Register to reply

Acceleration perpendicular to velocity

by negation
Tags: acceleration, perpendicular, velocity
Share this thread:
negation
#1
Dec27-13, 05:10 AM
P: 819
I'm at a loss. Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?
Phys.Org News Partner Physics news on Phys.org
Step lightly: All-optical transistor triggered by single photon promises advances in quantum applications
The unifying framework of symmetry reveals properties of a broad range of physical systems
What time is it in the universe?
Vanadium 50
#2
Dec27-13, 05:33 AM
Mentor
Vanadium 50's Avatar
P: 16,385
You're moving north but being pushed to the east.
negation
#3
Dec27-13, 05:34 AM
P: 819
Quote Quote by Vanadium 50 View Post
You're moving north but being pushed to the east.
Must it necessarily be for acceleration to be perpendicular to velocity? Could it also not be velocity 1 perpendicular to velocity 2?

Tanya Sharma
#4
Dec27-13, 05:41 AM
P: 1,320
Acceleration perpendicular to velocity

Quote Quote by negation View Post
I'm at a loss. Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?
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)
sophiecentaur
#5
Dec27-13, 06:18 AM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,186
Quote Quote by negation View Post
I'm at a loss. Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?
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.
Vanadium 50
#6
Dec27-13, 06:49 AM
Mentor
Vanadium 50's Avatar
P: 16,385
Quote Quote by negation View Post
Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?
Quote Quote by negation View Post
Must it necessarily be for acceleration to be perpendicular to velocity?
If you ask for acceleration perpendicular to velocity, you're going to get an example with acceleration perpendicular to velocity, yes.
Pythagorean
#7
Dec27-13, 06:56 AM
PF Gold
Pythagorean's Avatar
P: 4,287
negation, would you provide us with some context? What scenario caused you to ask this question?
rcgldr
#8
Dec27-13, 07:42 AM
HW Helper
P: 7,133
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.
negation
#9
Dec27-13, 08:48 AM
P: 819
Quote Quote by rcgldr View Post
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.
Do you think you could frame a geometric interpretation of this?
negation
#10
Dec27-13, 08:51 AM
P: 819
Quote Quote by Pythagorean View Post
negation, would you provide us with some context? What scenario caused you to ask this question?
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.
Pythagorean
#11
Dec27-13, 09:08 AM
PF Gold
Pythagorean's Avatar
P: 4,287
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.
A.T.
#12
Dec27-13, 09:29 AM
P: 4,064
Quote Quote by negation View Post
Do you think you could frame a geometric interpretation of this?
Malverin
#13
Dec27-13, 04:26 PM
P: 134
Quote Quote by negation View Post
I'm at a loss. Can someone explain to me the entire idea of what it means for acceleration to be perpendicular to velocity?
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/Ph.../Circular.html

D H
#14
Dec27-13, 05:48 PM
Mentor
P: 15,170
Quote Quote by negation View Post
Do you think you could frame a geometric interpretation of this?
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.
sophiecentaur
#15
Dec27-13, 06:45 PM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,186
Quote Quote by negation View Post
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.
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.
rcgldr
#16
Dec27-13, 07:40 PM
HW Helper
P: 7,133
Quote Quote by sophiecentaur View Post
For an elliptical orbit, the angle is only 90 degrees at apogee and perigee.
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, ... .
D H
#17
Dec27-13, 11:21 PM
Mentor
P: 15,170
Quote Quote by D H View Post
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.
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.


Register to reply

Related Discussions
At what times are the velocity and acceleration perpendicular? Calculus & Beyond Homework 5
Acceleration is perpendicular to velocity Special & General Relativity 11
Acceleration always perpendicular to velocity Introductory Physics Homework 6
Acceleration perpendicular to velocity in 2D General Physics 5
Acceleration of an object is always directed perpendicular to its velocity Introductory Physics Homework 1