# Proportions in circular motion don't make sense.

1. May 7, 2013

### Scorpius

Alright, here's the problem I'm have been given:

"A boy is riding a merry-go-round at a distance of 7.00 m from its center. The boy experiences a centripetal acceleration of 7.50 m/s2. What Centripetal acceleration is experienced by another person who is riding at a distance of 3.00 m from the center?"

After looking at this problem I started solving it using proportions. Here is what I did:

ac=Centripetal acceleration
v= velocity

ac$\propto$$\frac{v^2}{r}$

ac$\propto$$\frac{1}{7 m/3 m}$

$\frac{1}{7 m/3 m}$=$\frac{3}{7}$

$\frac{3}{7}$*7.5 m/s2=3.21 m/s2

According to my worksheet, this is the correct answer.

But something about the above proportion bothers me.

When it says ac$\propto$$\frac{1}{7m/3m}$ Why is Velocity one? In class we discussed that velocity of an object changes when you change "r" or the radius of the object from the center, but according to this proportion the velocity between the two boys is the same

I may need to remind you that this gave me the correct answer

What am I missing?

Last edited: May 7, 2013
2. May 7, 2013

### SteamKing

Staff Emeritus
Velocity is not = 1.

You are missing the fact that in circular motion, v = omega * r, where omega is the angular velocity in radians/sec. For the merry-go-round, omega = constant for uniform circular motion.

3. May 8, 2013

### haruspex

You treated the velocity as constant (wrong), but then you misapplied the proportionality, reversing it. Your method should have given you 7/3 * 7.5. Two errors cancelled out. What is constant is the angular speed, ω, so you should have used $a_c \propto r \omega^2$.