# help-calc problem

• Apr 17th 2006, 08:59 PM
WeirdRamenLove
Pre-calc problem
How do I begin, this is the math problem from hell in my opinion. We can use any method we please to solve this, but it's gotta' be in, and quick. My prof says that it's due tenatively around the end of the week. I'm feeling a little sick because I don't really know where to start. So if anyone has helpful advice or suggestions, it would be really appreciated.

-problem-

intro:
You may have seen or heard about hte circus act in which someone dives off a high platform into a small tub of water. Well, the interactive circus troupe has come up with a new wrinkle on this act. The have attached the divers platform to one of the seats on a ferris wheel, so that it sticks out horizontally, perpendicular to the plane of the Ferris wheel. The tub of water is on a moving cart that runs along a track, parallel to the plane of the Ferris wheel, and passes under the end of the platform. As the Ferris wheel turns, an assistant holds the diver by the ankles. The assistant must let go at exactly the right moment, so that the diver will land in the moving tub of water.

what you actually need to know:
-The ferris wheel has a radius of 50 feet
-The center of the Ferris wheel is 65 feet above the ground
-The ferris wheel turns counterclockwise at a constant rate, making a complete turn every 40 seconds.
-When the cart starts moving, it is 240 feet to the left of the base of the ferris wheel.
-The cart moves to the right along the track at a constant speed of 15 feet per second
-The water level in the cart is 8 feet above the ground
-The cart starts moving as the platform passes the 3 o'clock position.

Let t=0 represent the time when the Ferris wheel passes the 3o'clock position, which is also when the cart begins moving. Let W represent the number of seconds that elapse between t=0 and the moment when the diver is released.

the question:
For what choice of W will the diver land in the tub of water?
• Apr 18th 2006, 10:35 AM
ticbol
Quote:

Originally Posted by WeirdRamenLove
How do I begin, this is the math problem from hell in my opinion. We can use any method we please to solve this, but it's gotta' be in, and quick. My prof says that it's due tenatively around the end of the week. I'm feeling a little sick because I don't really know where to start. So if anyone has helpful advice or suggestions, it would be really appreciated.

-problem-

intro:
You may have seen or heard about hte circus act in which someone dives off a high platform into a small tub of water. Well, the interactive circus troupe has come up with a new wrinkle on this act. The have attached the divers platform to one of the seats on a ferris wheel, so that it sticks out horizontally, perpendicular to the plane of the Ferris wheel. The tub of water is on a moving cart that runs along a track, parallel to the plane of the Ferris wheel, and passes under the end of the platform. As the Ferris wheel turns, an assistant holds the diver by the ankles. The assistant must let go at exactly the right moment, so that the diver will land in the moving tub of water.

what you actually need to know:
-The ferris wheel has a radius of 50 feet
-The center of the Ferris wheel is 65 feet above the ground
-The ferris wheel turns counterclockwise at a constant rate, making a complete turn every 40 seconds.
-When the cart starts moving, it is 240 feet to the left of the base of the ferris wheel.
-The cart moves to the right along the track at a constant speed of 15 feet per second
-The water level in the cart is 8 feet above the ground
-The cart starts moving as the platform passes the 3 o'clock position.

Let t=0 represent the time when the Ferris wheel passes the 3o'clock position, which is also when the cart begins moving. Let W represent the number of seconds that elapse between t=0 and the moment when the diver is released.

the question:
For what choice of W will the diver land in the tub of water?

Are all the data as posted okay? Specifically, is that 240 feet okay? Or is that 15 feet per second correct?
• Apr 18th 2006, 12:04 PM
topsquark
Quote:

Originally Posted by WeirdRamenLove
How do I begin, this is the math problem from hell in my opinion. We can use any method we please to solve this, but it's gotta' be in, and quick. My prof says that it's due tenatively around the end of the week. I'm feeling a little sick because I don't really know where to start. So if anyone has helpful advice or suggestions, it would be really appreciated.

