided which confirm that the correct formula is KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2

 

[sophiecentaur]

Fantastist believes that the correct formula for KE is KE = \frac{1}{2} m v^2 even though multiple references and a derivation have been provided which confirm that the correct formula is KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2

What is there to disagree with about that? Except, perhaps that ω can be stated as v/R, which could be useful in this case because the v term would be in both parts. I can see that there could be a problem at the transition but isn't that a common situation in Physics? That's the beauty of the Impulse.

 

[A.T.]

This seems to be the root of the confusion:

The force of constraint can't do any work.

The force of constraint does no net work, so the total KE (linear + angular) doesn't change.

But the force of constraint can do negative linear work, and the same amount of positive angular work, so linear KE is converted to angular KE.

 

[Dale]

I think his confusion is deeper than that. I think that his confusion is that he believes that the mere presence of a constraint changes the KE.

In other words, he believes given two identical rigid bodies following identical motions wrt an inertial frame, if one is following the motion due to constraints and the other is following the identical motion freely or due to a potential then the KE of the two objects will be different.

It is clearly wrong, completely unjustified, contradicted by references, and disproven mathematically. I don't know what else can be done for Fantasist.

 

[Fantasist]

and R does not change at the moment when the rod hits the curved section

Yes it does. It is infinite before the curve and finite after.

$R$ should be infinite before the curve? $R$ is the radius vector from the origin of the coordinate system to the center of mass of the rod. And this can not change from something finite to something infinite (or vice versa) within an infinitesimal path length $ds$.
Also, you would then have an infinite angular momentum $l=m R\times v$ of the center of mass before the curve, which clearly can not be correct (whether you assume it changes or not).

 

[Fantasist]

This seems to be the root of the confusion:


The force of constraint does no net work, so the total KE (linear + angular) doesn't change.

But the force of constraint can do negative linear work, and the same amount of positive angular work, so linear KE is converted to angular KE.

Still, you have the problem to explain where the linear force parallel to the rail should be coming from. All forces of constraint here act only perpendicular to the rail.

 

[Dale]

$R$ should be infinite before the curve? $R$ is the radius vector from the origin of the coordinate system to the center of mass of the rod.

No, R is the radius of curvature of the path, sorry I wasn't clear about that. A straight path has an infinite radius of curvature, a curved path has a finite radius of curvature. Therefore R goes from infinite before the curve to finite on the curve.

 

[rcgldr]

Still, you have the problem to explain where the linear force parallel to the rail should be coming from. All forces of constraint here act only perpendicular to the rail.

As mentioned in a previous post, using a rod as the example object, the forces are perpendicular to the rail, but not perpendicular to the velocity (the path) of the center of mass of the rod.

 

[A.T.]

Still, you have the problem to explain where the linear force parallel to the rail should be coming from. All forces of constraint here act only perpendicular to the rail.

But they cannot act exactly at the mass center, because then the rod couldn't start rotating. In the curved part, off center forces acting perpendicular to the rail have a component parallel to the rail at the mass center.

 

[sophiecentaur]

But they cannot act exactly at the mass center, because then the rod couldn't start rotating. In the curved part, off center forces acting perpendicular to the rail have a component parallel to the rail at the mass center.

Why are you getting so wound up about 'practical details' here? The OP os about an ideal situation that assumes such practicalities have been taken care of. It's all taken care of by the word "constrained". Of course, you couldn't build the apparatus in the OP - it doesn't existt. But why pick out this one example of where strict theory demands impossible conditions but where the theory still predicts the result of a practical implementation. Physics is full of this sort of approach. Lens calculations, interference, impacts, all electrical circuit calculations etc. etc.

 

[A.T.]

Why are you getting so wound up about 'practical details' here?

I'm not getting wound up in anything. Just responding directly to a question.

 

[sophiecentaur]

Why bother to respond then? It just adds fuel to the nonsense element of a thread that was, for a few posts, quite interesting and fundamental.

 

[Fantasist]

No, R is the radius of curvature of the path, sorry I wasn't clear about that. A straight path has an infinite radius of curvature, a curved path has a finite radius of curvature. Therefore R goes from infinite before the curve to finite on the curve.

$R$ is only the radius of curvature if the path is in fact a circle around the origin.
The general expression for the linear kinetic energy in plane polar coordinates $R,\Theta$ is
$$\frac{m}{2}v^2 = \frac{m}{2}\dot{R}^2 + \frac{l^2}{2mR^2}$$
with $\vec{l}=m \vec{R}\times \vec{v} = mR^2\dot{\Theta}$ the angular momentum.
This expression is identical whether you have a mass wth velocity $v$ in a circular orbit with radius $R$, or whether you have a mass with velocity $v$ on a linear trajectory that happens to be at the point where its trajectory touches the circle with radius $R$ (in both cases will the radial velocity be $\dot{R}=0$ and $R$ and $l$ be identical, hence also the kinetic energy).

 

[Fantasist]

As mentioned in a previous post, using a rod as the example object, the forces are perpendicular to the rail, but not perpendicular to the velocity (the path) of the center of mass of the rod.

But the velocity of the center of mass has only a component parallel to the rail.

 

[Fantasist]

But they cannot act exactly at the mass center, because then the rod couldn't start rotating. In the curved part, off center forces acting perpendicular to the rail have a component parallel to the rail at the mass center.

The rotation is caused by a torque around the center of mass. And a pure torque doesn't affect the motion of the center of mass (see diagram below).

