Calculating Average Velocity and Acceleration of the Singapore Flyer

[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 \times r. The \times 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 \times v. It works out to a = -|ω|² r, so always pointing inwards. And rotating in sync.

^* a \times 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,

 

[gjh]

Started over. Back to vectors. v =.26[cos 60i+sin 60j) = .13i+.22j (this is instantaneous velocity). avg a = delta v/delta t = (.13i+.22j-.26i)/300 = (-.13i+.22j)/300 = (-4.3i+8.7j)10^-4 - this is closer, but magnitude of a would be 9.7(10)^-4 ... which is way off from 9(10)^-4. Almost there ... kinda/sorta. Comments?

 

[haruspex]

Started over. Back to vectors. v =.26[cos 60i+sin 60j) = .13i+.22j (this is instantaneous velocity). avg a = delta v/delta t = (.13i+.22j-.26i)/300 = (-.13i+.22j)/300 = (-4.3i+8.7j)10^-4 - this is closer, but magnitude of a would be 9.7(10)^-4 ... which is way off from 9(10)^-4. Almost there ... kinda/sorta. Comments?

Looks like rounding error.
Don't be so quick to plug in numbers. Keep everything algebraic as long as you can, only plugging in numbers at the end. You should find it simplifies a lot and gives an exact result,

 

[gjh]

Yes, plus I made a stupid computational error. .22j/300 = 7.33*(10)^4 not 8.7jx(10)-4. Thus, my answer would be a (average) =(-4.3i+7.3j)10^-4 in component form. mag a then is sqrt [(-4.3)^2+(7o.3)^2(10^-4)] = 8.9*10^-4 m/s^2. This is very close to book answer of -4.4i+7.6j)*10^-4 with mag 8.8 m/s^2. So rounding error was part of the problem - and carelessness. Thx for the tip!