Problem 1C-03*: Simple circular motion

A particle moves in a flat plane, and its rectangular coordinates are described as

```
```

(a) What kinds of trajectory does this motion generate?

(b) Calculate the velocity vector and the acceleration vector, and then describe their geometric orientations.

(c) Calculate the angular velocity and then describe the motion.

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Solutions:

(a) Let the position vector of the particle be denoted as r(t) = (x(t), y(t)). From Eqs.(1) we have

```
```
This implies that the trajectory is a circle with the center located at the origin of the coordinate system.

(b) Let the velocity vector and the acceleration vector be denoted as v(t) = (vx(t), vy(t)) and a(t) = (ax(t), ay(t)) respectively. From the definition of the velocity vector, we get

```
```
From Eqs.(1) the rectangular coordinate representation of the position vector of the particle can be written as

r(t) = (x(t), y(t)) = (Acos(ωt), Asin(ωt)).               (4)

The scalar product r(t)·v(t) can be calculated from Eqs.(4) and (3) as
```
```
This implies that the position vector and the velocity vector are perpendicular as shown in the figure.

The acceleration vector a(t) = (ax(t), ay(t)) is calculated according to the definition as

```
```
Then
```
```
and
```
Eq.(6) says the v(t) and a(t) are perpendicular, and Eq.(7) says
that a(t) and r(t) are antiparallel. Thus the geometric
orientation of the acceleration vector is shown as the green arrow in the above
figure.

(c) Consider the time interval from t to t+Δt. The particle travels from
point P to point Q along the circular trajectory with a speed |v(t)|
as shown in the second figure at the right. The length of the arc PQ is approximately
|v(t)|·Δt. Let the change of the angle during this
time interval Δt be denoted as Δθ, that is, ∠QOP = Δθ as
shown in the figure.

Then

where ε is +1 if the motion is counter clockwise, and -1 if the motion is
clockwise. The angular velocity is defined as

Substituting Eqs.(2) and (3) into the right-hand side of Eq.(9), we have

We note that θ(t) = ωt. If ω › 0, θ(t) increases as
t increases, thus the motion is counter
clockwisw and ε = +1. If ω ‹ 0, θ(t) decreases as t
increases, so the motion is clockwise and ε = -1. We can combine these
understandings into a simple form ε = ω/|ω|, and finally
we have

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