Calculating Average Velocity and Acceleration of the Singapore Flyer

[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 [itex]\ni v=r.ω = 0ms^-1[/itex]
at time t = 300s; r = 75m, ω = (1.05 radians/ 300s) [itex]\ni v = r.ω = 0.2625ms^-1[/itex]
[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 [itex]\ni v=r.ω = 0ms^-1[/itex]

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...

 

[negation]

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...

I have no problem spending days solving a problem. My life revolves around physics and math. I'm not satisfied leaving a problem unsolved. But indeed, this question is a real pain mainly because the definition of the terms comes across as rather chaotic to me.

the question asked about average acceleration, can I presume that are asking about [tangential velocity 2 - tangential velocity 1]/300s? What I am also confused is with the [v2+(-v1)]/300s that you stated. Why is there a negative?
As to your question, what do you mean by grab at the origin? I presume you meant for the radius to extends from (0,0)?
At t=0, my radius points straight north with tangential velocity 1 perpendicular to the radius and at t=300s, my radius points at 12.05 with tangential velocity 2 at perpendicular to this radius. The difference in radians between the radius is 1.05... Or pi/3 as you put it.
This change in angle is in similar ratio to the change in angle between the 2 tangential velocity vector.
This is based on the premise
delta r / r = delta v/v.
Am I right with the above? If I am, I will proceed to work further.

 

[BvU]

Your drawing in #14 shows two vectors. But they are attached at different points, so it's a little difficult to draw the difference vector. I tried to explain it stepwise: a-b = a+ (-b), even in the vector world. The word grab confused you, but I was afraid the words start or originate would also do that...

My impression is yes, they want the average acceleration going from vector velocity 1 to vector velocity 2. v2 minus v1. Better not use the term tangential velocity here to prevent someone from thinking it concerns a number.

And yes, in circular motion radius vector and velocity vector are perpendicular (otherwise it wouldn't be circular).

The red vector is v₂ - v₁. In other words, the average vector change in vector velocity. Divide by time and bingo.

 

[negation]

Your drawing in #14 shows two vectors. But they are attached at different points, so it's a little difficult to draw the difference vector. I tried to explain it stepwise: a-b = a+ (-b), even in the vector world. The word grab confused you, but I was afraid the words start or originate would also do that...

My impression is yes, they want the average acceleration going from vector velocity 1 to vector velocity 2. v2 minus v1. Better not use the term tangential velocity here to prevent someone from thinking it concerns a number.

And yes, in circular motion radius vector and velocity vector are perpendicular (otherwise it wouldn't be circular).

The red vector is v₂ - v₁. In other words, the average vector change in vector velocity. Divide by time and bingo.

The angle between v2 and v1 does look pi/3 so it makes sense. It would make more sense if [v2+v1]/300s.
What is the significance of -v1? I'm puzzled with the set up of the diagram.
The angle between v2 and -v1 does looks > pi/3 so I was wondering why do you take v2+(-v1)

 

[negation]

Edit

 

[negation]

Your drawing in #14 shows two vectors. But they are attached at different points, so it's a little difficult to draw the difference vector. I tried to explain it stepwise: a-b = a+ (-b), even in the vector world. The word grab confused you, but I was afraid the words start or originate would also do that...

My impression is yes, they want the average acceleration going from vector velocity 1 to vector velocity 2. v2 minus v1. Better not use the term tangential velocity here to prevent someone from thinking it concerns a number.

And yes, in circular motion radius vector and velocity vector are perpendicular (otherwise it wouldn't be circular).

