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Height and Weight

by Abarak
Tags: height, weight
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Abarak
#1
Jul12-07, 05:57 PM
P: 28
1. The problem statement, all variables and given/known data
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60s).

Part A:
Question: Find the speed of the passengers when the Ferris wheel is rotating at this rate.

Answer: 5.24 m/s This simple problem was found by calculating the circumference (314.16m) and dividing it my 60.

Part B:
A passenger weighs 724 N at the weight-guessing booth on the ground. What is his apparent weight at the highest point on the Ferris wheel?

Part C:
What is his apparent weight at the lowest point on the Ferris wheel?

Part D:
What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

part E:
What then would be the passenger's apparent weight at the lowest point?

2. Relevant equations
?????

To be quite honest, I have no idea how I would go about solving problems A - E....

-- Abarak
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Chi Meson
#2
Jul12-07, 08:31 PM
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In perfect circular motion, the net force on an object is centripetal force.

A the top of the circle, the centripetal force is provided by (+normal)+(-gravitational) forces.

The normal force is what we perceive as "apparent weight"
Abarak
#3
Jul13-07, 01:16 PM
P: 28
Newton’s Law of Gravitation states:

[tex]F_{gravity} = G\frac{m_1m_2}{r^2}[/tex]

(normal)+(-gravitational)
How can I find "apparent weight at the highest point on the Ferris wheel" from the gravitational force without knowing the second mass?

What I have so far
v = 5.24m/s
diameter = 100m
radius = 50m
Circumference = 314.16m

The wheel has 60 arms and makes 1 complete revolution every 60 seconds.

Thanks for the help so far,

Abarak

Doc Al
#4
Jul13-07, 01:56 PM
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Height and Weight

Newton's law of gravity is not needed here. The only gravitational force you care about is that due to the Earth--the passenger's weight, which is given.

As Chi Meson stated, the "apparent weight" equals the normal force. Identify the forces acting on the passenger at each point in question (Chi Meson told you what they are) and apply Newton's 2nd law of motion to solve for the normal force. What's the acceleration?
D H
#5
Jul13-07, 02:01 PM
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Quote Quote by Abarak View Post
Newton’s Law of Gravitation states:

[tex]F_{gravity} = G\frac{m_1m_2}{r^2}[/tex]
You are making this far too complicated. The difference in the gravitational force at the top and the bottom of the Ferris wheel is very, very small. Simply use [itex]g[/itex] as the acceleration due to gravity.
Abarak
#6
Jul13-07, 05:09 PM
P: 28
You are making this far too complicated.
Yeah, I tend to do that from time to time.

Let me just make sure I understand this correctly.

The passengers normal weight is 724 N

At the top of the Ferris Wheel all I need to do is add the normal weight to the gravitation weight. 724N+(-9.8[tex]m/s^2[/tex]) = 714.2N??

What about when the wheel is at the lowest point? I though no other forces would affect the person so I tried 724N and it did not work.

Again, thanks for all the help. I think I just need something to "click".

Abarak
Doc Al
#7
Jul13-07, 05:33 PM
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Quote Quote by Abarak View Post
The passengers normal weight is 724 N

At the top of the Ferris Wheel all I need to do is add the normal weight to the gravitation weight.
The passenger's weight is the gravitational weight. Two forces act on the passenger at each point:
(1) the normal force (also called, confusingly, the "apparent" weight); this is the force exerted by the seat on the person;
(2) the real weight (the gravitational weight), which you are given (it also equals mg).

Which direction do these forces act? (Hint: At the top of the Ferris wheel, the passenger is upside down.) Add them up (taking direction into account) to find the net force. Use that net force in Newton's 2nd law to solve for the normal force. (What's the passenger's acceleration?)



724N+(-9.8[tex]m/s^2[/tex]) = 714.2N??
Realize that this computation makes no sense. You are adding a force (724 N) to an acceleration: no can do! The units don't match.

What about when the wheel is at the lowest point?
The same two forces act on the passenger at the highest and lowest point. Of course the magnitude and direction of the normal force will be different--that's what you are trying to solve for.
Abarak
#8
Jul13-07, 07:30 PM
P: 28
Hey Doc Al,

Ok, this is what I did and I still got the wrong answer

real weight = 724N = m*g
724N = m * 9.8 m/s^2
m = 73.878kg

Because the seat needs to support the weight of the person I used
[tex]F = m * (\frac{v^2}{r})[/tex]
[tex]F = 73.878 * (\frac{5.24^2}{50}) = 40.57N[/tex]

normal force + real weight = Fnet

Part B:
40.57 + 724 = 764.57N

Part C:
724 - 40.57 = 683.43N

It stills says this answer is wrong? Am I calculating the normal force wrong?
Doc Al
#9
Jul13-07, 08:13 PM
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Your calculation of the mass and centripetal force is correct.

It turns out that this Ferris wheel rotates so slowly that at the top the seats must be right side up. (I was wrong when I said it was upside down at the top.) So that means the normal force acts upward at the top. (The weight, of course, acts down.)

So redo your summation of forces.
Abarak
#10
Jul13-07, 09:16 PM
P: 28
Hey Doc, DH and Chi!

I was able to solve every part with your help. Thank you very much for taking your time with me.

-- Abarak
Heat
#11
Oct8-07, 12:22 PM
P: 273
Hi, sorry to bring up this thread, but I have a similar problem.

Following the steps and reading the explanation I managed to get A-C.

But I could not determine D and E.

Part D:
What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

part E:
What then would be the passenger's apparent weight at the lowest point?
Hootenanny
#12
Oct8-07, 12:29 PM
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Quote Quote by Heat View Post
Hi, sorry to bring up this thread, but I have a similar problem.

Following the steps and reading the explanation I managed to get A-C.

But I could not determine D and E.

Part D:
What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
If the passanger's apparent weight it zero, what can you say about the passenger's weight with respect to the centripetal force?
Heat
#13
Oct8-07, 12:31 PM
P: 273
that they are equal.
Hootenanny
#14
Oct8-07, 12:33 PM
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Quote Quote by Heat View Post
that they are equal.
Correct! So how fast would the ferris wheel have to be turning for the centripetal force to be equal to the passenger's weight?


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