-problem-

intro:
You may have seen or heard about hte circus act in which someone dives off a high platform into a small tub of water. Well, the interactive circus troupe has come up with a new wrinkle on this act. The have attached the divers platform to one of the seats on a ferris wheel, so that it sticks out horizontally, perpendicular to the plane of the Ferris wheel. The tub of water is on a moving cart that runs along a track, parallel to the plane of the Ferris wheel, and passes under the end of the platform. As the Ferris wheel turns, an assistant holds the diver by the ankles. The assistant must let go at exactly the right moment, so that the diver will land in the moving tub of water.

what you actually need to know:
-The ferris wheel has a radius of 50 feet
-The center of the Ferris wheel is 65 feet above the ground
-The ferris wheel turns counterclockwise at a constant rate, making a complete turn every 40 seconds.
-When the cart starts moving, it is 240 feet to the left of the base of the ferris wheel.
-The cart moves to the right along the track at a constant speed of 15 feet per second
-The water level in the cart is 8 feet above the ground
-The cart starts moving as the platform passes the 3 o'clock position.

Let t=0 represent the time when the Ferris wheel passes the 3o'clock position, which is also when the cart begins moving. Let W represent the number of seconds that elapse between t=0 and the moment when the diver is released.

the question:
For what choice of W will the diver land in the tub of water?

Of course it's the Math problem from Hell! It's really a Physics problem. :)

Seriously, conceptually this isn't so bad, there's just a lot to keep track of. The problem winds up being tricky to solve for in the end, but I'll get you set up for it.

First, there are two objects to consider: the diver and the water bucket. Both of these objects need to have the same coordinates at time t = W. So I am going to derive expressions for each object as they move as a function of W.

Over all: I am going to choose an origin (where x and y are both 0) directly under the center of the ferris wheel on the ground. I will choose +x to the left and +y upward.

The bucket of water:
The y coordinate of the water level is easy: y = 8 (ft) is constant. (I'm going to drop the units from now on.) The x coordinate is almost as simple. The bucket starts at x0 = -240 and moves at a constant v = 15 (ft/s), so the x coordinate is x = -240 + 15t. So at t = W: x = -240 + 15W, y = 8.

The diver:
The diver is a bit trickier. The diver starts with rotational motion, then is released to go into free-fall motion. These are different motions. Let's start with the rotational motion.

The ferris wheel has a period of T = 40 s. Thus the angular frequency is $\displaystyle \omega = 2\pi /T=2\pi/40=0.15708$. (For the record this is 0.15708 rad/s.) The diver here is simply moving in a circle with radius 50 (ft), so his speed is a constant $\displaystyle v=r\omega=(50)(0.15708)=7.85398 \, ft/s$. Now, the speed is constant, but we need the components of the diver's velocity in the x and y directions. Again, the diver is simply moving in a circle so this is easy: $\displaystyle v_x=-r\omega cos(\omega t) = -(50)(0.15708)*cos(0.15708*t)$ and $\displaystyle v_y=r\omega sin(\omega t)=(50)(0.15708)*sin(0.15708*t)$. (Be sure your calculator is in radian mode!)

We will also need the x and y coordinates of the diver. This is, again, fairly simple to figure: $\displaystyle x = -50cos(\omega t)$ and $\displaystyle y = 65 + 50 sin( \omega t)$. (Recall the center of the ferris wheel is at (0, 65) and the diver starts at 3 o'clock and thus starts at (-50, 65).)

The diver is released sometime before t = W. Call this time t' for reference.

So when the diver is released, he is at $\displaystyle x = -50*cos(0.15708 t')$ and $\displaystyle y = 65+50*sin(0.15708 t')$ and has velocity components
$\displaystyle v_x=-(50)*(0.15708)*cos(0.15708 t')$
$\displaystyle =-7.85398*cos(0.15708 t')$
and
$\displaystyle v_y=7.85398*sin(0.15708 t')$.

Whew! Take a breather for a minute. We're getting there!

Now for the free-fall motion.

