Motion along a circle Each of the following equations in parts (a) (e) describes the motion of a

Motion along a circle Each of the following equations in...

Key Concepts

Parametric Equations of a Circle

This concept involves representing a circle using parametric equations, typically through functions such as cosine and sine to describe the x? and y-coordinates. Such representations simplify the study of motion along the circle by expressing both spatial coordinates as functions of a single parameter—often time—thereby facilitating the analysis of the particle's trajectory and dynamics.

Uniform and Nonuniform Circular Motion

Uniform circular motion refers to motion along a circular path with a constant speed, meaning the angular velocity remains constant over time. In contrast, nonuniform circular motion involves changes in speed (or angular velocity), even though the path remains a circle. Understanding whether the motion is uniform or not is essential for determining subsequent relationships between velocity and acceleration.

Velocity in Parametric Motion

The velocity vector of a particle moving along a parametric path is obtained by differentiating its position vector with respect to time. Its magnitude represents the instantaneous speed of the particle. This concept is key in analyzing whether the particle moves at constant speed (uniform motion) or experiences changes in speed (nonuniform motion).

Acceleration and Its Relationship to Velocity

Acceleration is the derivative of the velocity vector and characterizes how the velocity changes with time. In uniform circular motion, the acceleration is centripetal—directed towards the center of the circle—and therefore always orthogonal to the velocity vector. However, if the speed changes over time, there is an additional tangential component to the acceleration, which is not perpendicular to the velocity.

Direction of Rotation (Clockwise vs. Counterclockwise Motion)

The orientation of a particle's motion around a circle—whether it is clockwise or counterclockwise—is determined by the structure and signs of the parametric equations. By analyzing the order of the sine and cosine functions and their coefficients, one can deduce the direction of rotation as the parameter increases.

Initial Conditions in Parametric Trajectories

Initial conditions specify the starting position of the particle at a given time (commonly at t = 0), which significantly impacts the subsequent trajectory and dynamics of the motion. Even if different parameterizations yield the same path (for example, the same circle), the initial position determines the point from which the motion commences and can influence aspects like timing and direction with respect to that starting point.

Transcript

00:01 And this problem, we're going to be looking at different parameterizations that characterize a particle moving on the curve x squared plus y squared equals one.

00:12 So make more sense as we go through the first example.

00:15 So first we have r of t is equal to cosine of t plus sine of t, j.

00:27 And so the questions we want to answer are, does the particle have a constant speed? if so, what is its constant speed? is the particle's acceleration vector always orthogonal to its velocity vector? does the particle move clockwise or counterclockwise? and then lastly, does the particle go through the point zero one? so those are the things we want to answer.

00:52 So we'll go through it in this first one and it'll make more sense as we go.

00:56 So things we're going to need, we're going to want the velocity and the acceleration, because those are part of the question.

01:02 So for v of t, we take the derivative of rt.

01:07 So we'll have negative sine t .i plus cosine t .j.

01:14 And then for acceleration, we take the derivative of velocity with respect to time.

01:19 So we'll have negative cosine of ti minus sine of t .j.

01:29 Next.

01:30 What else are you going to need? so let's look at the speed.

01:34 How do we know if the first of all, what is the speed? we have v of t here, but this is in a vector.

01:44 It looks like it's changing with respect to time.

01:46 It is because we have a non -constant acceleration.

01:49 But so speed is defined as the magnitude of velocity.

01:54 So that is going to be, so we're going to sum the squares of the components.

02:01 So we'll have negative sine t squared plus cosine t squared, square root.

02:12 Right.

02:12 So i'll have the square root of sine squared t plus cosine squared t.

02:20 That by trigger entity is equal to 1.

02:24 And that's equal to 1.

02:25 And that is constant.

02:27 But we check this derivative with respect to t, we would get 0.

02:30 So it has a constant speed of 1.

02:34 Next, let's see, we want to know if the acceleration vector is always orthogonal to its velocity vector.

02:46 So how do we know two vectors are orthogonal? well, if we have a, vector a and vector b defined, let's write that a little better.

02:57 We have a is equal to a1i plus a2i, j, and vector b is equal to b1i, plus b2j, the, we know these vectors are orthogonal if and only if the dot product, a.

03:19 Dot b equal to a1b1 plus a2b2 is equal to zero.

03:28 So now i want to see if v .a .a is equal to zero.

03:35 So we have in the i component, we're going to have negative sine of t times negative cosine of t.

03:41 So that'll be sine of t, cosine of t.

03:46 And in the j component, we have negative sine of t times cosine of t.

03:50 So it'll be minus sine of t, cosine, of t.