The red vector is v₂ - v₁. In other words, the average vector change in vector velocity. Divide by time and bingo.

part(a)

[p(2)-p(1)]/300s
p(2) = [ 75m sin (pi/3) , 75m cos (pi/3)] = (65m , 38m)
p(1) = [75m sin(0) , 75m cos(0)] = (0m , 75m)
[p(2) - p(1)]/300 = (0.22i + 0.13j)ms^-1

part(b)

t = 0;
v 1 = r.ω = 75m [0 rad/ 75m] = 0 rad ms^-1

t = 5;

v2 = r.ω = 75m [(pi/3)/75m] = 0.261799386 rad ms^-1

v2 - v1 = 0.261799387m rad s^-1 = Δv

a→ = v→^2/r = (0.261799387m rad s^-1)^2 / 75 = 9.138...e-4

 

[BvU]

The angle doesn't just look like that, it IS. Remember the change in Θ = 2π/30 . 5 . The change in vector velocity v is what you need to change velocity vector v₁ into vector v₂. That is v₁ -v₂. That is the red vector: When you take add v₁ and add the red vector you end up at v₂. Now two equal length sides at π/3 make an equilateral triangle. So the length of the red vector is also 5 π m/min.

Average acceleration = (change in vector velocity) / time = (5 π m/min) / (5 min) = π m/min², with a direction as given by the red vector. Horizontal component -π/2 m/min², vertical component -1/2π √3 m/min². Divide by 3600 to get m/s².

---

Your drawing in #55 shows two equal length vectors on the left but on the right you draw v2 much too long, suggesting that delta v points straight down!

Then, in part (a), you calculate change in position / time. That gives you the average velocity vector. Not helpful for the average acceleration.
Some improvement is possible: we go from [0,75] to [65,37.5] so that gets us [0.216506351 -0.125] m/s on average . i.e. 0.25 m/s speed, less than the π/12 !

 

[negation]

The angle doesn't just look like that, it IS. Remember the change in Θ = 2π/30 . 5 . The change in vector velocity v is what you need to change velocity vector v₁ into vector v₂. That is v₁ -v₂. That is the red vector: When you take add v₁ and add the red vector you end up at v₂. Now two equal length sides at π/3 make an equilateral triangle. So the length of the red vector is also 5 π m/min.

Average acceleration = (change in vector velocity) / time = (5 π m/min) / (5 min) = π m/min², with a direction as given by the red vector. Horizontal component -π/2 m/min², vertical component -1/2π √3 m/min². Divide by 3600 to get m/s².

---

Your drawing in #55 shows two equal length vectors on the left but on the right you draw v2 much too long, suggesting that delta v points straight down!

Then, in part (a), you calculate change in position / time. That gives you the average velocity vector. Not helpful for the average acceleration.
Some improvement is possible: we go from [0,75] to [65,37.5] so that gets us [0.216506351 -0.125] m/s on average . i.e. 0.25 m/s speed, less than the π/12 !

How did you get 5pi? Infact, without knowledge of the length of v1 and v2, even knowing the change in angle between v1 and v2 does not help.

 

[vela]

View attachment 65887


part(a)

[p(2)-p(1)]/300s
p(2) = [ 75m sin (pi/3) , 75m cos (pi/3)] = (65m , 38m)
p(1) = [75m sin(0) , 75m cos(0)] = (0m , 75m)
[p(2) - p(1)]/300 = (0.22i + 0.13j)ms^-1

Oh, so part (a) was for only the five-minute interval, not a complete turn as you said before?

part(b)

t = 0;
v 1 = r.ω = 75m [0 rad/ 75m] = 0 rad ms^-1

Why do you keep plugging in 0 for ##\omega##? The angular speed is the rate at which ##\theta## changes. It's ##d\theta/dt##; it's not ##\theta/t##. Do you understand that when you say v=0, you're saying the wheel is not moving?

t = 5;

v2 = r.ω = 75m [(pi/3)/75m] = 0.261799386 rad ms^-1

v2 - v1 = 0.261799387m rad s^-1 = Δv

a→ = v→^2/r = (0.261799387m rad s^-1)^2 / 75 = 9.138...e-4

 

[BvU]

The 5 pi m/min is 2pi radians, a full circle, times the radius, divided by 30 minutes

 

[BvU]

Continuing with part (b). You are falling back to scalars again! Do I read a v1 = 0 here ? Of the kind rad m/s ?
But you divide by meters !?
That can't be a speed, velocity, whatever!