 

[Fantasist]

Why are you getting so wound up about 'practical details' here? The OP os about an ideal situation that assumes such practicalities have been taken care of. It's all taken care of by the word "constrained". Of course, you couldn't build the apparatus in the OP - it doesn't existt. But why pick out this one example of where strict theory demands impossible conditions but where the theory still predicts the result of a practical implementation. Physics is full of this sort of approach. Lens calculations, interference, impacts, all electrical circuit calculations etc. etc.

I don't know why you are saying the apparatus couldn't be built. It shouldn't be too difficult to demonstrate this. You might even be able to use your model railway for it. The motor would take care of the friction. Just put a straight and circular section of track together, mount some rod (with a suitable mass) to the train and see whether the speed changes when it hits the point where the track curvature changes (ufortunately, I am personally not in a position to do it myself).

This isn't a practical problem but just a theoretical issue (namely how to correctly handle the fact that the rotational motion is rigidly constrained to the linear motion).

 

[A.T.]

And a pure torque doesn't affect the motion of the center of mass (see diagram below).

Which means that in this case the constraint forces cannot exert a pure torque.

The above picture would violate conservation of KE, as I explained in post #54:

You need more KE to accelerate by Δv, than you gain by slowing down the same mass from the same v by the same Δv. You can consider the uniform rod as a collection of such symmetrical point mass pairs.

 

[Dale]

$R$ is only the radius of curvature if the path is in fact a circle around the origin.

No, the radius of curvature is the inverse of the curvature:

http://en.wikipedia.org/wiki/Radius_of_curvature_(mathematics)
http://en.wikipedia.org/wiki/Curvature

It is defined for arbitrary paths in arbitrary dimensions and is in no way restricted to a circle about the origin.

Again, R in the equations I posted above is not a coordinate (neither a generalized coordinate nor a polar coordinate). It is the radius of curvature of the path, a parameter of the path which changes along the path from infinite to finite.

 

[Dale]

And a pure torque doesn't affect the motion of the center of mass (see diagram below).

So what? Then it isn't a pure torque. Your constraints constrained the motion, they did not constrain the forces or torques. The forces and torques are whatever are required to produce the specified motion.

Again, you cannot both constrain the motion and constrain the forces. Once you specify one the other is determined. In this case, you specified the constraints on the motion, so the forces are whatever is needed to satisfy your specified constraints. You are attempting to add additional constraints (on the forces and torques) which are incompatible with your original constraints (on the translation and rotation).

 

[A.T.]

It shouldn't be too difficult to demonstrate this... unfortunately, I am personally not in a position to do it myself.

You should at least be able to draw a free body diagram of the simplest possible practical constraint mechanism required for this. When you analyze it at the transition, you will get a force that changes the linear speed of the rod center.

 

[AlephZero]

I don't know why you are saying the apparatus couldn't be built. It shouldn't be too difficult to demonstrate this. You might even be able to use your model railway for it. The motor would take care of the friction. Just put a straight and circular section of track together, mount some rod (with a suitable mass) to the train and see whether the speed changes when it hits the point where the track curvature changes (ufortunately, I am personally not in a position to do it myself).

That is completely missing the point. Of course you could analyze a physically realistic situation in enough detail to predict the outcome.

But the question as given in your OP is not physically realistic, assuming we take words like "rod" to mean what they usually mean in idealized questions about mechanical systems.

You can ignore the unreality and get a solution using energy methods (though for some reason you refuse to accept that strategy). That could be a good way to make a simple model of a realistic situation and use it to get a useful approximate result. But if you insist on the view that the situation in the OP is realistic, we might as well have a debate about what would happen in the track were shaped like a square circle.

 

[Fantasist]

Again, R in the equations I posted above is not a coordinate (neither a generalized coordinate nor a polar coordinate). It is the radius of curvature of the path, a parameter of the path which changes along the path from infinite to finite.

A circle with an infinite radius is still a circle, not a straight line ('infinite radius' just means 'for arbitrarily large finite radii').

Your equation for the kinetic energy of a point mass

\frac{m}{2}v^2 = \frac{m}{2} R^2 \omega^2

only holds for a circular motion. The correct equation should contain a radial kinetic energy as well

$$\frac{m}{2}v^2 = \frac{m}{2}\dot{R}^2 + \frac{m}{2} R^2 \omega^2$$

And as I mentioned before, the $R$ in the angular momentum $\vec{l}=m \vec{R}\times \vec{v}$ has to be consistent with the $R$ in the above equations, and if it would suddenly go from finite to infinite or vice versa it would violate angular momentum conservation.

 

[Fantasist]

You can ignore the unreality and get a solution using energy methods (though for some reason you refuse to accept that strategy). That could be a good way to make a simple model of a realistic situation and use it to get a useful approximate result. But if you insist on the view that the situation in the OP is realistic, we might as well have a debate about what would happen in the track were shaped like a square circle.

The square circle is a completely inappropriate comparison. There no kinks in the path here, Everything is smooth (differentiable). Only the second derivative of the path has a discontinuity. But this isn't a practical problem.

 

[mfb]

This thread got completely pointless.
Fantasist, we gave more than enough different methods and multiple descriptions how to solve the problem. If you continue to ignore them and stick to your errors, more repetitions won't help.

 

[Dale]

The correct equation should contain a radial kinetic energy as well

$$\frac{m}{2}v^2 = \frac{m}{2}\dot{R}^2 + \frac{m}{2} R^2 \omega^2$$

I derived the correct equation above.

And as I mentioned before, the $R$ in the angular momentum $\vec{l}=m \vec{R}\times \vec{v}$ has to be consistent with the $R$ in the above equations, and if it would suddenly go from finite to infinite or vice versa it would violate angular momentum conservation.

I already rebutted this argument. Again, the rod's momentum is obviously not conserved in this problem since it is not free from external forces.