Calculating Average Velocity and Acceleration of the Singapore Flyer

[negation]

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)

 

[vela]

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?

 

[negation]

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.

 

[vela]

Right. So you should know which of your answers for part (a) is correct.

 

[negation]

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

 

[vela]

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.

 

[negation]

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.

 

[vela]

Does that apply to part (a) too? If so, you need to redo that part too.

 

[negation]

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.

 

[negation]

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

 

[vela]

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.

 

[negation]

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

 

[vela]

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?

 

[negation]

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?

a_average→ = 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

a_average→ = Δv/Δt
∴4ms^-1 / 5 = 0.8ms^-2

 

[vela]

That's not correct. First, what is $\omega$, the angular speed of the Ferris wheel?

 

[negation]

That's not correct. First, what is $\omega$, the angular speed of the Ferris wheel?

ΔΘ/Δt

Why is it not?

 

[vela]

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.

 

[negation]

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?

 

[vela]

$\omega$

 

[negation]

I thought about it and had another go with the workings. It does make complete sense to me. Why is it flawed?

 

[vela]

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

 

[negation]

$\omega$

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

 

[vela]

Why do you think $2\pi = 60\text{ minutes}$?

 

[negation]

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?

 

[vela]

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 $8.73\times 10^{-5}\text{ m/s}^2$.

 

[negation]

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?

 

[vela]

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?

 

[negation]

v = wr is linear velocity isn't it? Why is it now speed?

 

[vela]

Velocity is a vector; $\omega r$ isn't.

 

[negation]

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?

 

[negation]

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?

 

[vela]

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.

 

[negation]

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

 

[vela]

so we know delta Θ = 0.523 radian

No, you don't.

 

[negation]

Edit

 

[negation]

a = v^2 / r = s/t = r . theta / t

delta theta = 1.05 rad

tangential velocity v = 1.05 rad x 75m / 300s = 0.263ms^-1

a = 0.263ms^-1 ^2 /75m = 9.138e-4ms^-2

 

[negation]

Can someone help me with this?

 

[haruspex]

a = v^2 / r = s/t = r . theta / t

delta theta = 1.05 rad

tangential velocity v = 1.05 rad x 75m / 300s = 0.263ms^-1

That's the tangential speed. (It's not a velocity since you're not quoting the direction, and anyway that changes. What's constant is the speed.)
You could have got this more easily: it travels πd = π150m in 30 minutes, so the speed is π*150/(30*60) m/s.

a = 0.263ms^-1 ^2 /75m = 9.138e-4ms^-2

What's your reasoning for that equation? You've divided the speed (constant) by the radius. That would give the angular speed, right?
You want the average acceleration, acceleration being a vector. For that, you need the change in velocity, also a vector. The speed is constant, v say, but the direction changes. At the bottom, the velocity is v horizontal. After 5 minutes, it's v in a different direction. What direction? What is the difference of the two velocities?

Velocity is a vector; ωr isn't.

At the risk of confusing negation further, let me put that differently.
If ω is angular speed and r is the magnitude of the radius then these are scalars and ωr gives the tangential speed, another scalar.
But angular velocity, $\vec\omega$ is a vector, and a specific radius can be a vector (from the axis to a given point on the rim) $\vec r$. Then the tangential velocity is the cross product $\vec \omega \times \vec r$, with the appropriate sign convention.

Back on part (a), I'm not sure I understand the question. Is it supposed to be the average angular velocity? The average velocity at the perimeter? The average velocity over the entire disc? If either of the last two, the answer would be zero, surely. But perhaps the "over a 5 minute interval" rider was supposed to apply to part (a) also?

 

[negation]

That's the tangential speed. (It's not a velocity since you're not quoting the direction, and anyway that changes. What's constant is the speed.)
You could have got this more easily: it travels πd = π150m in 30 minutes, so the speed is π*150/(30*60) m/s.

rω is scalar since r, theta and time are all scalar quantity and therefore v must be a scalar quantity too. My book must used the definition tangential/ linear velocity-I assume it was sloppy work by the author.
So what is the answer t part (a)? Is it 0ms^-1 or 0.262ms^-1?
Why am I getting conflicting answers?

What's your reasoning for that equation? You've divided the speed (constant) by the radius. That would give the angular speed, right?
You want the average acceleration, acceleration being a vector. For that, you need the change in velocity, also a vector. The speed is constant, v say, but the direction changes. At the bottom, the velocity is v horizontal. After 5 minutes, it's v in a different direction. What direction? What is the difference of the two velocities?

The question asked for average acceleration. a→ = v^2→/r is the instantaneous acceleration at anyone point of the rim. In uniform circular motion, the centripetal acceleration is the instantaneous acceleration is the average velocity since the change in velocity over any given time on the rim is the same. a→ = v^2/r is the centripetal acceleration isn't it? Why is it angular speed?
I'm getting really confused because I have been using my existing knowledge on problems I did and it was consistent.

