# Mechanics in a rotor problem

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1. May 26, 2015

### 0kelvin

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.

Last edited: May 26, 2015
2. May 26, 2015

### nrqed

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 $\mu_s n$)

3. May 26, 2015

### 0kelvin

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.

4. May 26, 2015

### nrqed

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 $F = \mu_s n$