Amusement park ride (circular motion)

In summary: Part c) is what I'm struggling with. The equation for the forces on the person is F= ma. What I'm not understanding is how to get from F=ma to W=mg. I've looked at the equation for the normal force of the person against the wall (F=mg) and I've also looked at the equation for the centrifugal force (F=mv^2/r). I've tried plugging in different values for m and v but I can't seem to get it to work. Can you help me with this part? Part d) is also something that I'm struggling with. I don't understand how the motion of the person
  • #1
~christina~
Gold Member
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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 [tex]\mu_s[/tex] 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[tex]\mu_s /g) ^1/2[/tex]

b) obtain a numerical value for T, taking R= 4.00m and [tex]\mu_s= 0.400.

[/tex]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 [tex]\mu_s= 0.400[/tex]

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

T= [tex]\sqrt{} (4 pi^2 (4.00m)(0.400) / 9.80m/s^2)[/tex] = 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 :smile:
 
Last edited:
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  • #2
Part a) requires a little bit of work but its not too difficult. The max force of static friction ([itex]f_{smax}=\mu_sN[/itex]) for the person to just stay stuck on the wall without moving needs to be equal in magnitude to the persons weight ([itex] W=mg[/itex]). The normal force of the person against the wall is just the centrifugal force.

[tex] F = \frac{mv^2}{r} [/tex]

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) :smile:
 
  • #3
Kurdt said:
Part a) requires a little bit of work but its not too difficult. The max force of static friction ([itex]f_{smax}=\mu_sN[/itex]) for the person to just stay stuck on the wall without moving needs to be equal in magnitude to the persons weight ([itex] W=mg[/itex]). The normal force of the person against the wall is just the centrifugal force.
[tex] F = \frac{mv^2}{r} [/tex]

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


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

I know that [tex] \sumFx= f_s max = \mu_s N = m(v^2/ r) = mg[/tex] (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...

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

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 centripital acceleration equation I can get

T= [tex]\sqrt{} 4pi^2 r/ ac[/tex] 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 :smile:
 
  • #4
~christina~ said:
a.)
I looked at what you said again and now I think...

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

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 [itex] T=\frac{2\pi r}{v} [/itex] for v and plug it into [itex] \mu_s (mv^2/r) = mg [/itex], then do a bit more rearranging to put it in the form T = ...
 

1. What is circular motion in amusement park rides?

Circular motion in amusement park rides refers to the movement of the ride in a circular path around a central point. This type of motion is typically seen in rides such as roller coasters, carousel, and spinning rides.

2. How does circular motion work in amusement park rides?

Circular motion in amusement park rides is achieved through a combination of centripetal force and inertia. The ride is designed to move in a circular path, while the riders experience a force towards the center of the circle, keeping them in their seats.

3. Is circular motion safe in amusement park rides?

When properly designed and maintained, circular motion in amusement park rides is safe. The rides undergo rigorous testing and inspections to ensure the safety of the riders. However, it is important for riders to follow all safety guidelines and restrictions.

4. What are the physics principles behind circular motion in amusement park rides?

Circular motion in amusement park rides involves the principles of centripetal force, inertia, and velocity. Centripetal force is responsible for keeping the ride on its circular path, while inertia causes the riders to continue moving in a straight line. The velocity of the ride also plays a role in creating the sensation of circular motion.

5. How do engineers design amusement park rides to achieve circular motion?

Engineers use a combination of mathematical calculations, computer simulations, and physical testing to design amusement park rides that achieve circular motion. They take into account factors such as the ride's speed, radius of the circular path, and the forces acting on the riders to create a safe and thrilling experience.

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