How Does Friction and Centripetal Force Affect a Person Inside a Rotating Rotor?

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Homework Help Overview

The discussion revolves around the dynamics of a person inside a rotating rotor, focusing on the roles of friction and centripetal force. Participants explore the forces acting on the person, including weight, friction, and normal force, and how these relate to the conditions for maintaining position against sliding down the wall of the rotor.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the free body diagram and the forces involved, questioning the relationship between friction force and centripetal force. They explore the conditions under which a person would slide down or stick to the wall, raising concerns about the implications of mass in their calculations.

Discussion Status

There is an ongoing exploration of the expressions for friction force and minimum velocity required to prevent sliding. Some participants express confusion over the clarity of the questions posed and the relationships between the forces, while others provide insights into the conditions for maximum friction and the role of mass in the equations.

Contextual Notes

Participants note that the friction force is defined differently when the person is about to slide versus when they are not sliding, indicating a need for clarity in understanding these conditions. There is also mention of the necessity for precise definitions and expressions to avoid confusion in the discussion.

0kelvin
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The free boy diagram of one person inside the rotor should have three vectors: weight down, friction up, normal to the center of the cylinder.

Is friction force given by centripetal force * static friction coefficient? Normal and centripetal are the same vector in this problem.

Is minimum velocity for a person to not slide down given by an equality between friction force and weight? Mass cancels out in both sides.

The previous question makes me think there is a contradiction: weight of a person is not required to explain why the person "sticks" to the wall when the rotor is rotating. But I did use mass to calculate the friction force. In the second question I've noticed that the mass cancels out, but the equation gives me the impression that if v is high enough, friction would become greater than weight and therefore, person would slide up.
 
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0kelvin said:
The free boy diagram of one person inside the rotor should have three vectors: weight down, friction up, normal to the center of the cylinder.

Is friction force given by centripetal force * static friction coefficient? Normal and centripetal are the same vector in this problem.

Is minimum velocity for a person to not slide down given by an equality between friction force and weight? Mass cancels out in both sides.

The previous question makes me think there is a contradiction: weight of a person is not required to explain why the person "sticks" to the wall when the rotor is rotating. But I did use mass to calculate the friction force. In the second question I've noticed that the mass cancels out, but the equation gives me the impression that if v is high enough, friction would become greater than weight and therefore, person would slide up.
Can you type the questions exactly as they are given? Written as they are now, the first question is incomprehensible. Even the second one is confusing.

Note that the friction force is given by \mu_s n only when the friction force is maximum, i.e. when the person is about to slide down. In all other situations, when the person is not sliding, the friction force is simply equal to mg (and it is then NOT equal to \mu_s n)
 
What's the expression that calculates the friction force?

My take is: ##F = \mu_s \frac{mv^2}{r}## Because centripetal is the Normal.

What's the expression that calculates the minimum velocity for the the person to "stick" to the wall?

My take is: ##mg = \mu_s \frac{mv^2}{r}## Mass cancels out in both sides.
 
0kelvin said:
What's the expression that calculates the friction force?

My take is: ##F = \mu_s \frac{mv^2}{r}## Because centripetal is the Normal.

What's the expression that calculates the minimum velocity for the the person to "stick" to the wall?

My take is: ##mg = \mu_s \frac{mv^2}{r}## Mass cancels out in both sides.
You are right IF we are considering the case when the speed is minimum and the person is just on the point of sliding down. If the speed is larger then we have instead

F_s= mg and n = m v^2/R

but we do not have anymore the relation F = \mu_s n
 

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