Ferris wheel rotation

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Homework Statement



The Singapore Flyer is te world's largest Ferris wheel. It's diameter is 150m and it rotates once every 30 min.
a) Find the Magnitude of the average velocity
b) Find the average acceleration at the wheel's rim.

The Attempt at a Solution



a) |v→| = Δs/Δt=75m.2π/30mins = 15.7m min^-1 (Could it be zero? p1 = (0,75) p2= (0,75); (p2-p1)/30mins = 0ms^-1)
b)
 
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Answers and Replies

  • #2
vela
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Homework Statement



The Singapore Flyer is te world's largest Ferris wheel. It's diameter is 150m and it rotates once every 30 min.
a) Find the Magnitude of the average velocity
b) Find the average acceleration at the wheel's rim.

The Attempt at a Solution



a) |v→| = Δs/Δt=75m.2π/30mins = 15.7m min^-1 (Could it be zero? p1 = (0,75) p2= (0,75); (p2-p1)/30mins = 0ms^-1)
b)
What's the definition of average velocity? Is it change in displacement over time or distance over time?
 
  • #3
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What's the definition of average velocity? Is it change in displacement over time or distance over time?

Yes, it's change in displacement/ change in time.
 
  • #4
vela
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Right. So you should know which of your answers for part (a) is correct.
 
  • #5
negation
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Right. So you should know which of your answers for part (a) is correct.

The answer to part(a) is 0ms^-1

b) a→ = v^2/r = [r^2Θ^2/t^2]/r = rΘ^2/t^2 = 75m(2π)^2/5^2 = 118.43m min^-1
 
  • #6
vela
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You found the instantaneous acceleration, but the problem asked for the average acceleration. (You didn't state it explicitly, but I assume the problem asks you for the average velocity and acceleration for one complete turn.)

Your answer for part (a) is correct.
 
  • #7
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You found the instantaneous acceleration, but the problem asked for the average acceleration. (You didn't state it explicitly, but I assume the problem asks you for the average velocity and acceleration for one complete turn.)

Your answer for part (a) is correct.

I made the mistake of leaving out a vital piece of information which was "over an interval of 5 mins"

b) Find the average acceleration at the wheel's rim over an interval of 5 mins.
 
  • #8
vela
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Does that apply to part (a) too? If so, you need to redo that part too.
 
  • #9
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Does that apply to part (a) too? If so, you need to redo that part too.

No, it applies only to part (b)

average acceleration = delta v/ delta t?

I'm having a little confusion here.
 
  • #10
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Here's another go:

a = Δv/Δt = (vf - vi )/Δt

vi = rω = 75m(0π/ 5) = 0m min^-1

vf = rω = 75m (2π/5) = 94.2m min^-1

94.2/5 = 18.8m min^-1
 
  • #11
vela
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Here's another go:

a = Δv/Δt = (vf - vi )/Δt

vi = rω = 75m(0π/ 5) = 0m min^-1
This result would say the ferris wheel is starting from rest (v=0).

vf = rω = 75m (2π/5) = 94.2m min^-1
And this one says that it's rotating at a rate of once every 5 minutes (##2\pi## per 5 minutes).

##v=\omega r## relates the linear speed ##v## of a point a distance ##r## away from the axis of rotation with the angular speed ##\omega##. If the ferris wheel is rotating at a constant rate, the speed will be constant. You need velocity, however, which is a vector. The speed gives you the magnitude of the vector, but there's still the direction of the vectors to worry about.
 
  • #12
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This result would say the ferris wheel is starting from rest (v=0).


And this one says that it's rotating at a rate of once every 5 minutes (##2\pi## per 5 minutes).

##v=\omega r## relates the linear speed ##v## of a point a distance ##r## away from the axis of rotation with the angular speed ##\omega##. If the ferris wheel is rotating at a constant rate, the speed will be constant. You need velocity, however, which is a vector. The speed gives you the magnitude of the vector, but there's still the direction of the vectors to worry about.

Have I manage to obtain the correct answer then?
I would have thought the average acceleration is the change in velocity/change in time
 
  • #13
vela
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No, you haven't done the problem correctly. You didn't determine the velocities correctly yet.

Draw a picture of the Ferris wheel. Label the point at the bottom as t=0. Which way does the velocity point and what is its magnitude? Now consider where that point is when t=5 min. The Ferris wheel takes 30 minutes to do a complete revolution, so after 5 minutes, by how many degrees will it have rotated? Which way does the velocity now point and what is its magnitude?
 
  • #14
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No, you haven't done the problem correctly. You didn't determine the velocities correctly yet.

Draw a picture of the Ferris wheel. Label the point at the bottom as t=0. Which way does the velocity point and what is its magnitude? Now consider where that point is when t=5 min. The Ferris wheel takes 30 minutes to do a complete revolution, so after 5 minutes, by how many degrees will it have rotated? Which way does the velocity now point and what is its magnitude?

Should the point at the top be labeled as t = 0?

