Minimum angular velocity formula

[edr2004]

Homework Statement

A ride in an amusement park consists of a cylindrical chamber that rotates around a vertical axis as shown in the diagram below. When the angular velocity is sufficiently high, a person leaning against the wall can take his or her feet off the floor and remain "stuck" to the wall without falling.

Construct an expression for the minimum angular velocity that the ride could rotate at such that the person remains stuck to the wall. Use the following Use the following when entering your symbolic expression:

m : for the mass of the person
g : for the gravitational field strength near the surface of the earth
r : the radius of the cylindrical chamber (from the center to the walls)
mu : for the coefficient of friction between the person's back and the wall
pi : for π = 3.141592654...

Homework Equations

|FN| = mv2/r the inward normal force
|Fs| = μs|FN|maximum force of static friction
Fs| = mg

The Attempt at a Solution

I thought this was the answer but it is not correct.
v = sqrt((g*r)/mu)

ANy help would be appreciated!

 

[Mentz114]

Setting the vertical forces to cancel out I get

m*g = mu*m*r*w^2

w = sqrt(g/(r*mu))

 

[m2003]

Your attempt at a solution is a good start, but it is missing a few key components. Let's break down the problem and see if we can come up with a correct expression for the minimum angular velocity.

First, let's consider the forces acting on the person as they lean against the wall of the cylindrical chamber. There are two main forces at play here: the normal force (FN) and the force of static friction (Fs). The normal force is the force pushing the person against the wall, while the force of static friction is the force preventing the person from sliding down the wall.

Next, we need to think about what conditions need to be met for the person to remain "stuck" to the wall. In other words, what is the minimum force of static friction required to counteract the person's weight and keep them in place? We can set up an equation to represent this:

Fs = mg

Where m is the mass of the person, g is the gravitational field strength, and Fs is the maximum force of static friction that the wall can provide.

Now, we can use the equation for the maximum force of static friction (|Fs| = μs|FN|) and substitute in our known values to get:

μsmg = mv^2/r

Where μs is the coefficient of friction, m is the mass of the person, g is the gravitational field strength, v is the linear velocity of the person (which is equal to the tangential velocity of the cylindrical chamber), and r is the radius of the cylindrical chamber.

Finally, we can rearrange this equation to solve for the minimum angular velocity (ω) that the chamber must rotate at in order for the person to remain "stuck" to the wall:

ω = v/r = sqrt(μsg/r)

Therefore, the correct expression for the minimum angular velocity is:

ω = sqrt(μsg/r)

I hope this helps! Remember, when tackling a physics problem, it's important to consider all the forces at play and the conditions that need to be met in order to come up with a complete and accurate solution. Keep up the good work!