The general formula for motion in the x-direction under constant acceleration is $\displaystyle x=x_0+v_{0x}(t-t_0)+1/2*a_x(t-t_0)^2$. (You don't normally see the t0 term since we usually set that to 0. We can't easily do that here, so I'm not going to try.) Since the diver is in free-fall, ax = 0. We know the position and velocity component when the diver was released (at t = t') so:
$\displaystyle x = (-50*cos(0.15708 t'))$
$\displaystyle -(7.85398*cos(0.15708 t'))*(t - t')$

The y formula works the same way, except now the acceleration is the acceleration due to gravity: ay = -32 ft/s/s. (Note the "-" sign. This is because the acceleration is downward.)
$\displaystyle y=(65+50*sin(0.15708 t'))$
$\displaystyle +(7.85398*sin(0.15708 t'))*(t - t') -16(t-t')^2$

Hold on! We're almost there!

So now we have coordinates for both the bucket and diver as functions of t. Recall that in order for the diver to get into the water these coordinates must be equal. So:
$\displaystyle x = -240 + 15W = (-50*cos(0.15708 t'))$
$\displaystyle - (7.85398*cos(0.15708 t'))*(W - t')$

and
$\displaystyle 8 = (65+50*sin(0.15708 t'))+$
$\displaystyle (7.85398*sin(0.15708 t'))*(W - t') -16(W-t')^2$

This is the Math problem you need to solve. I haven't done it yet. It's going to be tricky since it's a system of equations in two unknowns, t' and W, where the t' appears both inside and outside of a trig function. I'll work at it, but off-hand, I don't think I've ever managed to solve one of these. Perhaps one of the Math guru's here will be nice enough to lend a hand at this stage.
(Hopefully they've read this to the end! :))

Best of luck!
-Dan

PS Just had a thought. If we use a new time coordinate T such that T = W - t' and solve the system for t' and T we can keep the t' strictly inside the trig functions in the y equation. Might be helpful, might not be.
• Apr 18th 2006, 09:29 PM
WeirdRamenLove
Thank you so much, I didn't even think to look at it as a physics problem (I blame sleep deprevation for that.) I ended up going at this in this manner:

Basically, find the function of the ferris wheel using:
y=a sin b (x plus or minus theta) plus or minus c
then find the derivative of that function. That should give me any tangent off the wheel. Then progress from there.

I have no idea if I'm even going in marginally the right direction with this. I'm going to look at your way and see if I can manage it better. Thank you again for the help.
• Apr 18th 2006, 11:46 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Of course it's the Math problem from Hell! It's really a Physics problem. :)

Seriously, conceptually this isn't so bad, there's just a lot to keep track of. The problem winds up being tricky to solve for in the end, but I'll get you set up for it.

First, there are two objects to consider: the diver and the water bucket. Both of these objects need to have the same coordinates at time t = W. So I am going to derive expressions for each object as they move as a function of W.

Over all: I am going to choose an origin (where x and y are both 0) directly under the center of the ferris wheel on the ground. I will choose +x to the left and +y upward.

The bucket of water:
The y coordinate of the water level is easy: y = 8 (ft) is constant. (I'm going to drop the units from now on.) The x coordinate is almost as simple. The bucket starts at x0 = -240 and moves at a constant v = 15 (ft/s), so the x coordinate is x = -240 + 15t. So at t = W: x = -240 + 15W, y = 8.

The diver:
The diver is a bit trickier. The diver starts with rotational motion, then is released to go into free-fall motion. These are different motions. Let's start with the rotational motion.

The ferris wheel has a period of T = 40 s. Thus the angular frequency is $\displaystyle \omega = 2\pi /T=2\pi/40=0.15708$. (For the record this is 0.15708 rad/s.) The diver here is simply moving in a circle with radius 50 (ft), so his speed is a constant $\displaystyle v=r\omega=(50)(0.15708)=7.85398 \, ft/s$. Now, the speed is constant, but we need the components of the diver's velocity in the x and y directions. Again, the diver is simply moving in a circle so this is easy: $\displaystyle v_x=-r\omega cos(\omega t) = -(50)(0.15708)*cos(0.15708*t)$ and $\displaystyle v_y=r\omega sin(\omega t)=(50)(0.15708)*sin(0.15708*t)$. (Be sure your calculator is in radian mode!)

