# Acceleration perpendicular to velocity

1. Dec 27, 2013

### negation

I'm at a loss. Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?

2. Dec 27, 2013

### Vanadium 50

Staff Emeritus
You're moving north but being pushed to the east.

3. Dec 27, 2013

### negation

Must it necessarily be for acceleration to be perpendicular to velocity? Could it also not be velocity 1 perpendicular to velocity 2?

4. Dec 27, 2013

### Tanya Sharma

Think of a uniform circular motion (for ex. a car moving at a constant speed along a circular track).The velocity is tangential to the path ,but the acceleration is radial(towards the center)

5. Dec 27, 2013

### sophiecentaur

A force on an object will accelerate it, the acceleration is quite independent of the velocity of the object when the force is applied. The resulting velocity can be found by adding the vectors after the time you are interested in.

6. Dec 27, 2013

### Vanadium 50

Staff Emeritus
If you ask for acceleration perpendicular to velocity, you're going to get an example with acceleration perpendicular to velocity, yes.

7. Dec 27, 2013

### Pythagorean

negation, would you provide us with some context? What scenario caused you to ask this question?

8. Dec 27, 2013

### rcgldr

A common example of acceleration perpendicular to velocity would be a car traveling at constant speed on a winding or circular road. The path could be just about any curved shape. The acceleration would always be perpendicular to velocity, and the driver would only use enough throttle to maintain speed (zero tangental acceleration) and use steering inputs (centripetal acceleration) to turn the car.

9. Dec 27, 2013

### negation

Do you think you could frame a geometric interpretation of this?

10. Dec 27, 2013

### negation

It's just scattered information in my textbook. There's no context so I'm a little lost.
However, I've read up a bit on centripetal forces so I might have a rough idea but it's not sufficiently rigorous for me.

11. Dec 27, 2013

### Pythagorean

Ok, I think I see where you're coming from now

"Must it necessarily be for acceleration to be perpendicular to velocity? Could it also not be velocity 1 perpendicular to velocity 2?"

No, it's not necessary in general. It could also be that way, but in the case of centripetal force, this would likely cause the orbiting body to veer out of orbit unless it's perfectly controlled to a new orbit. For any particular orbit, you want to ensure that the acceleration in the radial direction is zero so that you maintain orbit.

12. Dec 27, 2013

### A.T.

http://nhn.nhn.ou.edu/~jeffery/astro/astlec/lec005/orbit_001_centripetal.png [Broken]

Last edited by a moderator: May 6, 2017
13. Dec 27, 2013

### Malverin

It depends on the point of view.

If you have rotational motion with constant speed and radius in polar coordinates, centripetal acceleration is perpendicular to the motion direction.

But if you project the velocity in Cartesian coordinate system, X an Y components of speed, change with rotation and it is easier to see where the acceleration comes from

http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Rotation/Circular.html

http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Rotation/Circular_clip_image001.jpg [Broken]

Last edited by a moderator: May 6, 2017
14. Dec 27, 2013

### D H

Staff Emeritus
He did: The path could be just about any curved shape.

All that one can deduce from an acceleration vector that is always perpendicular to the velocity vector is that speed is constant. The path that the object follows can be *any* curved shape. Conversely, a constant speed means that acceleration that is either zero or is always orthogonal to the velocity vector. This is easy to prove: Simply differentiate $||\vec v(t)||^2 = \vec v(t) \cdot \vec v(t)$. If the speed is constant, then so is $||\vec v(t)||$, and thus $\vec a(t) \cdot \vec v(t) \equiv 0$.

To get uniform circular motion you need to add some constraints to the acceleration vector. Uniform circular motion results if the curve is planar (i.e., has zero torsion) and if the acceleration vector is constant in magnitude and is always orthogonal to the velocity vector. The converse is also true.

15. Dec 27, 2013

### sophiecentaur

When you are first introduced to Centripetal Acceleration, it may not be made clear that you are dealing with a very special case of motion in a curve. The acceleration is only at right angles to the velocity for a circular path at constant speed. The angle between acceleration and velocity can be anything, in the general case. For an elliptical orbit, the angle is only 90 degrees at apogee and perigee. At other times it isn't. Acceleration is always towards the centre of the planet - that's all. The speed is changing all the time and so is the direction. In circular motion with just a central attractor, the acceleration is only radial. On a circular track, however, there can also be acceleration in a tangential direction, if you apply accelerator or brakes.

Did you see the film Gravity? People are moving through space in all sorts of directions and firing their thrusters and accelerating in other directions. They seemed to get the dynamics pretty convincing, aamof.

16. Dec 27, 2013

### rcgldr

But it would be possible to follow an elliptical path using only acceleration perpendicular to velocity; again, using the example of a car, a car traveling on an elliptical path at constant speed. As mentioned before, the path could be any curved shape, sine wave, parabola, hyperbola, spiral, ellipse, circle, ... .

17. Dec 27, 2013

### D H

Staff Emeritus
Not true. These are necessary but not sufficient conditions. If the acceleration vector switches sign on occasion you won't get circular motion. In addition to the above, the path has to be smooth (infinitely differentiable) to obtain uniform circular motion.

18. Mar 3, 2015

### AgentSmith

You could also think of a car moving north, then it gets hit by a car moving east. The car moving east accelerates the car moving north. So a is perpendicular to v. This assumes the same units, such a meters and seconds

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