Force at the Bottom of a Circular Amusement Park Ride

[artireiter]

Homework Statement

In an amusement park ride called The Roundup, passengers stand inside a 17.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

a. Suppose the ring rotates once every 4.50 s . If a rider's mass is 59.0 kg , with how much force does the ring push on her at the top of the ride?

b. Suppose the ring rotates once every 4.50 s . If a rider's mass is 59.0 kg , with how much force does the ring push on her at the bottom of the ride?

c. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Homework Equations

T_top = (m/r)* v^2_top - mg

T_bot = (m/r)* v^2_bottom+mg

The Attempt at a Solution

a. 400N
I finished this one and know the answer is correct.

b.
(1/2)mv_top²+mg2r= (1/2)mv_bottom²
T_bottom=((mv_bottom²)/r)+mg
T_bottom=((mv_top²)/r)+(4mgr/r)+mg
v_top=11.87
T_bottom=(50*(11.87²)/8.5)+(4*59*9.8*8.5/8.5)+(59*9.8)= 3868.7
So I am stuck on part B, thought I did it right but I was told it's wrong.

 

[kuruman]

Energy conservation is not an issue here, so your approach to (b) in inappropriate. The speed remains constant so you are not trading potential energy for kinetic energy. Why did you not use your relevant equation for T_bot?

 

[artireiter]

Energy conservation is not an issue here, so your approach to (b) in inappropriate. The speed remains constant so you are not trading potential energy for kinetic energy. Why did you not use your relevant equation for T_bot?

Do you mean why I didn't use Tbottom=((mvbottom²)/r)+mg ? I did at first and found the same thing I got using the other equation, sorry just didn't see the need to include both. Not sure what I'm doing wrong since I'm getting the same thing.

 

[kuruman]

You can't possibly get the same answer either way. The ring rotates once every 4.50 s. Its speed is constant and equal to 11.87 m/s as you found out. Just substitute in the equation you provided T_bottom=((mv_bottom²)/r)+mg where v_bottom=11.87 m/s.