If you want to do it analogous to what you did in part (a):


t = 0: v₁ = [ 2π/30 * 75 , 0 ] m/min
t = 5: v₂ = [ 2π/30 * 75 cos(π/3), -2π/30 * 75 sin(π/3) ] m/min

t = 0 to 5: average a = Δv/Δt = 2π/30 * 75 [1- cos(π/3), -sin(π/3) ] / 5 m/min²

Or, again, π [-1/2, -sin(π/3) ] m/min² magnitude: π

And, just like with the average velocity, the magnitude of the average acceleration vector is a little less than the instantaneous acceleration (which was π²/3).

---

you re-calculate v²/r which gives you the magnitude of the instantaneous acceleration.

 

[negation]

Oh, so part (a) was for only the five-minute interval, not a complete turn as you said before?

It's 5 mins. The question was ambiguous and I decided to work out an interval of 5 mins to determine if it fitted with the book's answer.

Why do you keep plugging in 0 for ##\omega##? The angular speed is the rate at which ##\theta## changes. It's ##d\theta/dt##; it's not ##\theta/t##. Do you understand that when you say v=0, you're saying the wheel is not moving?

Ok. so tangential speed, v = 0.262ms^-1

The 5 pi m/min is 2pi radians, a full circle, times the radius, divided by 30 minutes

2pi/1800s = 3.5 e-3 ms^-1 and is nothing close to 5pi.

 

[vela]

2pi/1800s = 3.5 e-3 ms^-1 and is nothing close to 5pi.

BvU calculated the speed as the distance a point moves in one revolution, which is the circumference of the circle, divided by the time it takes to go around the circle once, in units of meters per minute. You're comparing that to the angular speed, which is the angular displacement of one revolution divided by the time for one revolution, which should be in units of radians per second, not meter per second. They're two completely different quantities in different units. Of course they don't match numerically.

 

[BvU]

As we saw earlier on, after a full revolution the average velocity is 0. And the average acceleration is 0 too!

My 'learning suggestion' earlier on would demonstrate that average velocity and acceleration vectors go to instantaneous for smaller and smaller Δt

I am very sorry I can't come up with a little film or something (anybody have a link?) because there you would see a position vector r rotating at constant angular velocity ω. A vector!
ω is perpendicular to the plane in which the position vector rotates (into the plane for clockwise, i think. Someone correct me if I am wrong).
ω is constant, so the angular acceleration α (funny alpha!) is zero. Can't show...

You would also see a velocity vector v, perpendicular to the position vector r , in the plane in which the position vector rotates. Now comes the hard part: v = ω [itex]\times[/itex] r. The [itex]\times[/itex] is a vector outer product^* (as opposed to the dot product, the inner product). Since ω is constant and |r| is a constant too, |v| is constant, but the thing rotates in sync with r

You would also see (after a while, with a comment) the instantaneous acceleration a as well. Again somewhat toughly: a = ω [itex]\times[/itex] v. It works out to a = -|ω|² r, so always pointing inwards. And rotating in sync.

^* a [itex]\times[/itex] b is perpendicular to a and b (rotate a over the smallest angle to b and follow the corkscrew for the direction. The magnitude is |a| |b| sin(α) where α is the angle mentioned.

But the animation app would be very welcome. A link anyone?

 

[haruspex]

The entity moves cw direction.
I've established vi =0ms^-1 and vf = 0.262ms^-1.

No you have not. It is moving at constant speed: 0.262ms^-1 all the time.

 

[negation]

Continuing with part (b). You are falling back to scalars again! Do I read a v1 = 0 here ? Of the kind rad m/s ?
But you divide by meters !?
That can't be a speed, velocity, whatever!

If you want to do it analogous to what you did in part (a):t = 0: v₁ = [ 2π/30 * 75 , 0 ] m/min
t = 5: v₂ = [ 2π/30 * 75 cos(π/3), -2π/30 * 75 sin(π/3) ] m/min

t = 0 to 5: average a = Δv/Δt = 2π/30 * 75 [1- cos(π/3), -sin(π/3) ] / 5 m/min²

in t=0, you did r.ω for x and y. I understand how you arrive at 2pi/30*75 for x but I don't understand how you arrive at 0 for y. Is r=0 for y or is ω=0 for y? I'm incline to say r=0 for y since base on your attachment, v1 is on the x-axis. It has a radius in the x-component but not in the y-component.
If I break it down:
t = 0: [2pi/30mins * cos(0) , 2pi/30mins* sin(0)]

Why is equally puzzling is the negative in your y-component in t=5. Where did it came from?

