Good morning. We have already learned about the equation for centripetal acceleration
or the center-seeking acceleration acting on an object when it is moving along a circular path.
The magnitude of centripetal acceleration equals the magnitude of tangential velocity squared
over radius which also equals radius times the square of the magnitude of angular velocity.
We have even learned about the centripetal force or the net force acting in the in-direction
on an object when it is moving along a circular path. And we already know that,
according to Newton’s Second Law, the net force in the in-direction, or the centripetal force,
equals mass times the acceleration in the in-direction, or the centripetal acceleration.
What we have not learned, is where these equations come from. So, today, we derive the equation for
centripetal acceleration. In order to do that, we start with the equation for average linear
acceleration. Bobby, what is that equation? ♪ Flipping Physics ♪
Average linear acceleration equals change in linear velocity over change in time where both
acceleration and velocity are vectors. Why are we starting with average linear
acceleration? I thought we were trying to derive the equation for centripetal acceleration
and that is the acceleration of an object which is moving in a circular path, so why
are we not starting with angular acceleration? Billy, recall that centripetal acceleration is a
linear acceleration. It is the linear acceleration of an object which is moving along a circular
path, and it always acts inward, toward the center of the circle the object is moving through. So,
because centripetal acceleration is a linear acceleration, we start with the equation for
average linear acceleration. Now let’s look at a small yellow poker chip on a record where both
the chip and the record are moving with the same constant angular velocity. Bo, please determine
the angular acceleration of the yellow poker chip moving at a constant angular velocity.
(Wait, now you want the angular acceleration?) Yes Billy, please just hear me out.
Angular acceleration equals change in angular velocity over change in time.
Well, because the yellow chip is moving at a constant angular velocity,
the change in angular velocity equals zero, and its angular acceleration is also zero.
So, if the angular acceleration of the chip is zero, how would the chip
have a linear centripetal acceleration? That is a great question Billy. Let’s
add the vector which shows the linear velocity of the chip as it moves along the circle. Class,
what do we call this velocity? Tangential velocity.
Because it is always tangent to the circle described by the chip.
Correct. Now, Billy, is the tangential velocity of the chip constant?
Well, the chip is moving with a constant angular velocity, and tangential velocity equals radius
times angular velocity, so the tangential velocity of the chip is also constant?
Actually, no. The magnitude of the tangential velocity of the chip is constant,
however, the direction of the tangential velocity of the chip is always changing.
That means the tangential velocity of the chip is not constant, only the magnitude of
the chip’s tangential velocity is constant. That equation is actually the magnitude of tangential
velocity equals radius times the magnitude of angular velocity, so, technically, it is not
tangential velocity but rather tangential speed. Oh right. I get it. The length of the tangential
velocity arrow stays the same, however, the direction the velocity vector is pointing
changes as the chip moves around the circle. And, going back to the equation for average linear
acceleration, we know linear acceleration is caused by a change in linear velocity. Centripetal
acceleration must be the result of the change in the direction of the tangential velocity.
Okay. So, even though the magnitude of the tangential velocity is staying constant,
because the direction of the tangential velocity is changing, there must be a linear acceleration
and that linear acceleration is the centripetal acceleration.
And, if the acceleration of the chip were zero, the chip would not be moving in a circle,
it would be moving in a straight line. Right.
Correct everybody. That is the general theory. Now let’s go through and do the actual derivation.
To do that let’s change our illustration to show just the radius of the chip as it
moves along the circle. And we want to look at what happens when the object has gone through a
small change in angular position, delta theta. So, let’s identify this as our initial position
and this as our final position. Again, those are position initial and position
final for the chip after it has gone through a change in angular position, delta theta.
Now let’s add the displacement vector, delta r, to create a tip-to-tail vector diagram.
Now, it is subtle, but important, you realize what is shown here in this tip-to-tail vector
addition diagram is position initial plus change in position equals position final.
However, we can rearrange that equation to get change in position equals position final
minus position initial, so, this displacement vector, delta r, is, in fact, the change in
linear position of the chip when it goes through a change in angular position, delta theta.
