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Calculus proof of centripetal acceleration formula
Proving that a = v^2/r. Created by Sal Khan.
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- At1:51, why is p(t)=r? How did displacement become equal to the radius?(35 votes)
- 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(11 votes)
- Can this be done without engineering notation? I would like to see it.(17 votes)
- 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(28 votes)
- Hello. I was wondering where the angular velocity videos are as mentioned at6:00?(15 votes)
- 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.(9 votes)
- I got lost at the end. How or why do we know that w=v/r to be able to make that substitution?(7 votes)
- At6:05, how is d(theta)/dt equal to angular velocity? I thought angular velocity is v/r(6 votes)
- Those are equivalent formulations. Let's say you go around once per second. Then using the first formula you have 2pi/second. Using the second one you have 2*pi*r/(r*sec) and the r cancels out.(6 votes)
- What is actually a position vector? I always thought about the position as matter of coordinates. Can position have a direction?(7 votes)
- 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).(4 votes)
- 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 (1:25). 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 ...(3 votes)
- sal says that dO/dT it's omega but wasn't omega delta theta over delta time?or is the same in this context?(5 votes)
- 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)
- at5:00, 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?(3 votes)
- 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!!(2 votes)
- centripetal acceleration is a scalar or vector?, sal(2 votes)
- good question.
It is acceleration, so it must be a vector. Its direction is always changing (perpendicular to the velocity)(4 votes)
What I want to do in this video is a calculus proof of the famous centripetal acceleration formula that tells us the magnitude of centripetal acceleration, the actual direction will change it's always going to be pointing inwards, but the magnitude of centripetal acceleration is equal to the magnitude of the velocity-squared divided by the radius I want to be very clear, this is a scalar formula right over here; we're talking about the magnitude of the acceleration and the magnitude of the velocity. If these were vectors, we would have arrows drawn over it. So this really, I don't want people to get confused because this is a v, this is really referring to the speed-squared. and this is the magnitude, and these are all scalar quantities. So to do that, let's imagine some object, maybe it's some object in orbit around a planet or something, so let's say that's the planet, and you have some object that is in orbit around the planet, and it is going in a counter-clockwise direction, and so let's specify its position vector as a function as time. So that is its position vector and it is going to change as a function of time as this thing spins around We're going to assume, for the purposes of this proof. So that is our y-axis, and this is our x-axis. We're going to define theta as the angle between the positive x-axis and our vector. And we're going to assume this thing is in orbit with the radius of r. So the magnitude of our position vector, even though the direction is going to change the magnitude of our position vector is not going to change. It's always going to have length r. So this is going in a circle with radius r. The magnitude of our position vector, which is changing as a function of time, is going to be r. So how can we write the position vector in terms of its components at any given time? We can write the position vector, and I'll do it in engineering notation, and so you might want to review those videos if some of this looks foreign and I'll do a bit of basic trigonometry in breaking down the vector into its components and I encourage you to review some of those videos if some of that looks a bit daunting. If you take the position vector at any time, the magnitude is r, this angle is theta, its x-component, in blue, this vector right over here, the magnitude of the vector, I should say, is going to be r cosine of theta We learned that this came from basic trigonometry when we started two-dimensional projectile motion, we saw how to break these vectors down into its components, and the y-component of this vector is going to be r sine of theta So this is going to be r sine of theta. So the position vector at any time can be written as a sum of its x- and y-components. So it's the magnitude of its x-component, it's going to be r cosine of theta, and I could write theta as a function of time if I'd like, but I'm just going to write r cosine of theta actually, let me write it that way, so it shows theta is a function of time This thing is moving, and there's going to be that times the i-unit vector, we're in engineering notation over here. So that's the i-unit vector, it tells us that the x-component is going in the positive x direction. Plus the magnitude of the y-component, which is r sine of theta, which is going to be a function of time. So to be clear, the function of time applies to the theta. And that is going in the j-direction. So that is our j-unit vector. So now we have position as a function of theta which is actually a function of time. So let's take the derivative of this thing right here. So what is the derivative of our position vector with respect to time Well that's just going to be our velocity vector, as a function of time, and it's going to be equal to we just have to take the derivative of each of these parts with respect to time. And you just do the chain rule. So you're going to have the r sit outside cause that's just a constant. So you're going to have r the derivative of cosine of theta t with respect to theta t So I'm just doing the chain rule right over here. That's going to be negative sine of theta t and then as the chain rule, we also have to multiply that times the derivative of the theta of t with respect to t. So times d-theta, dt so this is just the chain rule right over here. So that's going to be how it's changing in the x-direction, and in the y-direction we do something very similar. In the y-direction we take the same derivative. We have the r scalar out front r, and then the derivative of sine of theta with respect to theta is going to be cosine of theta and I'll write it as a function of time, and then do