# Particle going in a circle with speed U problem

1. Jun 26, 2013

### hamilbr

1. The problem statement, all variables and given/known data
Hello all,
I am having some trouble with answering the problem below, mostly because I do not know what the letters stand for and what kind of graph is meant to be drawn. Any help on this would be greatly appreciated. Thanks

For a particle going in a circle with speed U, draw a graph showing:
x(t)
y(t)
u(t)=dx/dt
v(t)=dy/dt
du/dt
dv/dt

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jun 26, 2013

### Staff: Mentor

In your diagram you've indicated x and y axes. x(t) will be the x position with respect to time, y(t) the y position with respect to time. So to begin, draw two more graphs with t as the horizontal axis and place x on the vertical axis of one and y on the vertical axis of the other.

Clearly u(t) and v(t) are meant to be the x-velocity and y-velocity. du/dt is the derivative of y-velocity with respect to time (what do we usually call the derivative of a velocity?). You'll want to draw similar graphs for these items.

3. Jun 26, 2013

### HallsofIvy

Are you serious? You don't know what the letters stand for? That should have been the first thing you learned! x and y are the coordinates of that particle in some coordinate system. u and v, as is said in the problem are the derivatives (do you know what a derivative is?) of x and y and so are the x and y components of the velocity vector. du/dt and dv/dt are the x and y components of acceleration.

It helps to know that a circle, with center at (0, 0) and radius R has equation $x^2+ y^2= R^2$. And, of course, $sin^2(x)+ cos^2(x)= 1$. That means that parametric equations for the circle are $x= Rcos(\theta)$, $y= R sin(\theta)$. Now, the circumference of the circle is $2\pi R$ and if the particle is moving at constant speed U, it will complete the circle (moving through an angle of $2\pi$ radians) in time $\frac{2\pi R}{U}$ so has angular speed $\frac{U}{R}$. That means the angle, at time t, is $\theta= \frac{U}{R}t$.