Solving Circular Motion Help Homework: Find Speeds & Apparent Weight

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem related to circular motion, specifically focusing on a Ferris wheel's dynamics, including the calculation of speeds and apparent weights at different points during rotation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between apparent weight and normal force, questioning whether apparent weight is simply the product of mass and gravitational acceleration. There is also discussion about the differences in forces acting on a passenger at the lowest and highest points of the Ferris wheel.

Discussion Status

Some participants have provided clarifications regarding the definitions of apparent weight and normal force, which has aided in understanding the problem. However, there remains uncertainty about the calculations needed for the apparent weights at different points and the time for one revolution.

Contextual Notes

Participants express confusion about the application of formulas and the interpretation of the problem's requirements, indicating a need for further exploration of the concepts involved.

Santorican
Messages
10
Reaction score
0

Homework Statement



The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s.

Part A:Find the speed of the passengers when the Ferris wheel is rotating at this rate.
-I used V=(2piR)/T and got 5.24 which was correct.

Part B:A passenger weighs 862 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?
-This is where I am completely thrown off course. Is it asking for the normal force? I thought apparent weight was just m*g?

Part C:What is his apparent weight at the highest point on the Ferris wheel?
-Same issue with Part B

Part D:What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
-I would need the proper formula from Part B to find Part D, so I'm stuck here too.

Part E:What then would be the passenger's apparent weight at the lowest point?
-Same issue as Part D


Homework Equations



V=(2piR)/T
Sum of the forces in the y direction= N=m(g-(v^2/R))

The Attempt at a Solution



I attempted to use that formula but it said it was wrong so I have no clue what to do now. :frown:
 
Physics news on Phys.org
Santorican said:
Part B:A passenger weighs 862 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?
-This is where I am completely thrown off course. Is it asking for the normal force?
Yes, the apparent weight equals the normal force.
I thought apparent weight was just m*g?
No, m*g is the true weight.
 
Hi.

what do you know about the different between ferris wheel and a loop or circle.
Here is what i know At the lowest point in frerris wheel the
Normal force = (mv^2)/2 + mg
And At the highest point would be
Normal force = mg - (mv^2)/2

weight = mg ...
 
Doc Al said:
Yes, the apparent weight equals the normal force.

No, m*g is the true weight.

Doc Al you are the man! That little answer helped me so much thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 13 ·
Replies
13
Views
15K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K