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

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,
vx2-ux2=2axs---(I)
Fx=Max---(II)
And after some little algebra we get,
##\frac{1}{2}\ ##mv2x-##\frac{1}{2} \ m##u2x=Fx.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.

• vanhees71
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) .

• weirdoguy
Homework Helper
Gold Member
2022 Award
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) .

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?

• weirdoguy
Homework Helper
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.

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• hutchphd and Frigus
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

• jbriggs444
Gold Member
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.

• jbriggs444, hutchphd, Frigus and 1 other person
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.

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Gold Member
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
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
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