- #1

HangingOnByAPulley

- 4

- 3

- Homework Statement
- The amusement park ride shown in Figure 10 is a

large, rapidly spinning cylindrical room with a radius

of 3.0 m. The riders stand up against the wall, and the

room starts to spin. Once the room is spinning fast

enough, the riders stick to the wall. Then the floor

slowly lowers, but the riders do not slide down the

wall. Assume the coefficient of friction between

the wall and the riders is 0.40.

(b) Calculate the minimum speed of the rider

required to keep the person stuck to the wall

when lowering the floor.

- Relevant Equations
- Fc = mv^2/r

Fs = μsFn

Hi, I just had a question about this homework question.

I am not given the mass at all in any portion of the question. Fs = Fc because the static friction is the thing that keeps the rider stuck to the wall

My answer came out to about 3.4 m/s for the minimum speed that keeps the rider stuck to the wall, however, the textbook answer is 8.6 m/s and I am not entirely sure why. Am I supposed to factor in the fact that the floor is being lowered? I am not given the acceleration at which the floor is being lowered.

I isolated for V by doing Fs = Fc

μsmg = mv^2/r

v= √μs(g)(r)

v came out to 3.26 m/s. Where did I go wrong?

I am not given the mass at all in any portion of the question. Fs = Fc because the static friction is the thing that keeps the rider stuck to the wall

My answer came out to about 3.4 m/s for the minimum speed that keeps the rider stuck to the wall, however, the textbook answer is 8.6 m/s and I am not entirely sure why. Am I supposed to factor in the fact that the floor is being lowered? I am not given the acceleration at which the floor is being lowered.

I isolated for V by doing Fs = Fc

μsmg = mv^2/r

v= √μs(g)(r)

v came out to 3.26 m/s. Where did I go wrong?