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## Physics library

### Course: Physics library>Unit 4

Lesson 1: Circular motion and centripetal acceleration

# Calculus proof of centripetal acceleration formula

Proving that a = v^2/r. Created by Sal Khan.

## Want to join the conversation?

• At , why is p(t)=r? How did displacement become equal to the radius?
• p(t) is not the displacement vector, it is the position vector. Δp(t) i.e. the change in position vector would be the displacement. here, magnitude of p(t) would indeed be the radius as the centre if the circle is the origin
• Can this be done without engineering notation? I would like to see it.
• just look at the i part as the x component and the j part as the y component of the velocity vector, the operations stay the same.
The engineering notation is just a diffrent way of writing it down, it does nothing to the numbers
• Hello. I was wondering where the angular velocity videos are as mentioned at ?
• While there might be some mention of angular velocity in "Two-Dimensional Motion," I couldn't find it there. Instead, there is a video titled "Relationship between angular velocity and speed" under Physics > Torque, Moments, and Angular Momentum that explains what it actually is.
• I got lost at the end. How or why do we know that w=v/r to be able to make that substitution?
• What is actually a position vector? I always thought about the position as matter of coordinates. Can position have a direction?
• A point or position in a system of coordinates also corresponds to an arrow with length and direction. Instead of walking 5 km north then 3 km East it would be better to just walk immediately in the direction of the point. A scale drawing or a bit of trigonometry can tell you the precise angle from north to follow and also the distance which will be less than 5 plus 3 of course as you're cutting off the corner.
(1 vote)
• Sal says dθ/dt = ω. And ω is a vector, right? And t is scalar => dt is scalar. That must mean dθ is vector => θ is vector. What I would like to know is the direction of θ (and how this compares with the direction of ∆θ).
• No! ω is the angular velocity. And yes, usually, this is a vector quantity - but only if you use the "real" definition ( *ω* = r x v / |r|² ).
In your case, the argument has to be: dθ/dt = ω; Theta is obviously a scalar - cause θ and φ almost always are angles. Thus, ω is a scalar quantity in your case.
So, θ does not have a direction. It is the angle between the x-axis and the position vector (). And as always, it 'opens' in the positive (counterclockwise) direction of rotation.

In case, you'd calculate ω as a vector, it's direction would be perpendicular to the plane of rotation (only possible in 3 dimensions). Also, to be precise, it's a pseudovector ...
• sal says that dO/dT it's omega but wasn't omega delta theta over delta time?or is the same in this context?
• The notation Sal uses (dθ/dt) is essentially the same as ∆θ/∆t, but in the context of calculus (as done here), it is essentially looking at the small change in θ over the small change in time.
(1 vote)
• at , when Sal is using the chain rule to multiply by dθ/dt, shouldn't he have multiplied by dθ(t)/dt, or does it not matter? Why did the t go away?
• dθ/dt and dθ(t)/dt are the same thing. but adding t in the parentheses just helps us in understanding better what are we differentiating, in this case time . Hope i was able to clear the doubt!!