I'm confused. What is this emphasis about "bottom"? At t=0, isn't the position at (0,0)? and at t(5) the wheel is displaced at an angle theta equivalent to 2pi/30 x 5mins? -this is the interpretation I have been working on.In 30mins, the change in theta is 2pi so in 5 mins the change in theta is 1.05 radians
p(2) = (75sin 1.05, 75cos 1.05) = (65, 37)
p(1) = (75sin 0, 75 cos0) = (0,75)

therefore, [p(2)-p(1)]/1800s = (65,-38)/1800s = (0.036, -0.02)
average velocity is therefore (0.036, -0.02)

Are my steps correct up till here?

If it isn't, how do I find vf and vi?

 

[negation]

At the risk of confusing negation further, let me put that differently.
If ω is angular speed and r is the magnitude of the radius then these are scalars and ωr gives the tangential speed, another scalar.
But angular velocity, $\vec\omega$ is a vector, and a specific radius can be a vector (from the axis to a given point on the rim) $\vec r$. Then the tangential velocity is the cross product $\vec \omega \times \vec r$, with the appropriate sign convention.

In effect, ω→ is the angular velocity and ω is the angular speed, am I right?

Back on part (a), I'm not sure I understand the question. Is it supposed to be the average angular velocity? The average velocity at the perimeter? The average velocity over the entire disc? If either of the last two, the answer would be zero, surely. But perhaps the "over a 5 minute interval" rider was supposed to apply to part (a) also?

I can't answer this. The questions posed by the book can be severely ambiguous.

 

[negation]

Edit

 

[haruspex]

The question asked for average acceleration. a→ = v^2→/r is the instantaneous acceleration at anyone point of the rim.

That's the magnitude of the acceleration, yes. Technically, acceleration is a vector. The vector equation is $\vec a = \vec \omega \times (\vec r \times \vec \omega)$.

In uniform circular motion, the centripetal acceleration is the instantaneous acceleration is the average velocity

Not sure what you meant there. An acceleration is not a velocity, so it is not an average velocity either.

since the change in velocity over any given time on the rim is the same.

If the time period is fixed, yes. Over half a cycle, the change in velocity is from ωr in one direction to ωr in the opposite direction, a change of 2ωr in time 2π/ω, yielding an average acceleration of magnitude ω²r/π. Over a complete cycle the average acceleration is zero.

a→ = v^2/r is the centripetal acceleration isn't it? Why is it angular speed?

My mistake - I missed the ^2.

What is this emphasis about "bottom"?

Just picking an arbitrary point in the revolution for the start of the 5 minute interval. The point you pick won't matter because the question asks for the magnitude of the acceleration.

At t=0, isn't the position at (0,0)? and at t(5) the wheel is displaced at an angle theta equivalent to 2pi/30 x 5mins? -this is the interpretation I have been working on.

Fine.

In 30mins, the change in theta is 2pi so in 5 mins the change in theta is 1.05 radians

Better to keep it in terms of rational multiples of pi. π/3 in this case.

p(2) = (75sin 1.05, 75cos 1.05) = (65, 37)
p(1) = (75sin 0, 75 cos0) = (0,75)

therefore, [p(2)-p(1)]/1800s = (65,-38)/1800s = (0.036, -0.02)
average velocity is therefore (0.036, -0.02)

That's the sort of approach you need for part (a) if that question is also supposed to be only over a 5 minute interval. But I believe we're dealing with part (b) here.
The average velocity is of no interest in part (b). You want the actual velocity (as vectors) at two instants 5 minutes apart. You then take the difference of these as vectors (giving the change in velocity) divide by 5 minutes to get the acceleration (a vector), then take its magnitude.

 

[negation]

That's the sort of approach you need for part (a) if that question is also supposed to be only over a 5 minute interval. But I believe we're dealing with part (b) here.
The average velocity is of no interest in part (b). You want the actual velocity (as vectors) at two instants 5 minutes apart. You then take the difference of these as vectors (giving the change in velocity) divide by 5 minutes to get the acceleration (a vector), then take its magnitude.

Ok. I've overshot.

The issue now is how do I find vf and vi or in other words delta v?
and what is this velocity? Is it angular velocity or tangential velocity?

 

[negation]

What the answer gave for part (a) was (0.22i + 0.13j)ms^-1
What I got was 0.22i - 0.13j

if p(2) = (65m, 37m )and p(1) = (0m, 75m) then [p(2) - p(1)]/300s = 0.22i - 0.13j
so I'm going to take it that the answer sheet is flawed on this one. It's really really annoying to be honest. Also, the question asked for magnitude so I'm not sure why is it giving the answer in unit vector.

