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Acceleration perpendicular to velocity 
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#1
Dec2713, 05:10 AM

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



#3
Dec2713, 05:34 AM

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#4
Dec2713, 05:41 AM

P: 1,290

Acceleration perpendicular to velocity



#5
Dec2713, 06:18 AM

Sci Advisor
PF Gold
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#6
Dec2713, 06:49 AM

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#7
Dec2713, 06:56 AM

PF Gold
P: 4,262

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



#8
Dec2713, 07:42 AM

HW Helper
P: 7,055

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
Dec2713, 08:48 AM

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#10
Dec2713, 08:51 AM

P: 819

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
Dec2713, 09:08 AM

PF Gold
P: 4,262

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
Dec2713, 09:29 AM

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#13
Dec2713, 04:26 PM

P: 129

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 


#14
Dec2713, 05:48 PM

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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
Dec2713, 06:45 PM

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PF Gold
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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
Dec2713, 07:40 PM

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#17
Dec2713, 11:21 PM

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