(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away.

a. What is the source of the centripetal force acting on the riders?

b. How much centripetal force acts on a 55.0 kg rider?

c. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?

2. Relevant equations

Centripetal force = (mv^2)/r

Force of friction = (coefficient of static friction)*(normal force)

Normal force=m*g

3. The attempt at a solution

I have solved part B. Part A is more of a conceptual question that I have, and part C is the part where I am confused.

Part A attempt:

The reason I am confused on this question is that I am unsure of what thesourcerefers to. Does the source refer to thewallthat the people are "plastered" against? This would be my answer because if I were to picture myself in this ride, the wall would appear to be pushing me towards the center of the circle (which is what the centripetal force actually is).

Part B attempt:

I used the first formula I listed in the "relevant equations" section.

Centripetal force = ((55.0 kg)*(10.0 m/s)^2)/3.30 m

Centripetal force =1670 N

This is my answer.

Part C attempt:

I have to find the coefficient of static friction, so I know that I will have to use the second equation I listed in the "relevant equations" section. I will also need to somehow find the normal force and the force of friction.

To find the normal force,I am assuming that the rider's mass is 55.0 kg. No other mass is provided, and I can't think of another way I could calculate the normal force without the mass.

Normal force = m*g

Normal force = (55.0 kg)*(9.81 m/s^2)

Normal force =540 N

Thus, I have one part necessary to complete the problem.

This is the part where I am confused. I need to find the force of friction. However, the force of friction must be pointing upwards, to counterbalance the gravitational force pushing the riders downwards. Wouldn't the force of friction then be equal to the normal force? If I assume this, the coefficient of static friction would be one, which to me, doesn't seem like the correct answer.

Is there something I'm missing here?

Thanks for any help you can offer. This is my first post here as well.

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# Homework Help: Circular Motion and Friction on an amusement park ride

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