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

[Frigus]

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?

 

[Frigus]

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).

 

[Frigus]

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).

 

[Frigus]

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/

 

[Frigus]

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.

 

[Frigus]

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.

 

[Frigus]

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.

 

[Frigus]

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.

 

[Frigus]

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.

 

[Frigus]

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.

 

[Ibix]

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.

Some cultures put more emphasis on honorifics than others. Culture on PhysicsForums mostly follows current western forms, which means first names (or nicknames, for those of us not posting under our real names) for more or less everyone short of a head of state. Just say thank you at the end of the thread, and that's enough.

 

[Frigus]

I accept it but my this habit is not wrong like someone can point it and criticize me
And I also want to share a thought "only that person has right to nag someone who praises."

 

[Frigus]

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.

Firstly,I am not wasting someone time and if one thought that I am wasting his/her time then please just don't reply to my thread it's just that and secondly I thought a lot about a topic and sometimes I reply fast because I have seen or thought that explanation earlier so I tell them what is the place where I am stuck.

 

[jbriggs444]

I am not wasting someone time...

That run-on sentence could use a lot of punctuation. It accomplishes nothing but to disavow responsibility for your actions. A complete waste of electrons.

The poor formatting conveys an impression of disrespect toward the reader. The explicit message is also one of disrespect.

 

[phinds]

@Hemant, again, I assume that you are from a different culture and English is not your native language. While I believe you are doing your best to write in English, what people are telling you is that your best still needs some work. Your posts are sometimes difficult to understand. Being is a rush to post is not a good idea when you don't have a full grasp of English grammar and punctuation.

 

[Frigus]

I am really very sorry for my English and I accept my mistake, from now onwards I will do my best to explain what I write and also I want to tell that English is not my native language so I can't explain well in English.

 

[phinds]

I am really very sorry for my English and I accept my mistake, from now onwards I will do my best to explain what I write and also I want to tell that English is not my native language so I can't explain well in English.

You do MUCH better than most of us would do in whatever your native language is

 

[sophiecentaur]

So can you please tell me how can I give someone respect

That's an easy one. Just make sure your replies to what they say show that you have thought about what they have said - even if it seems to go against your preconceptions.
In these matters it is always a good idea to assume ( and make it clear) that you are wrong if your personal reasoning leads you away from accepted Physics. (Unless you have studied it all and are at the leading edge of research.) "Why am I wrong?" not "I must be right, despite what they are telling me" It usually gets much more helpful responses.
Sirs and Mams are not worth much on PF; you could well find you are talking to someone younger than you who happens to know the answer.

 

[Ibix]

@Hemant - @PeroK's post #11 gives you a correct proof, showing that the change of kinetic energy (proportional to speed squared) is zero if the acceleration and velocity are perpendicular. @jbriggs444's posts #17 and #19 explain why your original argument doesn't work, because what you were describing was a parabolic path, not a circular one.

 

[jbriggs444]

That's an easy one. Just make sure you replies to what they say show that you have thought about what they have said

Nicely put. Many of us just want to know that we are being listened to.

Let us go back and examine a response that did not measure up to such a standard.

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

There is no feedback here to indicate that anything has been taken on board. There is no indication of what has been found wanting in any of the previously provided correct answers. There is only a renewed demand for a final answer.

At the risk of reading more into a 14 word request than is actually there...

This is suggestive of a learning environment in which every question has a single "right answer" and in which the student's job is to produce the right answer on demand. In my upbringing, the preferred goal is understanding rather than on the parroting of "right answers". Understanding is a personal process. An explanation that works for one person may fail for someone else. We are trying to feed you an explanation which works for you. We cannot do that without feedback.

 

[Frigus]

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: $$\frac{dK}{dt} = \vec F \cdot \vec v$$

According to this derivation of work done which makes sense to me and which is as follow,
v_x²-u_x²=2a_xs---(I)
F_x=Ma_x---(II)
And after some little algebra we get,
$\frac{1}{2}\$mv²_x-$\frac{1}{2} \ m$u²_x=F_x.s

It means to me that if their is change in speed of object due to the force then it has done work as their is change in kinetic energy of object and if force doesn't changes the kinetic energy then their is no work done which is case of perpendicular force as it doesn't changes the speed but only direction. I know you have shown that as f and v are perpendicular then their dot product is zero but I can't make physical sense from the dot product and I try to understand it physically.

 

[Frigus]

What must its velocity have been a moment ago?