(mr.p?) Yes Bobby?
That is an isosceles triangle with an angle delta theta and two sides of equal magnitude,
r initial and r final, right mr.p? (Correct Bobby, but
how do you know position initial and position final are equal in magnitude?)
Because the radius of the circle the chip is going through is constant. So,
the magnitudes of r initial and r final are just r, the radius of the circle.
Very nice Bobby. Now, let’s add the tangential velocity of the chip back to our illustration.
And, because this tangential velocity is always tangent to the circle,
it is also always at a 90-degree angle to the radius. So, let’s add the initial velocity and
the final velocity which correspond to the initial and final positions
when the chip goes through the same small change in angular position delta theta.
Because both of these velocities are at right angles to r initial and r final,
we know the angle between velocity initial and velocity final is the same delta theta.
And because change in velocity equals velocity final minus velocity initial, velocity initial
plus change in velocity equals velocity final and we can create a similar isosceles triangle with
the same change in theta and two equal magnitude sides velocity initial and velocity final.
How do we know those two tangential velocities are equal in magnitude?
We showed it before. It’s because the angular velocity is constant and tangential velocity
equals radius times angular velocity. Actually, it’s tangential speed equals
radius times angular speed, right? Oh yeah, that’s right. Or the magnitude
of tangential velocity equals radius times the magnitude of angular velocity.
Right. That means we can label the magnitudes of the initial and final velocities as
tangential speed, but let’s still label it as v sub t just to be consistent.
Okay. Now, notice that the direction of the change in velocity of the chip is directed
inward towards the center of the circle. No matter where we had chosen our initial and
final positions to be, the change in tangential velocity of the chip is directed inward towards
the center of the circle. In other words, when an object is moving at a constant angular velocity,
the change in linear velocity of the object is always directed towards the center of
the circle. And, because linear acceleration equals change in velocity over change in time,
this is also the direction of the center seeking, centripetal acceleration of the object; always
in, towards the center of the circle! Nice. Even though the magnitude of the
tangential velocity is not changing, because the direction of the tangential velocity is changing,
that change in velocity is always directed in towards the center of the circle and is the
direction of the centripetal acceleration. Exactly Bo. Because both of these isosceles
triangles have the same angle delta theta, they are similar triangles.
(They fit one of the conditions for similar triangles, two pairs of corresponding sides
are proportional and the corresponding angles between those two pairs of sides are equal.)
Right. Because these are similar triangles, we know change in velocity over tangential speed
from either of the two sides of the triangle equals, looking at the other triangle, change in
r position over the radius from either of the two sides of the triangle. And we can solve for the
change in velocity. Notice that I have removed the vector symbols because we are only comparing the
magnitudes of these vectors which are represented by the lengths of the sides of these triangles.
And now we can go back to the original equation for linear acceleration and substitute the
quantity, change in position times tangential speed over radius, in for change in velocity.
And we get that the magnitude of the linear acceleration equals tangential speed over radius
times change in r position over change in time which is the magnitude of the average velocity.
And we can take the limit of the magnitude of the average velocity as change in time
decreases down to an infinitesimally small change in time. This is the definition of a derivative,
and we get that, if we consider the change in time to be infinitesimally small, the magnitude
of the average velocity becomes the magnitude of the instantaneous velocity of the chip
which is the magnitude of the tangential velocity of the chip. So, we can now replace change in
velocity over change in time with the magnitude of the tangential velocity and we get the magnitude
of centripetal acceleration equals the magnitude of tangential velocity squared over radius,
the equation for centripetal acceleration. Cool. Yeah. Very nice.
Again, for an object moving at a constant angular velocity, the magnitude of the tangential velocity
remains constant, however, the direction of the tangential velocity is changing,
and it is the change in the direction of the tangential velocity as a function of time which
is the acceleration in the in-direction and is called centripetal acceleration, which we just
derived. Thank you very much for learning with me today, I enjoy learning with you.