Centripetal force equation help

1. Dec 15, 2005

cheez

I have set up an equation like this:
v= r x omega
mu x m x g = m v^2 /r

but I didn't get the answer, which is 15.7.
Is the setting right? If not, can anyone show me the right steps?
thx!

2. Dec 15, 2005

vaishakh

3. Dec 15, 2005

cheez

v= r x omega

= (39 ft) x (0.022 rev/s) x ( 2 pi / 1 rev)

=5.391 ft/ s

mu x m x g = m x v^2 / r

mu x g = v^2/ r

(0.36) (32 ft/s^2) = (5.391 ft/s) ^2 /r

r= 2.5 ft

4. Dec 15, 2005

Staff: Mentor

Your equations are correct for finding the distance from the center where the centripetal force just equals the maximum static friction for a given angular speed. But note that the problem says:
You need to figure out how the angular speed increases as you walk towards the center. This requires additional information (the masses involved, for example); did you leave anything out of the problem statement?

5. Dec 15, 2005

cheez

no, I have typed the whole question.

6. Dec 15, 2005

Staff: Mentor

Then I don't see how you have enough info to solve the problem. If you assume that the turntable is massive enough that the person's walking does not affect its angular speed (which seems to contradict what was stated in the problem), then only by walking away from the center will you get to a point where the required centripetal force is greater than friction can provide.

Note that in your last post you calculated the speed at 39 ft and used it to solve for a different distance. That doesn't make sense. If anything, the angular speed would remain fixed, not the tangential speed (which depends on the distance from the center).

7. Dec 15, 2005

Tide

Perhaps the OP meant "massless turntable?"

8. Dec 15, 2005

Staff: Mentor

Good thinking, Tide! That will get the answer they want.