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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 5

Lesson 3: Rotational kinematics# Relating angular and regular motion variables

In this video David shows how to relate the angular displacement to the arc length, angular velocity to the speed, and angular acceleration to the tangential acceleration. Created by David SantoPietro.

## Want to join the conversation?

- At2:26, arc length can also be found using a degree.

so why do we use radians if we can find arc length by using

(Theta/360)*2pi*r? I mean why do we have to use radians?(13 votes)- radians have no units. This makes them much more convenient than degrees.(34 votes)

- How is arc length a distance and not a displacement?(11 votes)
- Displacement is the shortest distance from the initial position to the final position, while Distance is the actual path covered by a moving object. In this case, the tennis ball covers the arc as the actual path, therefore it is the distance. Displacement, in this case, is the line joining the initial position to the tennis ball. Vote up if it helped(40 votes)

- If tangential acceleration and centripetal acceleration are both components of the total acceleration, why tangential acceleration is not a vector whereas centripetal acceleration is a vector?(10 votes)
- You are correct, tangential acceleration is just as much a vector as centripetal acceleration. That's why at11:03in the video, he draws an arrow to represent it. Then, he goes on to say that you would add the tangential acceleration and the centripetal acceleration as vectors to get the total acceleration, just like we do with all vectors and their components. Did he say something somewhere that seemed to imply that tangential acceleration was not a vector?(11 votes)

- if v=r(omega) , then omega has units rad/s and r has units meters then what do we express v in?, (meter*rad)/s ?(4 votes)
- radians are unit-less because they just represent a ratio

omega has units of 1/s.

So v is m/s(5 votes)

- why cant we just call angular displacement 'angular distance' or angular velocity as 'angular speed'?(4 votes)
- Angular displacement is the displacement of an object from the starting point to a final point. If you are talking about the distance then it's all of distance that it took us to get to a final point from an initial point. This about this in this way; if you're in a circular road driving a car and you complete one lap i.e one revolution then your displacement would be 0 because you are at the same location from where you started but you distance traveled would be the circumference of that circular road track i.e the total distance that it took you to complete one lap.

For Angular velocity, it is the change in your angular displacement per time or in other words, how fast our angle is changing for every change in time(second), and since it is velocity, it must have a direction which are counter-clockwise or clockwise but the Angular speed is simply the magnitude of the angular velocity, not specifying the direction of the motion.(3 votes)

- why R(radius) was involved in every conversion, please?(3 votes)
- Because that's how you convert linear measures to angular.(4 votes)

- I am a bit confused about the usage of radians for angular measurements. Is pi(π) a unit of radians, or is it not? When converting different units, such as revolutions or degrees, to radians, should I solve for radians in terms of pi or terms of an integer?

For example:

[Question]: Convert 3 revolutions to radians.

To solve, one would have to do (3 rev.)(2π/1rev.)

So would the answer be 6π radians or 18.849 radians?(3 votes)- Radians are technically unitless since it is ratio of radius which is a distance to the circumference which is also a distance. This gives you meter/meter which cancel out leaving no units. π is not a unit it is just a constant of proportion.

How you report the answer depends on who you are reporting it to but in general an answer like 6π is more understandable that the decimal amount since using π you get a better idea of the proportion of the circle it is.(4 votes)

- If we're given numerical problem with degrees instead of radian, how do we convert it to radian?(2 votes)
- how do we relate 2d curviliner motion with angular variables(4 votes)
- Good question. What I understand there are three things: Centripetal Acceleration, Tangential Acceleration and Angular Acceleration. One can calculate tangential acceleration from angular acceleration. Then using tangential acceleration and Centripetal acceleration one can calculate Total acceleration....(2 votes)

- Is tangential acceleration equal to the angular acceleration?(3 votes)
- Tangential acceleration is not (numerically) equal to angular acceleration. If the motion is circular (radius is a constant) then they are related by the constant radius. alpha (angular acceleration) times the radius is equal to the tangential acceleration.(3 votes)