the chain rule you'll also have to multiply that by the rate at which theta is changing with respect to t, times d-theta, dt, and this is all going to be times our j-unit vector. Now, there's something you might already realize, and you should rewatch the video on angular velocity if this is foreign to you, but d-theta, dt, this is our angular velocity. That's why I said to rewatch that video. This right over here, the rate at which the angle changes with respect to time, that is angular velocity. So this right over here is angular velocity. And for the sake of this video, this is an assumption we'll have to make for this formula right over here we're going to assume, that omega, which is the rate of change of our angle with respect to time, we're going to assume that this is constant. So this is an assumption we're making for this proof. This is we are going to assume that omega is constant. And if omega is constant, then we can treat it as a constant and we can factor it out of this expression. So let's factor out a negative omega-r from this expression over here. So we can rewrite our velocity as a function of time is equal to I'm going to factor out a negative omega times r, and if you factor out a negative omega-r, what you're left with is this first term, sine of theta-t And I didn't have to make it explicit that theta is a function of t, but this makes it explicit that theta is a function of t, and then times our i-unit vector, plus, so if we're factoring out a negative omega-r, this becomes negative cosine of theta, which is a function of t, and that is times our j-unit vector. So we factored out a negative omega-r. Now let's take the derivative of this with respect to time. So if we take the derivative of velocity with respect to time, this is clearly just what the acceleration is as a function of time, and we're going to assume that the magnitude of this thing is constant, but the actual direction is changing, so this is the acceleration as a function of time, is going to be equal to this negative omega-r, so what is the derivative of this thing right over here? So the derivative of sine with respect to theta, we're just doing the chain rule here, the derivative of sine with respect to theta is going to be cosine of theta as a function of t. And then chain rule, we also have to take that and multiply it with the derivative of theta with respect to t. I could write d-theta, dt here But that once again, is just omega. So that is just omega, and that of course is in the i-direction. And from that, and next to that, we take the derivative of cosine of theta of t with respect to theta, so that's going to be that would be negative sine of theta, so we would have a negative out front so it becomes positive sine of theta as a function of t. And then we have to do the chain rule, the derivative of theta with respect to t. We have to multiply by this, and for that we could write d-theta, dt right there, but that again is the same thing as omega. And all that is being multiplied by the j-unit vector. So now let's factor out this other omega, and we get something interesting, we get the acceleration vector as a function of time is equal to, and if we factor out another omega, we get negative omega-squared r, I'm just factoring out another negative omega, times, and I'll write it in parentheses here, cosine of theta as a function of t times our i-unit vector plus sine of theta, which is a function of t times our j-unit vector. Now what is all of this business right over here? Just look at this part right over here, well r times this, especially if you distributed the r, that is exactly this thing right over here If you distribute the r, you get exactly r cosine theta as a function of t times our i-unit vector plus our sine theta as a function of t times the j-unit vector. So everything that I squared-off in orange right over here, this is our position vector as a function of time. So all that work we did, we just got a very interesting result. We got that our acceleration vector as a function of time is equal to the negative of our constant angular velocity-squared times our position vector And just to be clear, angular velocity is kind of the pseudo vector, it tends to be treated like a scalar, especially when you're dealing with two-dimensionals like this, it's really a pseudo scalar, but let's just go with this. We're assuming this right over here is a constant scalar quantity. Now, we're very very very very close here. Now what we want to do is to relate this this is essentially the scalar version of it, so if we wanted to take the magnitudes of both sides, so we're saying the acceleration vector is equal to this constant times the position vector, so let's take the magnitude of both sides of this thing So then we get the magnitude of the acceleration vector, which I'm just going to call a sub c, is going to be equal to you could say the magnitude of this negative omega-squared but when you take the magnitude, it's like taking the absolute value in fact, absolute value is just the one-dimensional version of magnitude, that's just going to be positive omega-squared we don't care about the direction, sign gives us the direction, we just care about the actual size. So this is going to be the magnitude of negative omega-squared times the magnitude of our position vector the magnitude of omega-squared is just going to be omega-squared you can get rid of the sign, and the magnitude of our position vector we saw at the beginning of this video, is just r, our radius so this right over here is just going to be equal to the radius of the circle that we're going around. Now, we also know the angular velocity, or the magnitude of the angular velocity, is equal to the magnitude of our velocity, or the speed of our object, divided by the radius of the circle that it is going around. So we could substitute that right over here. So if we square it, this is going to be (v over r)-squared, now we saw that in the video on angular velocity, times r and this is all going to be the magnitude of our acceleration, which is really our centripetal acceleration, our inward directed acceleration. So this is going to be equal to, and I think you see where this is going, This is equal to v-squared over r-squared times r, but this r cancels out with the r-squared, so you're just left with v-squared over r, and you're done! The magnitude of the centripetal acceleration is equal to your speed, the magnitude of your velocity, squared, divided by your radius. And, we are done.