The answer to part(b) as given by the sheet is (-4.4i +7.6j)ms^-2 which I have no clue as to how should I arrive.

 

[haruspex]

What the answer gave for part (a) was (0.22i + 0.13j)ms^-1
What I got was 0.22i - 0.13j

the question asked for magnitude so I'm not sure why is it giving the answer in unit vector.

Quite so. The question as posted only asks for the magnitude, so the two answers are the same.
If you want to compute the vector then you need more information. E.g. which way is it rotating?

The answer to part(b) as given by the sheet is (-4.4i +7.6j)ms^-2 which I have no clue as to how should I arrive.

I've already told you exactly what to do here.
You want the actual velocity (as vectors) at two instants 5 minutes apart.
Pick some arbitrary position, the bottom of the wheel, say. What is the tangential velocity of that point (a vector)?
5 minutes later, the wheel has rotated π/3. That point will have the same tangential speed but in a different direction. What is the new velocity vector? What is the difference of the two vectors?

 

[vela]

I'll note that the answer you're saying is for part (b) can't possibly be right. It's way too large given the information provided.

 

[negation]

Quite so. The question as posted only asks for the magnitude, so the two answers are the same.
If you want to compute the vector then you need more information. E.g. which way is it rotating?

It was not specified as to whether the direction of rotation is CCW or CW. My answer to part (a) was based on the assumption the entity rotates CW.
They're same if converted to magnitude since all magnitudes are positive scalars. But there clearly is a distinction if the answer remains in unit vector. I would assume the book made a blunder in this one. Can't see myself going wrong with my answer here.

I've already told you exactly what to do here.
You want the actual velocity (as vectors) at two instants 5 minutes apart.
Pick some arbitrary position, the bottom of the wheel, say. What is the tangential velocity of that point (a vector)?
5 minutes later, the wheel has rotated π/3. That point will have the same tangential speed but in a different direction. What is the new velocity vector? What is the difference of the two vectors?

I assume you meant tangential speed since r.ω are both scalar.
at time t = 0; r = 75m, ω = (0pi / 300s) = 0 \ni v=r.ω = 0ms^-1
at time t = 300s; r = 75m, ω = (1.05 radians/ 300s) \ni v = r.ω = 0.2625ms^-1
[0.2625 ms^-1 - 0ms^-1]/300s = 8.75e-4 ms^-2

 

[haruspex]

I assume you meant tangential speed since r.ω are both scalar.

No, I meant velocity, but you can get there by calculating the speed since the direction will be easy to determine.

at time t = 0; r = 75m, ω = (0pi / 300s) = 0 \ni v=r.ω = 0ms^-1

It is not getting faster. The wheel is turning at a constant rate. You already worked out what that rate is (ω) and the tangential speed (0.262ms^-1). Next, you need to find what direction a point on the periphery is moving at two points in time 5 minutes apart. That will give you two velocity vectors each of magnitude 0.262ms^-1 but in different directions. You want their vector difference.

 

[negation]

No, I meant velocity, but you can get there by calculating the speed since the direction will be easy to determine.

It is not getting faster. The wheel is turning at a constant rate. You already worked out what that rate is (ω) and the tangential speed (0.262ms^-1). Next, you need to find what direction a point on the periphery is moving at two points in time 5 minutes apart. That will give you two velocity vectors each of magnitude 0.262ms^-1 but in different directions. You want their vector difference.

The entity moves cw direction.
I've established vi =0ms^-1 and vf = 0.262ms^-1.
Average velocity is the difference between vf and vi over time 300s.

So isn't it just [vf-vi]/300s?

 

[BvU]

Hello negation. Appreciate your hard work. You 've got most of it but the ciphering creates mist and confusion. I don't think a little help is a spoiler at this point:

In your picture in #14 you have two vectors: v₂ and v₁. Draw them again in a new picture, where both grab at the origin. The angle between them is known, right? (Hint: don't use a calculator. Numbers like 0.523 -- which you later corrected to 1.047197551... -- are a lot less helpful than numbers like π / 3). Draw -v₁ and then v₂ + (-v₁). That is the change in the speed vector in 5 minutes. Note it is a vector and it doesn't point downwards. The magnitude is evident. Divide by 5 minutes and you get the average acceleration vector. Its length is π m/min². (Exactly. The 8.73... is only an approximation for π/3600. On the other hand, vela's 10^-5 seems to be a typo to me, so your -4 is better...).


If you want to learn something, repeat the exercise for 1 minute instead of 5 and so on, all the way to dt = 0. If all is well, you just might end up with the 9.138 * 10^-4. Or, more correctly ω² r being π²/3 m/min².

Don't let the hard work of calculating distract you from the insight: the instantaneous acceleration and the average acceleration are quite different beasts!

You'll know you understand it completely if you can swing a small weight at the end of a rope: what do you do to get it to go in a circle and what do you do to keep it circling...