Uniform circular motion and coefficient of friction

[KatieLynn]

Homework Statement

You are standing on a merry-go round spinning at 12 rpm. The radius of the merry-go round is 5 meters, and you have a mass of 57kg. What static frictional force between your shoes and the merry-go round is required to keep you from flying off?

Homework Equations

Vc=(2*pi*r)/T

Ac=(Vc^2)/R

Fnet=MA

The Attempt at a Solution

Alright here goes...

Vc=(2*pi*r)/T so Vc=(2*pi*5)/.2 = 157

Ac=(157^2)/5 =4934

here is where I get confused... I think I should use Fnet=ma

so it would be Fnet=(57)(4934)

I'm not sure what Fnet is composed of though...where does the coefficient of friction come in?

 

[pinkyjoshi65]

Use the equation Fs= Ms*Fn

Fs is the static force, Fn is the normal force and Ms is the coefficient of static friction..Is is better now?

 

[KatieLynn]

so it would be

Ms*Fn=(57)(4934)Fn = (57)(-9.81)

Ms*(57)(-9.81)=(57)(4934)

Ms=502.96 which isn't right did I calculate the acceleration or something wrong?

 

[pinkyjoshi65]

wait a minute, they are askig you to find the force, not Ms.

 

[KatieLynn]

So any words of advice for finding that? I'm honestly still pretty lost.

 

[hage567]

"Fnet" is your answer. Since that is the only force present and it what is keeping you on the merry-go-round. The problem isn't asking you to specify any further.

 

[KatieLynn]

My teacher gave us the answer its... Ff=450N required, Ff=45N available therefore you would slide off, how does Fnet equal that?

I'm not trying to be difficult, I just really do not understand.

 

[hage567]

Have you posted all of the information for this problem?

Also, your calculation of linear velocity is incorrect. You've divided by the frequency, not the period. T = 1/frequency, and is expressed in seconds. Do you know how to fix that?

It would be beneficial if you carried your units through your calculations, to see that they work out properly.

 

[KatieLynn]

!I think I've got it this time :)
Vc=(2*pi*r)/T so Vc=(2*pi*5)/(1/.2) = 6.283

Ac=(6.283^2)/5 =7.9

so now fnet=ma

fnet is only force of friction

Ff=(57)(7.9) =450N so that's how much force is required to keep you on the merry-go round.

Actually I did leave this part out because I wanted to figure out the first part first, which I see has made it confusing. So anyway the rest of the problem reads...If the coefficient of static friction between your shoes and merry-go round is 0.08, would you slide off the merry-go round?so...450N is what's needed to keep you on

Force of friction=(coefficient of friction)(force normal)

Ff=(.08)(57) = 45N

then 45.6N<450N so you slide off.

I'll try to keep track of my units better in the future.Is all the right now?

 

[hage567]

Actually I did leave this part out because I wanted to figure out the first part first, which I see has made it confusing. So anyway the rest of the problem reads...


If the coefficient of static friction between your shoes and merry-go round is 0.08, would you slide off the merry-go round?

Ah! That makes more sense.

the only little thing I'm not sure about is why when calculating the force of friction here Ff=(.8)(57) = 45.6N you do not also multiply it by -9.81. fnet=ma why do you leave out the a?

You do have to multiply by 9.81. You starting off by saying the coefficient of friction was 0.08, but you used 0.8 in your calculation. Your answer just happened to work out because you increased it by a factor of 10, which is approximately g.

Other than that, I think everything looks OK.

 

[KatieLynn]

Yep I figured that out shortly after I posted that, thanks for all your help :)

 

[hage567]

You're welcome.