Posted by Sarah on Monday, June 8, 2009 at 12:49pm.
A Ferris wheel with radius 14.0m is tuurning about a horizontal axis through its center. The linear speed of a passenger on the rin is constant and equal to 7.00 m/s. Whatare the magnitude and direction of the passengers acclerations as she passes through a) the lowest point in her circular motion and b) the highest point in her circular motion? c) how much time does it take the ferris wheel to make on revolution?

Physics  MathMate, Monday, June 8, 2009 at 1:04pm
(a), (b)
acceleration due to circular motion
= rω^{2} towards the centre
It is necessary to add/subtract the acceleration due to gravity at different points on the wheel
(c)
tangential speed, v= rω
v and r are know, so ω can be determined.

Physics  Sarah, Monday, June 8, 2009 at 1:12pm
For C i got .5 or 1/2 v=rw
7.0=14.0 w is that correct
im still a little confused on a and b

Physics  bobpursley, Monday, June 8, 2009 at 1:36pm
at the top, net acceleration=centacc9.8
at the bottom, add them to get net acceleration
centripetal acceleration=v^2/r r, v given.
distanceAround/time=v
solve for time.

Physics  MathMate, Monday, June 8, 2009 at 1:42pm
That's correct. The unit is radians / s.
For a and b, the acceleration I was referring to corresponds to the centripetal force required to keep the passenger in place.
Thus at the lowest point, the weight of the passenger adds to this centripetal force, and so the acceleration due to gravity, g, is additve to this acceleration.
On the other hand, at the highest point, the acceleration due to gravity tends to reduce the centripetal force required to keep the passenger in its place, so g must be subtracted from the centripetal acceleration that you calculate from the formula a=ω^{2}r.
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