Check out some similar questions!

dansbby · Oct 12, 2011, 04:16 PM

Some amusement parks have a double Ferris wheel, which consists of two
Vertically rotating wheels that are attached to each other by a bar that also
Rotates. There are eight gondolas equally spaced on each wheel. Riders
Experience a combination of two circular motions that provide a sensation
More thrilling than the classic single Ferris wheel. In particular, riders
Experience the greatest sensation when their rate of change in height is the
Greatest.

? Each of the two wheels is 6 m in
Diameter and revolves every 12 s.
? The rotating bar is 9 m long. The
Ends of the bar are attached to the
Centres of the wheels.
? The height from the ground to
The centre of the bar is 8 m. The
Bar makes a complete revolution
Every 20 s.
? A rider starts seated at the
Lowest position and moves
Counterclockwise.
? The bar starts in the vertical
Position.
Consider the height of a rider who begins the ride in the lowest car.

b) Explain how the sum of these two functions gives the rider?s height above
The ground after t seconds.
d) What is the maximum height reached by the rider? When does this occur?
e) What is the maximum vertical speed of the rider? When does this occur?

Unknown008 · Oct 13, 2011, 01:46 AM

Could you post what you got for parts a) and c)? For b), I don't know what functions it's talking about...

d) This most probably refer to the functions spoke in b) and for that, I sense that you will have to use differentiation and set to zero.

e) Without any calculation, this should occur when the rider is directly to the right of the centre of the wheel, and when the wheel itself is directly to the right of the centre of the rotating bar.

But given the restrictions, this can very well occur somewhere else, but it should clearly be close to that.

dansbby · Oct 13, 2011, 06:55 AM

For part a the equation for the small wheel is y=3sin(pi/6)t and thr big wheel is y=4.5sin((pi/10)(t-5))+8

Unknown008 · Oct 13, 2011, 07:25 AM

Okay, the first equation as you might have guessed, gives the position of the rider at any time t, relative to the center of the wheel. Except that when t = 0, in that equation, your rider is at the same height as the wheel, so it should be (t-3), just like the way you got (t-5) in the second equation.

The second equation on the other hand gives the position of the center of the wheel with respect to the ground, at any time t.

Do you see how, when they are added together, give the net height of the rider above the ground? Can you put that in your own words for part b)?

d) Okay, as I said earlier, there is differentiation for that part.

The net height of the rider is given by:

$h = 4.5\sin$\frac{\pi}{10}(t-5)$+8 + 3\sin$\frac{\pi}{6}(t-3)$$

Can you find the derivative of that and set the derivative to zero, solving for t? Then plug back t into the original h equation to get the maximum height?

e) For this one, get the second derivative of the height equation and solve for t. Then plug this into the first derivative.

If you realised it, the first derivative gives the rate of change of height (that is the speed the rider is going up and down) and the second derivative gives the acceleration of the rider with respect to the height.

Can you take it from here? :)

dansbby · Oct 13, 2011, 08:16 AM

Could you please Tell me the derivative for h I want to check if I did it right and also could you tell me the answer for part e. Also part c asks for the height at 2 minutes what would that be?

Unknown008 · Oct 13, 2011, 10:52 AM

Part c, you just put t = 2 min = 120 s since your equation is in seconds.

Okay, I'll do part (b), but you must show your answer for part (e) afterwards.

$h = 4.5\sin$\frac{\pi}{10}(t-5)$ + 8 + 3\sin$\frac{\pi}{6}(t-3)$$

$h = 4.5\sin$\frac{\pi t}{10}-\frac{\pi}{2}$ + 3\sin$\frac{\pi t}{6}-\frac{\pi}{2}$ + 8$

$h' = 4.5$\frac{\pi}{10}$\cos$\frac{\pi t}{10}-\frac{\pi}{2}$ + 3$\frac{\pi}{6}$\cos$\frac{\pi t}{6}-\frac{\pi}{2}$$

$h' = $\frac{9\pi}{20}$\cos$\frac{\pi t}{10}-\frac{\pi}{2}$ + $\frac{\pi}{2}$\cos$\frac{\pi t}{6}-\frac{\pi}{2}$$

Okay, that seems quite tough.

Did your instructor say anything about how many rounds the rotating bar goes? Because there are different points at which the rider can be said to reach the maximum height. By simple inspection, we see that the rotating bar gets at the maximum height every 10, then 30, then 50 seconds and so on. The wheel on the other hand, gets at its maximum height every 6, 18, 30, 42 seconds and so on and you can already see that at t = 30s, it is at its highest possible point.

ghghgh · Nov 17, 2011, 08:20 AM

Why do you find the derivative in part b?

ghghgh · Nov 17, 2011, 08:21 AM

Why do you find the derivative in part b?