Untitled.jpg


aaverage→ = Δv/Δt


change in Θ = 2π/60 . 5 = 0.523 radians

ratio of change in radial displacement to change in velocity
Δr/r = Δv/v

Δr: Δr^2 = r1^2 + r2^ - 2.r1.r2 cosΘ
Δr = 38.78m

velocity = Δp/Δt
p(2) = (75 sin 0.523 , 75 cos 0.523) = (37, 65)
p(1) = (75 sin 0 , 75 cos 0) = (0 , 75 )
Δt = 5

∴[p(2) - p(1)] / 5 = | [(37 , 65 ) - ( 0 , 75) ] | /5 = | (37, -10) |/5 = 7.7ms^-1

Δv = (Δr/r). v = 4ms^-1

aaverage→ = Δv/Δt
∴4ms^-1 / 5 = 0.8ms^-2
 
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  • #15
vela
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That's not correct. First, what is ##\omega##, the angular speed of the Ferris wheel?
 
  • #16
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That's not correct. First, what is ##\omega##, the angular speed of the Ferris wheel?

ΔΘ/Δt

Why is it not?
 
  • #17
vela
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Give me a number. You use that quantity in various places, but every time you do, it's a different number. It seems like you plug in numbers at random.
 
  • #18
negation
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Give me a number. You use that quantity in various places, but every time you do, it's a different number. It seems like you plug in numbers at random.

what is this number you want? Which variable is it for?
 
  • #19
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##\omega##
 
  • #20
negation
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I thought about it and had another go with the workings. It does make complete sense to me. Why is it flawed?
 
  • #21
vela
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I thought about it and had another go with the workings. It does make complete sense to me. Why is it flawed?

Well, for instance, the following isn't correct.
change in Θ = 2π/60 . 5 = 0.523 radians
 
  • #22
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##\omega##

If ω = ΔΘ/Δt
and ΔΘ = 0.523 radians since 2π = 60mins, therefore, 5 mins = 0.523 radian,
and Δt = 5 mins over the 0.523 radians
then
ω= 0.1046
 
  • #23
vela
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Why do you think ##2\pi = 60\text{ minutes}##?
 
  • #24
negation
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Why do you think ##2\pi = 60\text{ minutes}##?

I might have misread the question.
It should be ##2\pi = 30\text{ minutes}##

Is this the only blunder?
 
  • #25
vela
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No, you've made other mistakes; the biggest is using ##v_\text{avg}## for ##v##.

The magnitude of the average acceleration should turn out to be ##8.73\times 10^{-5}\text{ m/s}^2##.
 
  • #26
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No, you've made other mistakes; the biggest is using ##v_\text{avg}## for ##v##.

The magnitude of the average acceleration should turn out to be ##8.73\times 10^{-5}\text{ m/s}^2##.

How do I find v then? If it isn't v = rω nor v = dp/dt, then what other equations is there?
 
  • #27
vela
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The speed is given by ##v=\omega r##. This was your calculation:
Here's another go:

a = Δv/Δt = (vf - vi )/Δt

vi = rω = 75m(0π/ 5) = 0m min^-1

vf = rω = 75m (2π/5) = 94.2m min^-1

94.2/5 = 18.8m min^-1
Those aren't correct. Can you see why?
 
  • #28
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v = wr is linear velocity isn't it? Why is it now speed?
 
  • #29
vela
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Velocity is a vector; ##\omega r## isn't.
 
  • #30
negation
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The speed is given by ##v=\omega r##. This was your calculation:

Those aren't correct. Can you see why?

I can see why. I misread wrongly and used 2pi = 60 mins instead of 2pi = 30 mins. I should have used the latter and worked on that information instead. My frustration is, aside from this minor effort, is the mathematical reasoning correct then?
 
  • #31
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Velocity is a vector; ##\omega r## isn't.

I can see why ωr is a scalar quantity. But don't most book go by the definition rω = linear velocity?
 
  • #32
vela
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I can see why. I misread wrongly and used 2pi = 60 mins instead of 2pi = 30 mins. I should have used the latter and worked on that information instead.
That's not the error in those calculations.

My frustration is, aside from this minor effort, is the mathematical reasoning correct then?
No, I already told you you made other mistakes.

I can see why ωr is a scalar quantity. But don't most book go by the definition rω = linear velocity?
No. ##\omega r## is the speed or, equivalently, the magnitude of the instantaneous velocity, but it's not the velocity ##\vec{v}##, which is a vector. You will never see a book write ##\vec{v} = \omega r##. You will, however, see ##v = \omega r##. ##v## and ##\vec{v}## aren't the same thing. It's possible that a book may sloppily use the word velocity to mean speed, but careful authors won't do that to avoid misconceptions.
 
  • #33
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That's not the error in those calculations.


No, I already told you you made other mistakes.


No. ##\omega r## is the speed or, equivalently, the magnitude of the instantaneous velocity, but it's not the velocity ##\vec{v}##, which is a vector. You will never see a book write ##\vec{v} = \omega r##. You will, however, see ##v = \omega r##. ##v## and ##\vec{v}## aren't the same thing. It's possible that a book may sloppily use the word velocity to mean speed, but careful authors won't do that to avoid misconceptions.

so we know delta Θ = 0.523 radian
delta r = 75m based on law of cosine
r = 75m

and v = r.w
w = delta Θ/ delta t = (0.523rad - 0rad)/5 mins = 0.1046rad min^-1
v = 75m x 0.1046rad min^-1 = 7.84m min^-1

delta r/r x v = 7.845m min^-1 = delta v
a = delta v / delta t

a = 7.84m min^-1 / 5 mins = 1.569m min^-2

and it's wrong. I've exhausted all means
 
  • #34
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  • #35
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Edit
 
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