## Video transcript

- [Instructor] So in the previous video, we defined all the new
angular motion variables and we made an argument
that those are more useful in many cases to use than
the regular motion variables for things that are rotating in a circle. Since every point on the
string in tennis ball, let's say this is a tennis
ball you tied a string to and you're whirling around in a circle. Every point on the string
including the tennis ball will have the same angular displacement, angular velocity and angular acceleration. But even though using
angular motion variables is more convenient for these
rotational motion problems, it's also really important to know how to translate those
angular motion variables back into the regular motion variables. So that's what I wanna
show you in this video how to translate angular motion variables back into regular motion variables. So let's do this. The simplest possible
angular motion variable was the angular displacement because that just represented how much angle an object
has rotated through. So let's say it rotated through this much. We represented the angular
displacement with a delta theta and we call it the angular displacement. In Physics, we typically choose
to measure this in radians for a reason and I'll
show you in just a second. Now, how would we convert this into a regular motion variable? What regular motion
variable would that be? If I were to come at
this for the first time, I'd be like all right, this
is the angular displacement. Let's figure out how to relate it to the regular displacement but that would be weird. Because just think about it, the regular displacement for the ball that started over here
and made it over here would be from this point to that point, that would be the regular
displacement of the ball, the regular linear
displacement of the ball. That's a little weird. I don't wanna show you how to find that for one, you have to
use the law of cosines. That's a little more in depth than I'd wanna get to in this video. For two, the better reason this isn't all that useful in turns out. There's a much more useful quantity that would tell you how far the ball went. That's the arc length of the ball. So the ball traced out
of path through space around the circle. We call this the arc
length that it turns out this is much more useful
in a variety of problems. Good news is it's much easier to find than that regular displacement. So this is the arc length. People vary on what letter to use here. I've seen an l but most math books use s so we'll just use s as well. You might think this is
hard to find but it's not. In fact, if we use radians and
this is why we use radians, it's extremely easy to find. If we wanted to find the arc
length of this tennis ball, we're just gonna take the
radius of the circular path that tennis ball is tracing out. So in this case it'd be
the length of the string. We take that radius. If we're in radians, we just multiply by the
angular displacement. This is why the radians are so convenient. We just take that measurement in radians multiplied by the radius
of the circular path the object is tracing out. You get the arc length which
is the number of meters along this path that
the object has traveled. If that seems miraculous, it really isn't. I mean the reason why this works so well because this is how
the radian was defined. One radian is defined to be the angle through which you have to travel so that the arc length is equal to the radius of that circle. So this isn't a surprise. This was selected and
defined strategically so that we can use this unit and we get a really easy way to convert between the angular displacement, how many radians something
has rotated through, and how many meters it
has actually traveled through its arc. So this arc length is
gonna have units of meters as long as we measure
the radius in meters. All right, so that's one relationship between angular displacement, how much angles something
has rotated through, and how many meters it
has actually traveled. The next relationship I
wanna talk about relates the angular velocity to
the regular velocity. So remember in the previous video, we defined the angular velocity to be the angular displacement per times. So this is the rate at
which something is rotating through a certain amount of angle and the letter we use
to represent velocity was the Greek letter omega. So this angular velocity
represented the rate at which something in a circle. So it's rotating slowly. It's gonna have a small angular velocity. If it's rotating quickly, it's gonna have a large angular velocity. Obviously the speed and
the angular velocity are gonna be related because the higher the angular velocity,
the higher the speed. But, what is that relationship? How would we get from angular velocity to the regular velocity? Well it's not actually that hard at all because all we need to
do is turn this number of radians per second
into meters per second and I can do that. If multiply both sides of
this equation down here by R, I'll get R times omega is gonna equal R times delta theta and I still have to divide by delta t. So you just multiply both
sides of this equation by R. But look what I get. R times delta theta is
just the arc length. So this whole side over here is just how many meters that object has traveled around the edge of the circle divided by the time that it took. But that is just the speed. This arc length is just the
distance the object has traveled and the time is the time that it took and distance per time is just speed. So this is the speed of the object. I'm gonna write that as v even though it's not velocity. This is not a vector and it's not velocity because think about it, this arc length isn't displacement. This was the distance the object traveled. Distance per time is the speed. Displacement per time is the velocity. We didn't use displacement. Displacement was this weird one. We didn't wanna deal with that. So since we're choosing
to deal with arc length which is distance, what
we're gonna do is relate the angular velocity into the speed. Now we have that relationship. Look at this. This is R the radius
times the angular velocity equals the speed of the object. So this is the relationship between the angular
velocity and the speed. The speed of the object
is gonna equal the radius of the circular path the
object is traveling in times the angular velocity. I should box these. These are important. This arc length formula was how you relate the number of radians and
object has rotated through to how much arc length it's traveled, i.e. how much distance it's gone through. In this formula down here relates the angular velocity omega, the number of radians per second something has rotating with to how many meters per
second it's traveling. In other words, how many meters
per second it's tracing out along this arc length. So this is good. Now we know how to relate
the angular displacement to the distance the object has traveled and we know how to relate
the angular velocity to the speed of the object so you probably know what's coming next. We have to relate the angular acceleration to the regular acceleration. So we're called that
the angular acceleration which we represented
with a Greek letter alpha was defined to be the change in the angular velocity per times. It's the rate at which your
angular velocity was changing. So there's moving at the constant rate. You've got no angular acceleration because there's no change in omega. But if omega starts off slow and then it gets faster and faster, you do have angular acceleration. It's probably not a surprise that if you have angular acceleration, this ball is gonna have
regular acceleration too because it's speeding up