I have found that velocity a moment ago have greater magnitude from this it is concluded that something is wrong but now I can't figure it out what is wrong(as If I keep doing it and find velocity some moments ago and then again find the velocity of the moment from which I had started then my velocity from which I had started has increased value which is not possible) .
So can you please help me to figure it out

 

[PeroK]

I have found that velocity a moment ago have greater magnitude from this it is concluded that something is wrong but now I can't figure it out what is wrong(as If I keep doing it and find velocity some moments ago and then again find the velocity of the moment from which I had started then my velocity from which I had started has increased value which is not possible) .
So can you please help me to figure it out

You're main problem is continuing to use the SUVAT formulas (which only apply to constant acceleration) instead of a calculus-based approach (which is essential in this case).

The argument so far goes something like this:

Q: Let's assume the acceleration is constant for some finite moment of time, then the speed changes. What am I doing wrong?

A: The acceleration is not constant for any finite period of time. It's changing continuously. You need to use calculus.

Q: I don't like calculus. Let's assume the acceleration is constant for some finite moment of time, then the speed changes. What am I doing wrong?

Repeat ad nauseam.

 

[A.T.]

I can't make physical sense from the dot product and I try to understand it physically.

The dot product separates a component of a vector that is parallel to the other vector, and multiplies the magnitudes of only these parallel components.

But I don't think you need the concept of work here. See post #15:
https://www.physicsforums.com/threa...esnt-change-the-magnitude.982912/post-6285641

 

[jbriggs444]

Physics is a quantitative science. Let us go back to the first post in this thread. It is not quantitative. It is qualitative. It asks the question: "is velocity increasing" but does not answer the question "how fast".

With luck, the following process will justify the result from differential calculus.

So let us refine the setup a bit. We have an object moving rightward at speed $V_0$. It is being accelerated upward at an acceleration rate of $a$. We define a unit time increment $V_0/a$. Obviously, this is the time that it would take for an acceleration of magnitude a to bring the object to a halt. We can refer to time $t$ expressed in those units.

What is the tangential speed v(t) at t=0?

That's an easy one: $v(0) = V_0$

What is the tangential speed v(t) at t=1?

That is not much harder. In these units, it is clear that $v_y(t) = V_0t$ and that $v_x(t) = V_0$. The vector sum of the two is $\sqrt{V_0^2+V_0^2} = \sqrt{2}V_0$. This is approximately 1.4 $V_0$.

What is the tangential speed v(t) at t=0.1?

Easy. $v_y(t) = V_0t = 0.1V_0$. $v_x(t) = V_0$. The vector sum is $\sqrt{V_0^2+(0.1V_0)^2} = \sqrt{1.01} V_0$ This is approximately 1.005 $V_0$.

What is the tangential speed v(t) at t=0.01?

We can carry out the calculation again. This time it's $\sqrt{1.0001} V_0$. That's approximately 1.00005 $V_0$

See the pattern?

What tangential acceleration do each of these results represent?

With T=1, we got a $0.4V_0$ velocity increment in 1 unit of time. In these units that's $0.4 a$.

With T=0.1 we got a $0.005 V_0$ velocity increment in 0.1 units of time. In these units, that's $0.05 a$.

With T=0.01 we got a $0.00005 V_0$ velocity increment in 0.01 units of time. In these units, that's $0.005 a$.

See the pattern?

What's the limit of tangential acceleration as you decrease the step size asymptotically toward zero?

How does one interpret this result?

Each time one does the calculation with a particular step size, no matter how small, there is a bit of error involved with the assumption that the acceleration is constant and upward. In truth, the object is traversing a circular path and the acceleration is constantly changing in direction.

The small tangential acceleration that is calculated with a particular small step size is the difference between the average tangential acceleration that would have been seen on the assumed parabolic path and the actual tangential acceleration that is present on the actual circular path.

As step size decreases toward zero, the discrepancy (error) between the calculation and the reality becomes smaller and smaller. The limit that is approached is what the result would be with no remaining error or discrepancy.

That is to say that the true and correct result is zero tangential acceleration.

 

[Frigus]

Physics is a quantitative science. Let us go back to the first post in this thread. It is not quantitative. It is qualitative. It asks the question: "is velocity increasing" but does not answer the question "how fast".

With luck, the following process will justify the result from differential calculus.

So let us refine the setup a bit. We have an object moving rightward at speed $V_0$. It is being accelerated upward at an acceleration rate of $a$. We define a unit time increment $V_0/a$. Obviously, this is the time that it would take for an acceleration of magnitude a to bring the object to a halt. We can refer to time $t$ expressed in those units.

What is the tangential speed v(t) at t=0?

That's an easy one: $v(0) = V_0$

What is the tangential speed v(t) at t=1?

That is not much harder. In these units, it is clear that $v_y(t) = V_0t$ and that $v_x(t) = V_0$. The vector sum of the two is $\sqrt{V_0^2+V_0^2} = \sqrt{2}V_0$. This is approximately 1.4 $V_0$.

