# &quot;webassign&quot; problem, CENTRIPETAL ACCEL

1. Nov 6, 2007

### lettertwelve

[SOLVED] &quot;webassign&quot; problem, CENTRIPETAL ACCEL

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

The Emerald Suite, a revolving restaurant at the top of the Space Needle in Seattle, Washington, makes a complete turn once every hour. What is the magnitude of the centripetal acceleration of the customer sitting 14.5 m from the restaurant's center?

2. Relevant equations

where T=time

3. The attempt at a solution

so then it would be

4*pi*14.5/3600^2, which gives me: 1.41E-5

but it's showing me that its incorrect...but i dont see how!

EDIT: also, is there a difference between the magnitude of centripetal acceleration as opposed to mere centripetal acceleration?

Last edited: Nov 6, 2007
2. Nov 6, 2007

### hotcommodity

The tangential velocity "v" is $$v = \frac{2 \pi r}{T}$$ if I remember correctly. Can you tell me what equation you used for the centripetal acceleration?

3. Nov 6, 2007

### lettertwelve

i used 4*pi*R/T^2, as my teacher directed in class for this particular problem.

for T i converted it to seconds

4. Nov 6, 2007

### hotcommodity

The equation you posted will give you the units for acceleration, but I've never seen that equation before. Try using the equation for tangential velocity that I gave you, and plug it into $$a_{centrip} = \frac{v^2}{r}$$.

5. Nov 6, 2007

### lettertwelve

I tried your equation, and got 4.41E-5, which gave me the correct answer. i don't understand why my teacher gave me that equation though...

i actually used your equation yesterday, and got the same answer. it must have been a webassign glitch?

thank you anyways.

my mistake was that i didnt do the tangential velocity first.

6. Nov 6, 2007

### hotcommodity

You're welcome. It'd be interesting to know where your teacher got that equation, but I'm glad it all worked out :)

7. Nov 6, 2007

### hage567

There is an error here, the pi term should be squared. That's why you didn't get the correct answer with it.

That equation is just the result of substituting $$v = \frac{2 \pi R}{T}$$ into $$a=\frac{v^2}{r}$$