Edit: I got v2 - v1 = (0.22672, -0.131)e-4 ms^-1
divide it by 5 and I get
(7.5573, -4.366)e-4 ms^-1

 

[haruspex]

in t=0, you did r.ω for x and y. I understand how you arrive at 2pi/30*75 for x but I don't understand how you arrive at 0 for y. Is r=0 for y or is ω=0 for y?

The speed is rω. At t=0, BvU is considering the point at the bottom of the wheel. That point is moving horizontally, so the x component is rω and the y component is zero.
If you want to see how those components derive directly from r and ω then you have to do it in vectors: ##\vec v_0 = \vec r_0 \times \vec \omega##. Taking z to be towards you and taking the centre of the wheel to be the origin, ##\vec r_0 = <0, -r, 0>##, ##\vec \omega = <0, 0, ω>## (or maybe it's ##\vec \omega = <0, 0, -ω>##).
More generally, what will ##r_t## be? (Hint it will involve trig functions of ωt.)
At t = 5 minutes, that point of the wheel will be at ##\vec r_5##. So what will its velocity ##\vec v_5## be?

 

[negation]

The speed is rω. At t=0, BvU is considering the point at the bottom of the wheel. That point is moving horizontally, so the x component is rω and the y component is zero.
If you want to see how those components derive directly from r and ω then you have to do it in vectors: ##\vec v_0 = \vec r_0 \times \vec \omega##. Taking z to be towards you and taking the centre of the wheel to be the origin, ##\vec r_0 = <0, -r, 0>##, ##\vec \omega = <0, 0, ω>## (or maybe it's ##\vec \omega = <0, 0, -ω>##).
More generally, what will ##r_t## be? (Hint it will involve trig functions of ωt.)
At t = 5 minutes, that point of the wheel will be at ##\vec r_5##. So what will its velocity ##\vec v_5## be?

I have solved it I believe. answer is (7.5573, -4.366)e-4 ms^-1 although the book has the x and y value in reverse, i believe they set the diagram differently from mine. In the end, the magnitude would be the same as my answer above.

 

[haruspex]

I have solved it I believe. answer is (7.5573, -4.366)e-4 ms^-1 although the book has the x and y value in reverse, i believe they set the diagram differently from mine. In the end, the magnitude would be the same as my answer above.

Yes, those are the right numbers, near enough. If you took the initial position at the bottom, and x as horizontal, positive in the direction of movement, and y upwards positive, then the average acceleration over 5 minutes would be (-4.363, +7.557).

 

[gjh]

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)

I grappled with this problem also: my approach was simply to set coordinate system at the center of the wheel. The initial position vector would be -75j. After 5 minutes, the wheel will have completed (5/30)*360 degrees or 60 degrees. The final position vector (after 5 mins) would be R = r (sin(theta)i +cos(theta)j). Substituting theta = 60 degrees and r = 75 m, you get R= 65 i -.37.5j. The average v = delta displacement/delta time. Substituting (final position vector - initial position vector)/ delta time = 65i-37.5j-(-37.5j) = (65i+37.5j)m/300 s = .22i+.13j m/s. Mag v is then ((.22)^2+(.13)^2)^.5 = .262 m/s. [This agreed with answer given in odd numbered solutions in the text - so think it correct].

The centripetal acceleration at 5 minutes is (Mag v)^2/R = (.262)^2/75 = 9(10^-4) m/s^2. [This also agree with the text for magnitude, but answer was also given in vector components as [-4.5i+7.8j]10^-4 - haven't been able to figure out how they got that. I tried accel(avg) = delta v/delta t, but can't get close to book answer for components of acceleration - maybe mixing avg and instantaneous accel? Not sure. Comments please!

 

[haruspex]

can't get close to book answer for components of acceleration

Please post your attempt,