# Why are changes in direction considered accelerating

1. Apr 18, 2015

### physdoc

I've seen this question come up before and I have an intuitive explanation: In circular motion, the object is being pulled toward the center, so some of its velocity is being imparted from its tangential path toward the center. If its position with reference towards the center is changing, meaning if its continuously being pulled from its tangential path, then the tangential velocity is always increasing, but this increase is never seen. Its like if an object is moving in a straight line and some of its motion starts to be imparted towards the side: in order to maintain the same speed in the original direction it would have to be accelerating in that direction; this is what I meant when I said it is not seen.
Just to take an extreme example for purposes of clarity: suppose we move back and forth in a straight line; in order to move in the opposite direction, we must stop, then accelerate. if we take the limit over shorter and shorter time intervals and over shorter and shorter distances, then this back and forth motion will approach a point where the object is ALWAYS speeding up.

2. Apr 18, 2015

### Simon Bridge

It is a mathematical consequence of the definition of acceleration as the rate of change of velocity.
Set up an equation where the velocity changes only in direction, then work out what the rate of change of that velocity is.

Your discussion shows some possible misconceptions I'd like to address.

In circular motion, the tangential velocity is not always increasing ... the magnitude is always the same, but the direction of the tangent changes. It is not a case of there really being an increase but you don't see it - there really is not an increase. The word "increase" is not something you can usefully apply to the change in direction like you can to the change in magnitude.

The final "clarifying example" makes no sense.
Please provide a concrete example where a decreasing velocity is always increasing when you take the limit of small time intervals.

It looks like you are conflating common-use and technical definitions of words.

Last edited: Apr 18, 2015
3. Apr 18, 2015

### Staff: Mentor

As Simon said, it's a consequence of the definition of acceleration being a rate of change change of velocity. Velocity is composed of a magnitude (speed) and a direction. So if the direction changes, even if the speed doesn't, you have acceleration.

4. Apr 18, 2015

### rumborak

I always thought it to be particularly peculiar that, while an electron flying through a magnetic field will be constantly accelerated to fly in a circle, no work is done at all. But I guess it's the same scenario as a satellite orbiting Earth; there too no work is done.

5. Apr 18, 2015

### Simon Bridge

That is because Energy is not a vector. Changing direction at constant speed (and constant potential) means the energy remains the same ... so no change in energy means no work.

6. Apr 18, 2015

### Staff: Mentor

Indeed. A planet in a perfectly circular orbit has no work done on it by gravity, even though it is constantly accelerating.

7. Apr 19, 2015

### Staff: Mentor

You have to apply a force to change an object's direction of motion and f=ma. Also, if it makes you feel better you can apply x, y and z velocity components to the vector and they will change with the acceleration even if it is circular.

8. Apr 19, 2015

### physdoc

Because of its inertia. It resists change in speed, and resistance is acceleration. I don't understand what you mean by "The word "increase" is not something you can usefully apply to the change in direction like you can to the change in magnitude." You imply that you can't relate two different ideas.
Its inertia is what causes change in direction to effect change in magnitude.
Please explain the "technical definition" of the word increase, and how is it different from its common use definition.
"increase" is only a word, and you can use it any way you want, i.e. to explain the relation between two different things. Didn't Maxwell use the word acceleration to imply a relation between two increases, like the relation between the magnetic and electric fields, two seemingly different things?

Last edited: Apr 19, 2015
9. Apr 19, 2015

### physdoc

This makes perfect sense. That force that you're applying is such because it goes against the objects inertia. It works against it's inertia to maintain its tangential path because the object, in its tangential path, is also resisting.

10. Apr 20, 2015

### rumborak

Yeah, a magnetic field certainly does not do any work on an electron when it sends it into circles. Not that Wikipedia is the best source for this, but it's what came up first:

http://en.wikipedia.org/wiki/Magnetic_field#Force_on_a_charged_particle

And the "holding an object requires work" analogy, that is a grave (yet common) misunderstanding. That is simply a inadequacy of human muscles. A chair holding a ball performs no work at all.

11. Jul 22, 2016

### physdoc

It follows from Newton's third law that an object will resist changes in its direction of motion; so the sun pulls on the earth, then the earth resists this pull with a force, which is an acceleration, and the earth and sun are continually in "tug of war". This is also Newton's first law; an object resists changes in its direction of motion, with its own force, which is another word for resistance; this force is applied toward maintaining its direction of motion, or in other words, it applies a force in its original direction of motion to resist the force of another object which is acting to change the direction of the first object. It's not like the original object lets the "mover" somehow "piggy-back" off its momentum; there is no motion in the new direction until it is created; to create motion where there is none requires an acceleration; the original "moved" object is not moving at all in the new direction; if we take a freeze frame, it is stationary in the new direction.

