# Does the particle have constant speed? If so, what is it constant speed?

1. Jun 25, 2011

### Easy_as_Pi

1. The problem statement, all variables and given/known data
Each of the following equations in parts (a)-(e) describes the motion of a particle having the same path, namely the unit circle x2 + y2= 1. Although the path of each particle in part (a)-(e) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions.

i. Does the particle have constant speed? If so, what is its constant speed?
ii. Is the particle's acceleration vector always orthogonal to it velocity vector?
iii. Does the particle move clockwise or counterclockwise around the circle?
iv. Does the particle begin at the point (1,0)?

a. r(t) = cos(t)i + sin(t)j, t$\geq$0

2. Relevant equations
v = dr/dt
a = d2r/dt
s=|v|

3. The attempt at a solution
Well, I could scan the page of my textbook that this problem is on, but I'll just say it instead: there is not a single example, be it in the text or the practice problems, that deals with anything like this. So, the only attempt I could make was to find the first derivative of r(t), which represents velocity, and then take the second derivative of r(t), which represents acceleration. r'(t) = -sin(t)i + cos(t)j, r"(t) = -cos(t)i + -sin(t)j. I know that, for speed to be constant, velocity must be constant, which means acceleration must equal zero. I have no idea how to show or determine if acceleration equals zero for every value t$\geq$ 0. My book says the answer is yes, it does have a constant speed equal to 1. However, acceleration does not remain constantly equal to 0. If t = pi, accel is -1, if t = pi/4, accel is -sqrt2. I know I must be something wrong here, but I'm very lost.

2. Jun 25, 2011

well, I see only part (a) of your question.
i)because speed is a vector and when your particle is moving on a circle it keeps changing direction the speed vector is not constant therefore there is an acceleration that exerts a centripetal force on the particle. but the "length" of the vector stays constant everywhere and is equal to 1 m/s (in SI).
ii) Yes. because in a circular motion the centripetal acceleration is always toward the center and is parallel to the radius and by a theorem in geometry you can show that the radius of a circle is perpendicular to the tangent line that touches the circle at that point.
iii) in part a, it goes counterclockwise. in this case It's very simple to tell. You can tell that by drawing a figure for yourself and give two different values to it and see what direction It's going. in more complicated cases You'll need to study the derivatives as well.
iv) t = 0 yields r(0) = cos(0)i + sin(0)j = (1,0). so the answer is Yes.

I know I'm not supposed to give you the answers, but I just gave you the answers for the part a and I couldn't do any better to guide you but not give you the answers. Try to solve the other parts yourself.

Last edited: Jun 25, 2011
3. Jun 25, 2011

### Easy_as_Pi

I have all the answers in the back of the book. So, getting them isn't the hard part. It's solving the problems that is. I only posted part a, because I like to do this on my own. I figured if someone could explain how to solve for the answers to part a, I could apply that work to b-e. I understand part iv's answer, but not the rest. How would I, with calculus, because it's for my calc 3 class, prove the other 3? My book said speed is constant, but you say it's not. Do you mean the velocity vector is not constant due to the change in direction?

4. Jun 25, 2011

My bad. speed is the length of the velocity vector. I thought of the terminology the other way. I thought that velocity is the length of the speed vector. anyway, the velocity vector is NOT constant, because in a circular motion the direction of the velocity is changed, but the length of the velocity vector which is the speed remains constant at every point and you can easily show that by calculating the length of the velocity vector.
about the 2nd part, well, Do you know anything about the dot product of two vectors? I explained to you why the acceleration vector is perpendicular to the velocity vector using high-school geometry. It's because the velocity is tangent to the path, and the path is on a circle. According to a theorem in geometry, the radius of a circle is always perpendicular to the tangent line passing through some point a which is in the intersection of the radius and the circle.
the 3rd part is a little bit hard. In this case It's simple to show that the movement of the particle is counter-clockwise. take two arbitrary numbers like a and b such that a>b. plug them in and draw a picture for yourself. you'll see how the particle is moving. but if the particle changes direction, then this method will fail. then you'll have to study the derivatives of the parameter t as well.

Last edited: Jun 25, 2011
5. Jun 25, 2011

### Easy_as_Pi

Alright, part ii makes a lot more sense, now, too. I took a break from the problem and came, and used the dot product like you suggested. I can't believe I didn't think to do that already. By calculating the magnitude of velocity, speed, I get:
speed = sqrt(cos^2(t)+sin^2(t)). Now, I see why, mathematically, this must always be 1. Now, say I had something that wasn't sqrt(sin^2 + cos^2) for the speed, how would I prove or disprove constant speed?

6. Jun 25, 2011