Why a force perpendicular to the velocity doesn't change the magnitude?

[Hemant]

Main Question or Discussion Point

As I have shown in the picture even if their is minimal change but shouldn't it increase after a long time as minimal changes will keep accumulating.

 

[phinds]

Is this a homework problem?

 

[Hemant]

Is this a homework problem?

No sir.

 

[phinds]

OK. Well, your question is completely unclear. I can't figure out whether there is a small offset applied to the motion and then it is left alone over time (which is what your picture looks like) or a continuous small offset being applied constantly over time (as your question implies).

 

[Hemant]

OK. Well, your question is completely unclear. I can't figure out whether there is a small offset applied to the motion and then it is left alone over time (which is what your picture looks like) or a continuous small offset being applied constantly over time (as your question implies).

Sir force is applied continuously perpendicular to the velocity as in case we rotates a stone using string and sorry I missed showing the force in my picture.

 

[Bandersnatch]

In force diagram you drew, you allow the acceleration vector to be no longer perpendicular to the instantenouos velocity for the entire period Δt when it acts along the direction of the V2 vector, apart from the very first moment t1. It doesn't matter how short you make that period - the acceleration is not perpendicular to V1 at any time apart from the first instant.
For a change in direction only (as with e.g. circular motion) the acceleration vector must remain perpendicular to velocity at all times.

 

[Ibix]

As @Bandersnatch notes, the problem is that you are allowing your velocity vector to change in your small time interval, but not the acceleration vector. If there's enough time for the velocity vector to change (possibly only infinitesimally) then there's enough time for the acceleration vector to change (possibly only infinitesimally). So the total acceleration in your $\Delta t$ won't be purely vertical, but will have a leftward component which will lead to it satisfying $|\vec v_1+\vec v_2|=|\vec v_1|$ (in the limit as $\Delta t$ goes to zero, anyway).

 

[Hemant]

As @Bandersnatch notes, the problem is that you are allowing your velocity vector to change in your small time interval, but not the acceleration vector. If there's enough time for the velocity vector to change (possibly only infinitesimally) then there's enough time for the acceleration vector to change (possibly only infinitesimally). So the total acceleration in your $\Delta t$ won't be purely vertical, but will have a leftward component which will lead to it satisfying $|\vec v_1+\vec v_2|=|\vec v_1|$ (in the limit as $\Delta t$ goes to zero, anyway).

Sir can you please tell me from where can I get this proof.

 

[PeroK]

Sir can you please tell me from where can I get this proof.

Someone else had the same question recently:

https://www.physicsforums.com/threads/a-proof-that-magnetic-forces-do-no-work.981895/

 

[Hemant]

Sir I can't figure it from that thread can you please give me an proof of this.

 

[PeroK]

Sir I can't figure it from that thread can you please give me an proof of this.

Let $K = \frac 1 2 m v^2 = \frac 1 2 m \vec{v} \cdot \vec{v}$ be the kinetic energy of a particle. $$\frac{dK}{dt} = \frac 1 2 m \frac{d}{dt}(\vec{v} \cdot \vec{v}) = m (\frac{d\vec{v}}{dt} \cdot \vec{v}) = m\vec{a} \cdot \vec v = \vec F \cdot \vec v$$

Hence, if the force is perpendicular to the velocity, then $\frac{dK}{dt} = 0$, which means the kinetic energy of the particle is constant, hence the speed is constant.

 

[Hemant]

Let $K = \frac 1 2 m v^2 = \frac 1 2 m \vec{v} \cdot \vec{v}$ be the kinetic energy of a particle. $$\frac{dK}{dt} = \frac 1 2 m \frac{d}{dt}(\vec{v} \cdot \vec{v}) = m (\frac{d\vec{v}}{dt} \cdot \vec{v}) = m\vec{a} \cdot \vec v = \vec F \cdot \vec v$$

Hence, if the force is perpendicular to the velocity, then $\frac{dK}{dt} = 0$, which means the kinetic energy of the particle is constant, hence the speed is constant.

Sir but in this derivation we have used the result of derivation"which proves that perpendicular force cannot change the magnitude of velocity" which is my question?
Sorry to argue sir.

 

[PeroK]

Sir but in this derivation we have used the result of derivation"which proves that perpendicular force cannot change the magnitude of velocity" which is my question?
Sorry to argue sir.

