Calculating Minimum Angular Velocity for Rotor Carnival Ride | Dynamics Problem

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Homework Help Overview

The problem involves calculating the minimum angular velocity required for a carnival ride, specifically the Rotor, to keep riders adhered to the wall as the floor drops. It is situated within the dynamics subject area, focusing on forces acting on a person standing against a rotating cylindrical wall.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the necessity of knowing the mass of the person to find the angular velocity, with some suggesting that the mass may not be needed as it cancels out in the equations. There is mention of using a free body diagram to analyze the forces acting on the rider.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of a free body diagram and the concept that mass may not be necessary for the solution. However, there is no explicit consensus on the approach to take.

Contextual Notes

Participants are considering the implications of the mass variable in their calculations and whether it can be treated as a constant that cancels out. There is uncertainty about how the teacher may evaluate answers that include variables instead of numerical values.

Kudo Shinichi
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HELP!A dynamics problem

Homework Statement


In a carnival ride called the Rotor, people stand against the inside wall of cylindrical chamber. when the chamber rotates with a sufficiently high angular velocity, the floor of the chamber is lowered and the people "stick" to the wall. the coefficient of static friction between a person and the wall is mu(s)=0.5 and the radius of the chamber is r=3.5m.
what is the minimum angular velocity, omega, of the chamber (in revolutions per minute) that will keep the people from sliding down the wall?

The Attempt at a Solution



a=(omega)^2*radius
two unknown variables
therefore, we need to find the acceleration
Frictional force=mu(s)*m*g
mu(s)=0.5 g=9.8
frictional force=0.5*m*9.8
and
frictional force=ma
mu(s)*m*g=ma
a=mu(s)*g
sub the a I got into a=(omega)^2*radius to find omega

however, while i was trying to find the frictional force, there are two unknown variables, which are frictional force and the mass. I think that i can solve this question as soon as I know the mass for the person...can anyone teach me how to find the mass of the person?

Thank you for helping me.
 
Last edited:
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You are correct that you need the mass of the person. If you can't find the mass of the person , give your answer in terms of "m" and mention that "m" is the mass of the person.
 


JoAuSc said:
You are correct that you need the mass of the person. If you can't find the mass of the person , give your answer in terms of "m" and mention that "m" is the mass of the person.
What you are saying is that I will not get a numerical answer for this question, instead i will have a variable in my answer, right?
I wish my teacher will not mark this as a wrong answer^^thanks
by the way,so there is no way to find the mass, right?
 


I don't think you need the mass of rider to solve this problem because it cancels out. This is a statics problem i.e. the rider is not moving. Do a free body diagram on the rider with all the forces defined (hint: I see 4 forces). Sum of the forces in X-direction is 0 and sum of all the forces in Y-direction is 0. Solve for u.
 


RTW69 said:
I don't think you need the mass of rider to solve this problem because it cancels out. This is a statics problem i.e. the rider is not moving. Do a free body diagram on the rider with all the forces defined (hint: I see 4 forces). Sum of the forces in X-direction is 0 and sum of all the forces in Y-direction is 0. Solve for u.

Yes, I see it now. The mass should cancel out because the gravitational force, the centripetal force, and the force of friction should all depend linearly on mass.
 

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