Amusement park ride (circular motion)

[~christina~]

Homework Statement

An amusement park ride consists of a large vertical cylinder that spins about it's axis fast

enough that any person held up against the wall when the floor drops away. The

coeficient of static friction between the person and the wall is \mu_s and the

radius of the cylinder is R.

a) show that the maximum period of relvolution necessary to kep the person from falling is
T= (4 pi^2 R\mu_s /g) ^1/2

b) obtain a numerical value for T, taking R= 4.00m and \mu_s= 0.400. <br /> <br />How many revolutions per minute does the cylinder make?

c) If the rate of revolution of the cylinder is made to be somewhat larger, what happens

to the magnitude of each one of the forces acting on the person?

What happens to the motion of the person?

d) If instead the cylinder's rate of revolution is made to be somewhat smaller, what

happens to the magnitude of each of the forces acting on the person?

What happens in the motion of the person?


picture: http://img337.imageshack.us/img337/3765/19178363wb7.th.jpg

Homework Equations

F= ma= m(v^2/r) ?

The Attempt at a Solution

I have no idea how to explain a person's motion in this ammusement ride according to the forces..

a) I need help in this part

b)
R= 4.00m \mu_s= 0.400

T= (4 pi ^2 R \mu_s / g)^1/2

T= \sqrt{} (4 pi^2 (4.00m)(0.400) / 9.80m/s^2) = 6.44

Revolutions per min? I'm not sure how to get that


I think I'll tackle the the previous before I answer the rest
c)
d)



Help pleaase

Thank You very much

 

[Kurdt]

Part a) requires a little bit of work but its not too difficult. The max force of static friction (f_{smax}=\mu_sN) for the person to just stay stuck on the wall without moving needs to be equal in magnitude to the persons weight (W=mg). The normal force of the person against the wall is just the centrifugal force.

F = \frac{mv^2}{r}

For part b) you've forgotten to take the square root of 6.44.

And i'll wait till you've posted your atempts to c) and d)

 

[~christina~]

Part a) requires a little bit of work but its not too difficult. The max force of static friction (f_{smax}=\mu_sN) for the person to just stay stuck on the wall without moving needs to be equal in magnitude to the persons weight (W=mg). The normal force of the person against the wall is just the centrifugal force.
F = \frac{mv^2}{r}

a.) show that the maximum period of relvolution necessary to keep the person from falling is T= (4 pi^2 R\mu_s/g)^1/2


how do I relate what you said to get...T= (4 pi^2 R\mu_s/g)^1/2

I know that \sumFx= f_s max = \mu_s N = m(v^2/ r) = mg (you said the max force of static friction for the person to stay stuck on the wall without moving needs to be equal in magnitude to the person's weight mg)

I looked at what you said again and now I think...

f_s= \mu_s N = \mu_s (mv^2/r)= mg

I have no idea which is alright or if both are incorrect but I still don't see how I'd show that the maximum period of revolution necessary to keep the person from falling is that equation given when ..generally

T= 2 pi r/ v

ac= v^2/ r

I see from my book in a example that plugging the velocity after rearranging the T equation and plugging into the centripetal acceleration equation I can get

T= \sqrt{} 4pi^2 r/ ac however I still don't see how I can get the equation given for this particular problem from that..

For part b) you've forgotten to take the square root of 6.44.

oops.. 2.54s


I really really need help in the equation for a)

Thanks

 

[Kurdt]

a.)
I looked at what you said again and now I think...

f_s= \mu_s N = \mu_s (mv^2/r)= mg

I have no idea which is alright or if both are incorrect but I still don't see how I'd show that the maximum period of revolution necessary to keep the person from falling is that equation given when ..generally

T= 2 pi r/ v

You're on the right lines here. If you rearrange T=\frac{2\pi r}{v} for v and plug it into \mu_s (mv^2/r) = mg, then do a bit more rearranging to put it in the form T = ...