Circular Motion Physics

[EvoN1020v]

In a "Rotor-ride" at a carnival, riders are pressed against the inside wall of a vertical cylinder 2.0 m in radius rotating 1.1 revolutions per sec when the floor drops out. What minimum coefficent of friction is needed so a person would not slip out?

 

Collecting the information from the question:

[math]f = 1.1[/math]

[math]r = 2.0 m[/math]

[math]\mu = ?[/math]

 

First, I used this formula: [math]a_c = 4\pi^2rf^2[/math] and it yielded [math] 95.5377706 m/s^2[/math].

 

To find the velocity you use: [math]v = \frac{2\pi2}{T}[/math]. Therefore, you need the value of [math]T[/math]. [math]T=\frac{1}{f}[/math], yielded [math]13.82300768 m/s[/math].

 

Next step, I know this is wrong, because [math]\mu[/math] can't be over 1.

 

What I did: [math]F_f = F_c[/math]

[math]\mu = \frac{v^2}{rg}[/math]

[math]\mu = 9.738814536[/math]

 

Can anyone tell me what I did wrong? And how to continue the solving, because I know that you need [math]F_f[/math] to be greater than [math]F_c[/math], so the person won't drop through the floor.

 

[5614]

Ff = friction

and

Fc = centripetal

right?

 

 

(will look at this in more detail tomorrow... need sleep now!)

[EvoN1020v]

Yes, right.

[swansont]

What I did: [math]F_f = F_c[/math]

 

Why?

 

What is the direction of the force of friction, and what is the direction of the centripetal force? Should they be equal? What force is the friction countering to keep the people from sliding out?

[s pepperchin]

The problem is that you are using the wrong equation.

 

[math]F_c=ma[/math]

 

[math]F_c=ma_c[/math]

 

is correct but to figure out [math]\mu[/math] you use:

 

[math]F_f=\mu F_N[/math] where [math]F_N=-F_c[/math]

 

and in order for the person not to slide down the frictioanl force must be equal and opposite to the force of gravity.

 

[math]F_g=\mu (-F_c)[/math]

 

[math]m g=\mu (-ma_c)[/math]

 

[math]g=-\mu a_c[/math]

 

[math]\mu = \frac{-g}{a_c}[/math]

 

plug in the numbers

 

[math]\mu = \frac{-(-9.8\frac{m}{s^2})}{95.5377706\frac{m}{s^2}}[/math]

 

[math]\mu = .1026[/math]

 

It is easier to see if you draw it out as you go though.

[swansont]

And the point is to let the poster work it out, rather than do the work for him or her. And besides, you have a sign error in your third equation line.

[s pepperchin]

I don't think so the normal force is in the opposite direction as the centripital force.

[swansont]

I don't think so the normal force is in the opposite direction as the centripital force.

 

What direction does the normal force point?

[EvoN1020v]

I drew a FBD (Free Body Diagram), and [math]F_N[/math] points to the north, because the person is standing inside the rotor-ride. As [math]F_N[/math] points to the north, [math]F_g[/math] points to the south, and [math]F_a[/math] points toward to the center of the circle.

 

I have a question: Does both [math]F_c[/math] and [math]F_a[/math] points toward the center of the center?

 

Another question: Is it really necessary to put negative sign for the gravity? I know it points down, but sometimes, it's not required to put negative sign all the time. At least, what I have seen so far in my Physics class.

[5614]

Sorry, I never really got around to looking at this thread.

 

So s pepperchin said that [math]F_N=-F_c[/math] and it doesn't because they act in the same direction... However whilst being incorrect, it doesn't have a chain effect because he made another mistake too:

 

in order for the person not to slide down the frictioanl force must be equal and opposite to the force of gravity.

 

[math]F_g=\mu (-F_c)[/math]

 

Now the correct answer uses:

Ff = -Fg

and

Fn = Fc

so by substituting the two you should have got

[math]-F_g = \mu F_c [/math]

but using his incorrect signs he ended up with

[math]F_g=\mu (-F_c)[/math]

but it's exactly the same thing so all of his calculations are correct, except he made two errors with signs which cancelled each other out? Right?

 

The Fn should have been replaced with Fc (not negative) and the Ff should have been replaced with -Fg (not positive). s pepperchin did both of the nots so it cancelled each other out... can someone just confim that?

 

=====

 

EvoN1020v: the signs are very important. Occasionally they don't matter, but generally they do. Imagine this situation: if I have a speed call it 5m/s and I accelerate at 1m/s^2 after 1 second how fast am I going? 6m/s? Maybe, but what if I was accelerating in the opposite direction to the way I was originally moving at 5m/s... then I would be in effect decelrating, but deceleration is a form of acceleration. You must specify a direction in which you are accelerating, or moving, or anything. And you do this by defining a direction as positive and if something is in the opposite direction it is negative.