Can someone help me figure this out?

  • Thread starter Thread starter curiousgeorge99
  • Start date Start date
  • Tags Tags
    Figure
AI Thread Summary
To determine the revolutions per minute (rpm) for a 15m diameter Ferris wheel to create a sensation of weightlessness at the top, the key equations involve centripetal force and angular velocity. The initial calculation yields a linear velocity of 8.57 m/s, which is then used to find the period (T) of the wheel's rotation, resulting in T=5.49 seconds. To convert this to rpm, the relationship between linear velocity and angular velocity (v = rω) is crucial, along with the conversion factor that relates radians per second to revolutions per minute. The discussion emphasizes the importance of understanding these relationships and suggests using a free body diagram to visualize the forces involved. The final goal is to confirm that the required rpm is 11.
curiousgeorge99
Messages
16
Reaction score
0

Homework Statement


How many revolutions per minute would a 15m diameter Ferris wheel need to make for a person to feel 'weightless' at the topmost point?


Homework Equations


F= mv~2/r - mg
v = 2pi r/ T


The Attempt at a Solution


I get 8.57 m/s using the equation F= mv~2/r - mg
I think I need to use v = 2pi r/ T to figure out T. If I do this, I get T=5.49. I have no idea if this is right. If it is, then I'm not sure what 5.49 represents and how to turn it into rev/minute. The answer is supposed to be 11 rev/min.
 
Physics news on Phys.org
do you know what the relationship between V and \omega is?
 
V = displacement / time elapsed,

W = number of radians / second

Is that right?

Acceleration is A = change in velocity / change in time
Angular accel. A = change in # of radians / change in time.

This is about all I know. I'm not sure how they are related or how you get rev/min from here.

Its probably staring me right in the face but I still can't see it
 
You are on the right track with mv^2/r = mg, but that solves for the linear/tangential velocity (in m/s). Remember to convert that to angular velocity (rad/s) on your way to finding rpm.
 
v=r\omega

do you know the relationship between \omega and rev/min?
 
Is the relationship between W and rev/min: 0.1047 rad/s = 1 rev/min.
 
curiousgeorge99 said:
Is the relationship between W and rev/min: 0.1047 rad/s = 1 rev/min.

yes, do you know why?
 
No, no idea why. I just have that at the front of my textbook.
 
curiousgeorge99 said:
No, no idea why. I just have that at the front of my textbook.

2 * \pirad/s = 1rev/s right? cause 2 * \pi is the whole circle
 
Last edited:
  • #10
ok, that makes sense!

thanks

I'm going to see if I can solve this problem now.
 
  • #11
nope, still lost!

I think I'm stuck on where to start.

I keep trying to figure out the velocity in m/s then convert to rev/min.

Is there a better way to do it?
 
  • #12
i have the same problem...
 
  • #13
can any1 help?
 
  • #14
try to do a free body diagram showing all forces when the person is at the top of the ferris wheel
then equate them
one of them will have what u want in it
and u are provided with what u don't know in the question

let us know if u still can't get it
 
Back
Top