Centripetal acceleration of fairgrounds ride

plowboy15
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A fairgrounds ride spins its occupants inside a flying-saucer-shaped container. If the horizontal circular path the riders follow has a 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 2.60 times that of gravity?
 
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Welcome to PF plowboy! How far have you gotten on your end? Do you understand the physics and forces involved? Have you formulated a mathematical approach?
 
I get you have to use ac=Vt/r and that r will equal 8 and ac= 2.60. I think you will change the formlua to ac*r=Vt
 
plowboy15 said:
I get you have to use ac=Vt/r and that r will equal 8 and ac= 2.60. I think you will change the formlua to ac*r=Vt
I think you're close and on the right track. Take a look at the wiki on Uniform Circular Motion here:
http://en.wikipedia.org/wiki/Circular_motion#Uniform

The equations mentioned after these texts can be simplified/adapted:
The acceleration due to change in the direction is:
and
The centripetal and centrifugal force can also be found out using acceleration:

If we assume a mass of '1', can you see how the second equation for centripetal force simplifies to the first equation for acceleration? You mentioned an acceleration 2.60 times that of gravity. Do you know how to compute 2.6g?

I hope we're both on the right track! :w
 
This thread really belongs in the homework forum, so I'm moving it there.
 
okay
 

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