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### Course: High school physics > Unit 4

Lesson 1: Uniform circular motion introduction- Angular motion variables
- Distance or arc length from angular displacement
- Angular velocity and speed
- Connecting period and frequency to angular velocity
- Radius comparison from velocity and angular velocity: Worked example
- Linear velocity comparison from radius and angular velocity: Worked example
- Change in period and frequency from change in angular velocity: Worked examples
- Circular motion basics: Angular velocity, period, and frequency

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# Distance or arc length from angular displacement

Relating angular displacement to distance traveled (or arc length) for a ball traveling in a circle. Derivation of formula for arc length.

## Want to join the conversation?

- shouldn't the angular displacement in the first example be zero?(11 votes)
- OK so don't confuse angular displacement with regular displacement, that is why Sal told in the video that displacement is zero, but the angular displacement is 2pi radians and for two complete revolutions it is 4pi radians. I would suggest you to watch the video on Angular motion again to clear your doubts.(15 votes)

- At7:44the last problem why did you subtract pi/3-pi/6 if it went in a positive direction? Shouldn't it be pi/3+pi/6= pi/2 (work: 2pi/6+pi/6= 3pi/6= pi/2).(2 votes)
- One thing that Sal mentions in the early part of the video is that pi/3 and pi/6 are positions in units of radians.

When he tries to find the**angular**displacement, he tries to find how much the ball was**rotated in total**. The ball went from pi/6 radians to pi/3 radians, so Sal took pi/3-pi/6 radians to be the total.*To help visualize why, try this:*

Imagine trying to find the displacement if a man moved from 4 on the number line to 8 on the number line. The answer is obviously**8-4=4**.**Now let us try to solve the original problem**. Remember that with angular displacement, counterclockwise is positive and clockwise is negative (just like right is positive and left is negative in the example above). The final position is pi/3. The initial position is pi/6. So, the angular displacement is pi/3-pi/6.*Key things to remember:*

The formula for angular displacement is**not**P_f - P_s, where P_f is final position and P_s is starting position. That is because angular displacement is**how**much has the**angle**moved. So, the angular displacement of completing a full rotation counterclockwise (2 pi radians) may appear to be 0. After all, the position of the ball didn't change, right? But we are**not**looking for the change in the ball. We are looking for the change in**angle**. So the answer for the displacement of a complete rotation (2 pi radians) is 2 pi radians!(8 votes)

- So take for a example a wheel is rotated through angles of 30degrees, 30rads and 30rev, respectively which one will be my angular displacement when im looking for path length and my radius is 4.1m?(4 votes)
- At1:00how did you get Δθf=5π/2?(3 votes)
- The original position was π/2, then there is a rotation of 2π, so by adding those up, you get the final position of 5π/2.(2 votes)

- Why is theta intial pi/2?(1 vote)
- To know this, a background in algebra 2 is helpful (because that is when most people learn about the unit circle). Here is the explination:

Radians (which are used here) are like degrees, where they measure angles. In terms of circles, like are in this problem, 360 degrees is all the way around a circle once, 180 degrees is half way around the circle, and 90 degrees is a quarter around the circle. Radians are similar, however, they are measured differently. 2π is all the way around a circle, π is half way around a circle, and 1/2π (or π/2) is a quarter around the circle. So basically:

-360 degrees=2π radians

-180 degrees=π radians

-90 degrees=π/2 radians

In circular motion problems like this, here's another thing to keep in mind: the point of 0 degrees (0 radians), AKA the starting point, is the positive x axis. So when the ball is at its initial point (θ initial), it has rotated a quarter around the circle (since positive rotation is counter-clockwise). Since a quarter around the circle is π/2 radians, theta initial is π/2 radians.

P.S.- radians will is a very conventional unit in math and physics at high school/college levels. Here are some lessons explaining radians and the unit circle if you want to learn more in depth:

Unit circle-https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1

Radians- https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:radians/v/introduction-to-radians

I hope that this helps!(2 votes)

- At2:40, Sal says that the units for S are meters.

I thought that Δθ was measured in radians, so why are the units for S not (meters)(radians)?(1 vote)- One radian is the length of the radius of the same circle. So the length of 2π radians is really just the length of 2π radiuses. Therefore, if your radius was 6 meters, multiplying 2π radians with the radius would give you 12π meters.(1 vote)

- how does pi over 3 minus pi over 6 equal pi over 6? Wouldn't that equal like, 2pi minus pi over 6?(1 vote)
- I have two questions:

1. For the first diagram, why is the angular displacement 2pi? Shouldn’t it be 0 by the formula?

2. For the second diagram, if we go the other way(counter clockwise) 3pi/2 angular displacement then we would get an angular displacement of -pi/2 by the formula, therefore we would get the same result as in the second diagram by the formula, although the distance is about 3 times the one on the second diagram, how does this make sense?(1 vote) - 1:05

why is the pie final = 5 pie / 2 ?(1 vote) - displacement is the change in r vector so why we are more interested in the length(1 vote)