in its angular rotation. It's gonna be changing
its velocity as well. So how do we this? How do we relate the angular acceleration to the regular acceleration? Well the simplest thing to
try is we go work down here. We multiply both sides of
our equation by the radius and we found the relationship
the related speed to angular velocity. So let's try it again. Let's multiply both sides
of this equation by radius and see what we get. On the left hand side,
we can get the radius times the angular acceleration. That's gonna equal the
radius times the change in angular velocity
over the change in time. So all I've done here is multiply both sides of this equation. This definition of angular
acceleration by the radius. So let's see what we get
on the right hand side. We got R times delta omega. So this is really R times
the change in omega. Well that's just omega
final minus omega initial and then divide it by delta t so I can distribute this R and get that. This would equal R times omega final minus R times omega initial divided by the time that it took. But now look what happens. We've got R times omega final
and R times omega initial. We know what R times omega is. It's the speed, not the
velocity, but the speed. So I could rewrite this. I could say that this is
really the final speed minus the initial speed over the time that it took to change by that much speed. So this is, if I just
rewrite the left hand side, this is what R times alpha is equal to. Now if I were you, I'd be
tempted to just be like, oh look, we did it. That's the acceleration which
change in speed over time. But you gotta be careful. Acceleration, the true acceleration vector is the change in velocity per time, but these are not velocity vectors. These were speeds. So this isn't the true
acceleration vector. This is something different. This is the change in speed per time. So that's still an acceleration but it's not necessarily
the entire acceleration because there's two ways to accelerate. You can change your speed
or change your direction. Basically this acceleration we just found doesn't take into account any acceleration that's coming from
changing your direction. This is only the acceleration that's gonna be changing your speed. If I were you, I'd probably
be confused at this point. So let me try to show you what this means. If this ball is rotating in a circle just by the mere fact that the
ball is rotating in a circle, it has to be accelerating even if the ball isn't
speeding up or slowing down. There's got to be an acceleration because this ball is changing the direction of its velocity. These are gonna be a force. It's centripetal force, in this
case it would be the tension. There's gotta be a
centripetal acceleration that's changing the
direction of the velocity. That is not this acceleration over here. This is a different acceleration. We know the centripetal
acceleration is directed inward. We already know how to find
this centripetal acceleration. Remember the formula for
centripetal acceleration is the speed squared
divided by the radius. This component, this
centripetal acceleration is the component of the acceleration that changes the
direction of the velocity. So I'm gonna say that again
because this is important. The centripetal acceleration
which you can find with v squared over R is the
component of the acceleration that changes the
direction of the velocity. If something is going in a circle, it must have centripetal acceleration. But this acceleration component that we found down here is different. This is what's changing the speed. You don't have to have this
if you're going in a circle. You could imagine
something going in a circle at a constant rate. If that's happening, it's
got centripetal acceleration but it doesn't have this thing down here because this thing we found R times alpha is the change in the speed
of the object per time. How would I draw that up here if I wanted to represent this a that we found down here visually up here? I'd draw tangential to
the direction of motion, i.e. tangential to the circle because components of acceleration that are directed
perpendicular to the velocity change the direction of the velocity but components of
acceleration that are directed parallel to the direction
of the velocity change the magnitude of the
velocity, i.e. the speed, to change the magnitude of the velocity, in other words to speed
something up or to slow it down. You need a component of that acceleration that's either in the directional motion or opposite of the directional motion. If it was opposite the directional motion, the acceleration would be
slowing the object down. If the component of
acceleration is in the direction of motion, then it's
speeding the object up. That's what we found down here. That's what this component
of acceleration is, R alpha which is why it's often called the tangential acceleration. So I'm gonna write that up here. The tangential acceleration
which is equal to R times alpha, the radius times the angular acceleration is the component of the acceleration that's changing the
magnitude of the velocity, i.e. it's changing the speed. In order to do that,
it's gotta be directed tangential to the direction of motion. That's what this R times alpha represents. So let me box that. That's important. This is the formula to find
the tangential acceleration. It doesn't give you
the total acceleration. We know that there's always
a component of acceleration that's acting centripetally if an object is going in a circle and you could that with v squared over R. But if the object going in
a circle is also speeding up not only going in a circle
but changing its speed, it's gonna also have this
component of acceleration which is the tangential acceleration. So at this point you might be confused like wait, we've got
tangential acceleration, we've got centripetal. Which one is the acceleration? Well, they're both just components
of the total acceleration which you could find if
you really wanted to. You could say that the
total acceleration squared. You could use the Pythagorean theorem because these are the two
perpendicular components of the total acceleration. We said the total acceleration squared would equal the tangential
acceleration squared plus the centripetal acceleration squared. What we would be finding is
the total acceleration squared which if you wanted a direction, the direction of that total acceleration would point this way somewhere. If you had centripetal acceleration inward and let's say the object was speeding up. So let's say it wasn't slowing down. So you didn't have this component. You've got this forwards
component and inwards component. The total acceleration would
be directed here somewhere. Since you could form a
right triangle out of this with these two sides,
you can imagine moving this centripetal acceleration
over to this side and you could find the hypothenuse which would be the total acceleration by just taking the tangential
acceleration squared plus the centripetal acceleration squared and then taking a square
root would give you the magnitude of the total acceleration. So recapping, there's two
components of acceleration, the tangential acceleration
which is R times alpha, either speeds an object
up or slows it down, the centripetal
acceleration works to change the direction of the motion of the object. You can relate the speed of an object to the angular velocity
by multiplying by R. You could relate the arc
length, i.e. the distance the object traveled around
this edge of the circle to the angular displacement
by also multiplying by R. So these three equations or how you relate the angular motion variable
to its linear counterpart.