03:54 And these are the same, but with opposite sign, so that equals zero.

03:58 So yes, that is orthogonal.

04:01 And then lastly, oh, no, we want to know if the particle is moving clockwise or counterclockwise.

04:07 So a really easy way to do this is we will plot the vector v of 0.

04:15 On our graph right here.

04:19 So v .0, we have a sine of zero in the i component.

04:23 So that's going to be zero i, and we have one in the j component.

04:28 So one j.

04:30 That looks like that.

04:33 That looks like counterclockwise to me.

04:36 So last we have, does the particle begin at the point one zero? so all we need to do to find that is take r of zero, which is going to be one, 1i and 0j plus 0 j and if we write that as a point we'll get 1 comma 0.

05:02 Okay, so that was the first parameterization.

05:07 So next we have r of t is equal to cosine of 2 t plus sine of 2 t.

05:23 So again we want to find v of t, a of t.

05:29 So v of t, that'll be 2, negative 2 sine of 2ti plus 2 cosine of 2tj.

05:40 And then for a of t, we'll have negative 4 cosine of 2ti minus 4 sine of 2tj.

05:51 So what were our questions again? does it have a constant speed? so we need to find the magnitude of v of t.

05:58 So that's going to be square root of negative 2 sine of 2t quantity squared plus 2 cosine of 2t quantity squared.

06:14 That's equal to the square root of 4 sine squared t plus 4 cosine squared t.

06:27 Don't know why i put parentheses in there.

06:29 That's equal to square root of 4, which is equal to 2.

06:32 So it has a constant of two units per unit time.

06:36 Well, two unit distance per unit time.

06:38 Okay.

06:39 So next, we want to know if the acceleration vector is orthogonal to its velocity vector.

06:46 So we have b .a.

06:49 So that's going to get us positive 8, sine of 2t, cosine of 2t, and minus 8 cosine of 2t, sine of 2t.

07:06 These are the same terms, but with an opposite sign, so that's zero.

07:12 So that means that the velocity is orthogonal to the acceleration.

07:17 All right, next.

07:20 Does the particle move clockwise or counterclockwise? so we have, do this in blue again, because it looks nice.

07:27 So we have a v of zero is going to be zero i plus 2j.

07:35 Now i want to plot that on my unit circle.

07:39 There's my very sloppy looking circle.

07:43 And that vector is going to be right here.

07:49 So yes, that is moving counterclockwise.

07:53 All right.

07:53 And we want to know what is r of zero.

07:59 And that's going to be equal to 0.

08:02 It's going to be equal to 1i plus 0 .j.

08:06 So that does go through 0 .1.

08:08 Well, it does begin at the point 0 .1.

08:11 Okay.

08:12 So next, we have the parameterization.

08:15 R of t equal to, what do we have here? cosine of t minus pi over two, i plus sine of t minus pi over two, j.

08:43 So v of t is going to be, so if you notice, the derivative of the inside is always going to be one, so we don't have to do a chain rule each time.

08:51 So i'll have negative sine of t minus pi over two i plus cosine of t minus pi over two j.

09:03 And for a of t, we'll have negative cosine of t minus pi minus pi minus pi minus pi over two.

09:18 All right, we want to find the magnitude of v of t.

09:27 That's going to equal square root of, and we'll have a sine squared of t minus pi over two plus cosine squared of t minus pi over two.

09:41 The angles inside of these trig functions are the same.

09:46 So we use the trigger identity, so that's equal to root one, which is equal to one.

09:49 So that's constant speed of one.

09:53 V .a, that's going to be, so sign of t minus, pi over 2, cosine of t minus pi over 2, minus sine of t minus pi over 2, cosine of t minus pi over 2.

10:15 That's equal to 0, so they're orthogonal.

10:18 So now we want r of 0, which is going to be, so we'll have cosine of negative pi over 2.

10:29 To i plus sign of negative pi over to j that's going to be zero i plus one j and we need v of zero v of zero which is so we're going to have okay so this should be negative sorry about that so now we're going to have negative negative one so that's one i plus zero j okay okay okay, so we want to plot that on the unit circle.

11:10 It's my beautiful circle.

11:12 Don't make fun of it.

11:14 Don't laugh.

11:14 Okay, so r of 0 is going to be this point here.

11:20 And the vector is going to look like this.

11:25 There's my velocity.

11:27 Okay, so it's going counterclockwise, and it does not start at 1 -0.

11:34 So those answer those two questions.

11:37 Right.

11:38 And lastly, we have the weirdest of them all.

11:45 Let me make sure.

11:52 Looks like there's a typo.