Circular Motion (10 points)?

Question

Spin Out! An interesting amusement park activity involves a cylindrical room that spins about a vertical axis (figure below). Participants in the "ride" are in contact with the wall of the room, and the circular motion of the room results in a normal force from the wall on the riders. When the room spins sufficiently fast, the floor is retracted and the frictional force from the wall keeps the people "stuck" to the wall. Assume the room has a radius of 2.2 m, and the coefficient of static friction between the people and the wall is μs = 0.44

Picture: http://www.webassign.net/giocp1/5-p-026.gif

Question: Apply Newton's second law along both the vertical and the radial directions. Find the minimum rotation rate for which the riders do not slip down the wall.

The answer is 0.507 rev/s
Please explain to me how to do this problem. Thank you.

odu83 · Accepted Answer

consider a fbd of a rider who is stuck, therefore zero acceleration in the vertical:

m*g-f=0  where f is the friction force

friction is N*μs where N is the normal force
In this case, N=m*r*w^2
or
therefore the minimum w is found by
m*g-m*r*μs*w^2=0
solve for w
w=sqrt(g/(r*μs))

Convert the units to rev/s
1rev=2*pi radians

0.507 rev/s

klipfel · Answer

All objects on the carousel can have the equal angular velocity, denoted as ω. ω∙radius = tangential speed. The extra the item is from the center the higher the worth for the radius will probably be. For this reason the object extra from the middle will have a larger tangential speed.

Tom M · Answer

Here's how to do this problem:

1)  Attend class and pay attention and ask questions when you don't understand things.

2)  If you completed step 1, then the solution is trivial.

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Circular Motion (10 points)?

3 Answers