# Constant radial and angular velocity; find magnitude of acceleration.

1. Sep 12, 2013

### oddjobmj

1. The problem statement, all variables and given/known data

(a) A particle moves in a plane with constant radial velocity [(r)\dot] and constant angular velocity [θ\dot]. When the particle is at distance r from the origin, determine the speed of the particle.
Data: [(r)\dot] = 3.2 m/s; [θ\dot] = 2.3 rad/s; r = 3.2 m.

(b) When the particle is at distance r from the origin, determine the magnitude of the acceleration.

2. Relevant equations
a=v^2/r

3. The attempt at a solution
This question was posted on these forums before which I found using the search function. The answer to part (a) was derived after the help from that thread. The result is 8.026 m/s.

I am not sure about part (b), however. In my search I have found many suggesting that a=v^2/r. Why is this the case? Also, I haven't been able to get that to work.

What I tried is this:

Find the circumference of the circle with radius 3.2m which is 20.1062 m. Find the period of the particle which is 2*pi/2.3 which is 2.73182 seconds. Using that I was thinking I could find the angular velocity in m/s which I calculated to be 7.447 m/s. If I use that in the equation a=v^2/r I get a=7.447^2/3.2=16.93 m/s^2 which is incorrect.

What am I missing? Also, where does a=v^2/r come from? Thanks in advance for the help!

2. Sep 12, 2013

### TSny

This is a problem that apparently assumes that you are familiar with expressing velocity and acceleration in polar coordinates. See for example this link.

The expression $v^2/r$ is the formula for "centripetal acceleration" for a particle moving on a circle. But, in your problem the particle is not moving on a circle. In your problem the particle will have two components of acceleration (radial and tangential) as shown near the bottom of the page in the link above.

Your calculation of 16.93 m/s2 does in fact happen to give the correct answer for the magnitude of the radial component of acceleration for this problem. (It would not give the correct radial acceleration if $\ddot{r} \neq 0$.) To find the total acceleration, you will also need to find the tangential component of acceleration.

3. Sep 13, 2013

### oddjobmj

Thank you for your help TSny.

I am confused about how to find r, r(dot), etc. As you suggest it is not moving in a circle but since the radial and angular velocity are constant it is hard for me to imagine the radius changing. I might just be tired but I can't even seem to remember what r dot and theta dot are. I was under the impression that these were just derivatives but I thought r was constant so the derivative would be 0?

I have no idea how to find the tangential component of acceleration. The link you provided suggests a formula in two parts for acceleration; one is radial and the other tangential. As you noted I have solved for the radial portion but can't seem to figure out how to solve the other portion. If I understood how to find rdot and theta dot I would probably be better suited to take the next step.

Is it really just: 2*3.2*2.3+3.2*0=14.72 for the tangential portion of acceleration? i.e. 2*r(dot)*theta(dot)+r*theta'

EDIT: Apparently that worked. If you take the above value of 14.72 and find the hypotenuse formed with the other component of acceleration you get 22.43 m/s^2 which is correct.

Last edited: Sep 13, 2013
4. Sep 13, 2013

### TSny

Yes, $\dot{r}$ and $\dot{\theta}$ are just the time derivatives of $r$ and $\theta$. As $\theta$ increases, so does $r$. The particle is moving along a spiral of Archimedes

That's it. Good work.