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Ferris wheel rotation

  • Thread starter negation
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  • #51
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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.
 
  • #52
BvU
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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.
 

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  • #53
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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)
 
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  • #54
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  • #55
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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.

Untitled.jpg



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
 
  • #56
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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 !
 
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  • #57
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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.
 
  • #58
vela
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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
 
  • #59
BvU
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The 5 pi m/min is 2pi radians, a full circle, times the radius, divided by 30 minutes
 
  • #60
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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.
 
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  • #61
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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.
 
  • #62
vela
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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.
 
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  • #63
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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?
 
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  • #64
haruspex
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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.
 
  • #65
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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
 
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  • #66
haruspex
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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?
 
  • #67
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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.
 
  • #68
haruspex
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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).
 
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