What is the tangential speed v(t) at t=0.1?

Easy. $v_y(t) = V_0t = 0.1V_0$. $v_x(t) = V_0$. The vector sum is $\sqrt{V_0^2+(0.1V_0)^2} = \sqrt{1.01} V_0$ This is approximately 1.005 $V_0$.

What is the tangential speed v(t) at t=0.01?

We can carry out the calculation again. This time it's $\sqrt{1.0001} V_0$. That's approximately 1.00005 $V_0$

See the pattern?

What tangential acceleration do each of these results represent?

With T=1, we got a $0.4V_0$ velocity increment in 1 unit of time. In these units that's $0.4 a$.

With T=0.1 we got a $0.005 V_0$ velocity increment in 0.1 units of time. In these units, that's $0.05 a$.

With T=0.01 we got a $0.00005 V_0$ velocity increment in 0.01 units of time. In these units, that's $0.005 a$.

See the pattern?

What's the limit of tangential acceleration as you decrease the step size asymptotically toward zero?

How does one interpret this result?

Each time one does the calculation with a particular step size, no matter how small, there is a bit of error involved with the assumption that the acceleration is constant and upward. In truth, the object is traversing a circular path and the acceleration is constantly changing in direction.

The small tangential acceleration that is calculated with a particular small step size is the difference between the average tangential acceleration that would have been seen on the assumed parabolic path and the actual tangential acceleration that is present on the actual circular path.

As step size decreases toward zero, the discrepancy (error) between the calculation and the reality becomes smaller and smaller. The limit that is approached is what the result would be with no remaining error or discrepancy.

That is to say that the true and correct result is zero tangential acceleration.

Aha! I got it
Thanks a lot

 

[sophiecentaur]

but I can't make physical sense from the dot product and I try to understand it physically

Many people confuse the terms "physically" and "familiar". If we are exposed to something enough times then it becomes familiar and a physical 'feeling' will often follow. It is easy for something to make sense when you don't need to make a significant course change in your picture of the world. If, as a result of hammering out the dot product, you get comfortable with it then you will find that 'physical' feeling in totally different contexts. That's something that can be very satisfying.

 

[Frigus]

Many people confuse the terms "physically" and "familiar". If we are exposed to something enough times then it becomes familiar and a physical 'feeling' will often follow. It is easy for something to make sense when you don't need to make a significant course change in your picture of the world. If, as a result of hammering out the dot product, you get comfortable with it then you will find that 'physical' feeling in totally different contexts. That's something that can be very satisfying.

I am irritated with dot product and cross product too much as whenever I try to understand it I only see rules to use it.i have seen it on YouTube,books, course material e.t.c. I tried many times to understand it but it doesn't makes sense to me, sometimes it seems like I will not be able to make sense of it but that feeling of satisfaction which I get when I understands something derives me to do it.i hope your opinion will help me.

Thanks.

 

[sophiecentaur]

whenever I try to understand it I only see rules to use it

I have a feeling that you found the same problems (well forgotten by now) when you were first learning about speed, density, areas etc. etc. Those are all now a part of your 'mental / physical' model of the world. I seriously suggest that using the dot and cross products in real situations will allow them to slip noiselessly into your unconscious models, along with other processes. Just follow the 'rules' often enough.
When you think about the problems that the ancients had with all the maths associated with mechanical phenomena we all take for granted I think you would acknowledge that nothing comes naturally in Physics.

 

[Frigus]

I have a feeling that you found the same problems (well forgotten by now) when you were first learning about speed, density, areas etc. etc. Those are all now a part of your 'mental / physical' model of the world. I seriously suggest that using the dot and cross products in real situations will allow them to slip noiselessly into your unconscious models, along with other processes. Just follow the 'rules' often enough.
When you think about the problems that the ancients had with all the maths associated with mechanical phenomena we all take for granted I think you would acknowledge that nothing comes naturally in Physics.

I will now do tonnes of problems related to cross and dot product and will definitely the bird sight view and will tell you when I get it.
Thanks.

 

[sophiecentaur]

cross and dot

The massive difference between the scalar and vector results are very significant. It's no wonder that the dot product makes so much sense - for instance the work done, related to magnitudes and angle between.

 

[Frigus]

The massive difference between the scalar and vector results are very significant. It's no wonder that the dot product makes so much sense - for instance the work done, related to magnitudes and angle between.

True,I thought about dot product for a while and then I realized that it is just a name given to a the process of finding the vector along the direction of another vector but in case of cross product the thing which irritates me most is that when I study about it then for making sense of it most websites,article,books e.t.c gives example of torque and then while studying torque they tell that torque direction is perpendicular to the plane because we do it's cross product.