## Video transcript

- [Instructor] What we're
going to do in this video is try to draw connections
between angular displacement and notions of arc length
or distance traveled. Right over here, let's
imagine I have some type of a tennis ball or something, and it is tethered with a
rope to some type of a nail. If you were to try to
move this tennis ball, it would just rotate around that nail. It would go along this blue circular path. Let's just say for the sake of argument the radius of this blue
circle right over here, let's say it is six meters. You could view that as
the length of the string. We know what's going on here. Our initial angle, theta
initial, the convention is to measure it relative
to the positive x-axis. Our theta initial is pi over two. Let's say we were to then
rotate it by two pi radians, positive two pi radians. We would rotate in the
counterclockwise direction two pi. Then the ball would end up
where it started before. Theta final would just
be this plus two pi, so that would be five pi over two. Of course, we just said that we rotated in the positive
counterclockwise direction. We said we did it by two pi radians, so I kind of gave you the answer of what the angular displacement is. The angular displacement
in this situation, delta theta, is going to
be equal to two pi radians. The next question I'm going to ask you is what is the distance the
ball would have traveled? Remember, distance where we
actually care about the path. The distance traveled would be essentially the circumference of this circle. Think about what that is, and actually while you pause
the video and think about what the distance the ball traveled is, also think about what
would be the displacement that that ball travels. Well the easier answer, I'm
assuming you've had a go at it, is the displacement. The ball ends up where it
starts, so the displacement in this situation is as. We're not talking about
angular displacement. We're just talking about regular
displacement would be zero. The angular displacement
was two pi radians, but what about the distance. We'll denote the distance by S. As we can see, as we'll see, we can also view that as arc length. Here the arc is the entire circle. What is that going to be equal to? Well, we know from
earlier geometry classes that this is just going to be the circumference of the circle which is going to be equal
to two pi times the radius. It's going to be equal to
two pi times six meters which in this case is going to be 12 pi. Our units, in this case,
is going to be 12 pi meters would be the distance that we've traveled. Now what's interesting right over here is at least for this particular
case to figure out the distance traveled to
figure out that arc length, it looks like what we did is we took our angular displacement, in fact we took the magnitude
of our angular displacement, you could just view that as
the absolute value of it, and we multiplied it times
the radius of our circle. If you view that as the
length of that string as R, we just multiply that times R. We said our arc length,
in this case, was equal to the magnitude of our
change in displacement times our radius. Let's see if that is always true. In this situation, I have
a ball, and let's say this is a shorter string. Let's say this string
is only three meters, and its initial angle, theta
initial, is pi radians. We see that as measured
from the positive x-axis. Let's say we were to rotate it clockwise. Let's say our theta final
is pi over two radians. Theta final is equal
to pi over two radians. Pause this video, and see if you can figure out the angular displacement. The angular displacement
in this situation is going to be equal to theta
final which is pi over two minus theta initial which
is pi which is going to be equal to negative pi over two. Does this make sense? Yes, because we went clockwise. Clockwise rotations by convention
are going to be negative. That makes sense. We have a clockwise rotation
of pi over two radians. Now based on the information
we've just figured out, see if you can figure out the arc length or the distance that this tennis ball at then end of the
string actually travels. This tennis ball at the end
of the three meter string. What is this distance
actually going to be? Well, there's a couple of
ways you could think about it. You could say, hey
look, this is one fourth of the circumference of the circle. You could just say,
hey, this arc length, S, is just gonna be one fourth times two pi times the radius of three
meters, times three. This indeed would give
you the right answer. This would be, what, two
over four is one half, so you get three pi over two,
and we're dealing with meters, so this would be three pi over two. But let's see if this is
consistent with this formula we just had over here. Is this the same thing if we were to take the absolute value of our displacement. Let's do that. If we were to take the
absolute value of our, I'd should say our angular displacement, so we take the absolute value
of angular displacement, and we were to multiply it by our radius. Well, our radius is three meters. These would indeed be equal because this is just going to be positive
pi over two times three which is indeed three pi over two. It looks like this formula
is holding up pretty well. It makes sense because
what you're really doing is you're saying, look, you have your traditional
circumference of a circle. Then, you're thinking
about, well, what proportion of the circle is this arc length? If you were to say, well,
the proportion is going to be the magnitude of your
angular displacement. This is the proportion of the
circumference of the circle that you're going over. If this was two pi, you'd
be the entire circle. If it's pi, then you're
going half of the circle. Notice, these two things cancel out. This would actually give
you your arc length. Let's do one more example. Let's say in this situation, the string that is tethering our
ball is five meters long, and let's say our initial
angle is pi over six radians right over here. Let's say that our final angle is, we end up right over there. Our theta final is equal to pi over three. Based on everything
we've just talked about, what is going to be the
distance that the ball travels? What is going to be the arc length? Well, we could, first, figure
out our angular displacement. Our angular displacement is
going to be equal to theta final which is pi over three minus theta initial which is pi over six which is going to be equal to pi over six,
and that makes sense. This angle right over
here, we just went through a positive pi over six radian rotation. We went in the counterclockwise direction. Then, we just want to
figure out the arc length. We just multiply that times the radius. Our arc length is going
to be equal to pi over six times our radius times five meters. This would get us to
five pi over six meters, and we are done. Once again, no magic here. It comes straight out of the idea of the circumference of a circle.