Calculus proof of centripetal acceleration formula Proving that a = v^2/r
Calculus proof of centripetal acceleration formula
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- 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
- 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
- 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
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