We will also need the x and y coordinates of the diver. This is, again, fairly simple to figure: $\displaystyle x = -50cos(\omega t)$ and $\displaystyle y = 65 + 50 sin( \omega t)$. (Recall the center of the ferris wheel is at (0, 65) and the diver starts at 3 o'clock and thus starts at (-50, 65).)

The diver is released sometime before t = W. Call this time t' for reference.

So when the diver is released, he is at $\displaystyle x = -50*cos(0.15708 t')$ and $\displaystyle y = 65+50*sin(0.15708 t')$ and has velocity components
$\displaystyle v_x=-(50)*(0.15708)*cos(0.15708 t')$
$\displaystyle =-7.85398*cos(0.15708 t')$
and
$\displaystyle v_y=7.85398*sin(0.15708 t')$.

Whew! Take a breather for a minute. We're getting there!

Now for the free-fall motion.

The general formula for motion in the x-direction under constant acceleration is $\displaystyle x=x_0+v_{0x}(t-t_0)+1/2*a_x(t-t_0)^2$. (You don't normally see the t0 term since we usually set that to 0. We can't easily do that here, so I'm not going to try.) Since the diver is in free-fall, ax = 0. We know the position and velocity component when the diver was released (at t = t') so:
$\displaystyle x = (-50*cos(0.15708 t'))$
$\displaystyle -(7.85398*cos(0.15708 t'))*(t - t')$

The y formula works the same way, except now the acceleration is the acceleration due to gravity: ay = -32 ft/s/s. (Note the "-" sign. This is because the acceleration is downward.)
$\displaystyle y=(65+50*sin(0.15708 t'))$
$\displaystyle +(7.85398*sin(0.15708 t'))*(t - t') -16(t-t')^2$

Hold on! We're almost there!

So now we have coordinates for both the bucket and diver as functions of t. Recall that in order for the diver to get into the water these coordinates must be equal. So:
$\displaystyle x = -240 + 15W = (-50*cos(0.15708 t'))$
$\displaystyle - (7.85398*cos(0.15708 t'))*(W - t')$

and
$\displaystyle 8 = (65+50*sin(0.15708 t'))+$
$\displaystyle (7.85398*sin(0.15708 t'))*(W - t') -16(W-t')^2$

This is the Math problem you need to solve. I haven't done it yet. It's going to be tricky since it's a system of equations in two unknowns, t' and W, where the t' appears both inside and outside of a trig function. I'll work at it, but off-hand, I don't think I've ever managed to solve one of these. Perhaps one of the Math guru's here will be nice enough to lend a hand at this stage.
(Hopefully they've read this to the end! :))

Best of luck!
-Dan

PS Just had a thought. If we use a new time coordinate T such that T = W - t' and solve the system for t' and T we can keep the t' strictly inside the trig functions in the y equation. Might be helpful, might not be.

This is going to have to be solved numericaly. In which case you have
probably carried the analysis further that is necessary or convienient.

$\displaystyle \tau(W)$ - the time that the diver reaches the 8' level when releast at time $\displaystyle W$.

$\displaystyle \chi(W,\tau(W))$ - the x-coordinate of the diver at time $\displaystyle \tau(W)$ when released at time $\displaystyle W$.

$\displaystyle d(\tau(W))=-240+15 \tau(W)$ - the x-coordinate of the cart at time $\displaystyle \tau(W)$.

We are now interested in finding $\displaystyle W$ such that:

$\displaystyle d(\tau(W))=\chi(W,\tau(W))$

which can be done by binary search (try assuming the $\displaystyle W \in [0,20]$ for starters).

RonL