Not at all. $$\vec F \cdot \vec v = 0 \ \Rightarrow \frac{dK}{dt} = 0 \ \Rightarrow \frac{dv}{dt} = 0$$

PS Note that in post #11 I showed that: $$\frac{dK}{dt} = \vec F \cdot \vec v$$

 

[jbriggs444]

As I have shown in the picture even if their is minimal change but shouldn't it increase after a long time as minimal changes will keep accumulating.

Whenever I see students with this misconception, they always look at the situation evolving forward in time. That is, they look at the velocity vector now, apply the [assumed to be] constant acceleration and derive that the velocity vector later has increased in magnitude. From this they conclude that speed must be increasing.

None of them ever bother to do the same calculation going backward in time to determine the velocity a moment ago. If they did, they would see that the speed a moment ago must also have been higher. So the same [mistaken] argument proves with equal force both that speed is increasing and that it is decreasing.

Edit: Of course, that's the difference between a parabolic path and a circular path. For a parabolic path (constant acceleration), the speed really does increase both forward and backward in time. For a circular path (always perpendicular acceleration) the speed stays the same both ways.

 

[A.T.]

As I have shown in the picture even if their is minimal change but shouldn't it increase after a long time as minimal changes will keep accumulating.

To decrease velocity magnitude, you need an acceleration component anti-parallel to velocity.
To increase velocity magnitude, you need an acceleration component parallel to velocity.
If the acceleration has neither component, then velocity magnitude cannot change.

 

[Hemant]

None of them ever bother to do the same calculation going backward in time to determine the velocity a moment ago. If they did, they would see that the speed a moment ago must also have been higher. So the same [mistaken] argument proves with equal force both that speed is increasing and that it is decreasing.

Sir can you please explain me this point I can't understand what you want to say.

 

[jbriggs444]

Sir can you please explain me this point I can't understand what you want to say.

The SUVAT equations work equally well in predicting how a system behaves going forward in time and going backward in time. You can trace the trajectory of a stone on string or a planet in its orbit into the future or into the past.

You have presented a calculation which purports to show velocity increasing into the future. How about applying that calculation to see how velocity behaves going into the past?

The laws of classical Newtonian physics are invariant under time reversal. If you can take a symmetric situation and derive an asymmetry, you've goofed.

 

[Hemant]

How about applying that calculation to see how velocity behaves going into the past?

Sir how can I apply it?

 

[jbriggs444]

Sir how can I apply it?

Look at your original post. You evolved the velocity forward in time as the object moved to the right under a constant upward acceleration. Go back and do it again. But now the object is arriving from the left under constant upward acceleration and arrives at the center with rightward velocity v. What must its velocity have been a moment ago?

 

[PeroK]

Sir how can I apply it?

I'm not sure how helpful this is. SUVAT equations assume constant force and constant acceleration, which cannot be the case if the force remains perpendicular to the velocity.

It's a pity you've been distracted from the mathematics in post #11.

 

[Hemant]

Please help me get out of this problem, which is correct explanation of this.

 

[jbriggs444]

Please help me get out of this problem, which is correct explanation of this.

Your misunderstandings seem to go deep. Do you understand how a first derivative is defined? Do you understand limits? Tangent lines to a curve?

 

[Vanadium 50]

Hemant, first, enough with the "Sir, if for no other reason than some of the "sirs" on PF are women.

Second, you are wasting the time of the people trying to help you as well as your own by not thinking about the answers you are gettng. How do I know this? Because your responses occur only minutes after someone else's post. You're not leaving enough time to think about it - and you just turn around and demand answers.

I don't think this is the first time this has been pointed out to you.

If you don't understand something, think about it. If you still don't understand, think some more. If you still don't understand, write a carefully composed message showing you have thought about it and discussing exactly what you don't understand. Don't just demand answers of us.

 

[Hemant]

Hemant, first, enough with the "Sir, if for no other reason than some of the "sirs" on PF are women.

So can you please tell me how can I give someone respect and as we can see most of the people on this thread are male so to whom may I say mam.

 

[phinds]

So can you please tell me how can I give someone respect and as we can see most of the people on this thread are male so to whom may I say mam.

It isn't necessary. Just talk to us like you would talk to a friend.

I see you have not bothered to put any information about yourself on in your profile so I can't tell for sure but I think it is a safe assumption that you come from a culture where it is very important to show overt respect for your elders and superiors. That's not necessary on this forum, just be polite and that's plenty.