Here is a simple, but more extreme, example to illustrate this: if we have an object moving back and forth, in opposite directions, does it not have to stop, then change direction in order to go in the opposite direction? But if it does this, it would have to speed up. This is where the limit comes in: objects moving in a circle are basically constantly turning, but because that turn is not in the exact opposite direction we have to take some kind of limit. when an object, in a circular path arrives at the opposite point, or on the other side, in relation to the sun, it has effectively turned in the opposite direction; in order to do this it must have accelerated, or stopped completely, turned around, and accelerated to gain motion again.

12. Jul 22, 2016

### physdoc

My original premise is that associated with a change in direction is a change in speed

13. Jul 22, 2016

### Staff: Mentor

That premise is incorrect, although it is correct to say that a change in direction is associated with a change in velocity.

Getting this situation right requires a clear understanding of the difference between speed (a number, the magnitude of the velocity vector) and velocity (a vector).

14. Jul 22, 2016

### Staff: Mentor

If it makes you feel better you can apply x, y and z velocity components to the vector and they will change with the acceleration even if it is circular.

15. Jul 22, 2016

### sophiecentaur

This is not correct. In order for linear motion to reverse, the speed must decrease to zero and then increase in the opposite sense. We have to assume continuous variation in position and speed if we are to avoid infinite forces and discontinuities in any acceptable model.
The terms Speed and Velocity are very well defined. If you stick with those definitions then there need be no confusion.
To stop, implies that negative acceleration is applied for some time. If the same acceleration (magnitude and direction) is applied then the object will slow, stop and then go faster in the other direction. No step changes in any of the variables. What inconsistency are you looking for?

16. Jul 22, 2016

### rcgldr

A a simple example, imagine a car moving at constant speed on a winding road. The car changes direction, but not speed. The occupants of the car can get a sense of the acceleration via the centripetal force that the car exerts onto them. (The pavement exerts a centripetal force onto the tires which exert an equal but outward force onto the pavement, a Newton third law pair of forces).

Last edited: Jul 22, 2016
17. Jul 22, 2016

### Simon Bridge

In post 11, "it applies a force in the original direction of motion" directly contradicts the 3rd law statement (since a 3rd law reaction force is directed opposite to its pair) in the first sentence and is not how the 1st law works. Besides, it is the 2nd law that defines force. Other issues are covered above.

Sticking by your guns no matter what is an admirable trait in many situations, but not in science.
In this case the problem appears to be mainly a language issue. Physics uses many familiar words as technical terms with a definition somewhat different to the common use. To understand physics, you have to understand this. Insisting on applying the common use to these terms will continue to result in inconsistancies. Use the technical physics definitions and the inconsistancies vanish.

18. Jul 22, 2016

### rcgldr

Take the case of circular motion on a 2d plane where the only acceleration is centripetal acceleration. The components of velocity $v_x$ and $v_y$ change over time but the speed $\sqrt{v_x^2 + v_y^2}$ remains constant.

Or consider the case of a particle moving along a parabola at constant speed, so the only acceleration is centripetal.
Define the equations for position as:
$x = \pm \sqrt{2\ y}$
$y = \frac{1}{2} \ x^2$

The equations for velocity are:
$v_x = \frac{c}{\sqrt{x^2 + 1}}$
$v_y = \frac{c \ x}{\sqrt{x^2 + 1}}$
$speed = \sqrt{v_x^2 + v_y^2} = \sqrt{\frac{c^2}{x^2+1} + \frac{c^2 \ x^2}{x^2+1}} = \sqrt{\frac{(c^2) \ (x^2 + 1)}{x^2+1}} = c$

Taking derivatives results in accelerations, the equations are:

$a_x = \frac{d(v_x)}{dt} = \frac{v_x \ d(v_x)}{dx} = \frac{- \ c^2 \ x}{(x^2+1)^2}$
$a_y = \frac{d(v_y)}{dt} = \frac{v_y \ d(v_y)}{dy} = \frac{\ c^2}{(x^2+1)^2}$

To confirm that acceleration is perpendicular to velocity:
$\frac{-a_x}{a_y} = \frac{v_y}{v_x} = x$

Last edited: Jul 23, 2016