An amusement park ride consits of a large verticle cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8m/s^2. Given g=9.8m/s^2, the coefficient of static friction between the wall and himan is 0.612, and the radius of the cylinder R=6.6m. For simplicity, neglect the person's depth and assume he or she is just a physical point on the wall. The person's speed is v=2piR/T. Where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum ro

Question

An amusement park ride consits of a large verticle cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8m/s^2. Given g=9.8m/s^2, the coefficient of static friction between the wall and himan is 0.612, and the radius of the cylinder R=6.6m. For simplicity, neglect the person's depth and assume he or she is just a physical point on the wall. The person's speed is v=2piR/T. Where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum ro

anonymous · Answer

maximum ro...? Though did you attempt to draw a free body diagram?

anonymous · Answer

It actually is partiall drawn for me but i just labeled the forces onto the diagram

anonymous · Answer

find the maximum rotation i'm sorry.

anonymous · Answer

\[2\pi \sqrt{(coefficient of static friction)(radius) divided by 9.8m/s^2

anonymous · Answer

Well, for it to be just able to stick on the wall, that means there is no reaction force from the wall.
|dw:1352772336185:dw| 
Solve the forces and find the speed on it. i.e $F_c = \frac{v^2}{r}$

anonymous · Answer

The velocity will slowly decrease because of friction.

anonymous · Answer

|dw:1352772692497:dw| this is the equation i first used but i got the wrong answer