Ferris Wheel Physics Problem: Finding the Landing Spot for Dropped Keys

In summary, a passenger on a Ferris wheel with a radius of 5.00m drops his keys at the 10 o'clock position while the wheel is 1.75m above the ground. The keys will land 58.0cm to the right of the base of the ride. To find the velocity at the time the keys were dropped, the circumference formula for a circle should be used to calculate the distance traveled in one revolution. The velocity at this time will be V0 for the purposes of projectile motion. The angle of V0 can be found using basic geometric properties, with an angle of 60 degrees being a possible answer.
  • #1
skateza
45
0

Homework Statement


A passenger on the ferris wheel described in problem 18 (Problem 18: Fairgoers ride a Ferris wheel with a radius of 5.00m The wheel completes one revolution every 32.0s) drops his keys when he is on he way up and at the 10 o'clock position. Where do the keys land relative to the base of the ride?

Also: the diagram reveals the ferris wheel is 1.75m above the ground, and the center of the wheel is = base of the ride.

Homework Equations


all projectile motion equations


The Attempt at a Solution



I found the solution to be 58.0cm but I am pretty sure i didn't do it right. my steps were
1)Setup a diagram using the ten o'clock position to be 30degrees with a hypotenuse of 5m.
2)Found dx
3)t=(1/12)<rotation of wheel*(32s)<How long for the whole wheel to rotate
4) Found Vox
5) Found Voy
6) Used Voy in y(t) equation, where y(t)=0, gave me a quadratic i solved for t=3.44s (Supposing 0 = ground)
7) found x(3.44)
8) Subtracted the distance from center of the wheel from the radius

=58.0cm to the right of the base.

My intuition tells me i did something wrong at the start. My teacher told me today, the radius vector is perpinduclar to the Vo vector, so i think i can use that with pythag to find Vo somehow, any insight?
 
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  • #2
How did you obtain your current v0? I don't see it in your procedure.

I don't think you can obtain v0x and v0y without that.
 
  • #3
i used dx and dy (part 2) divided by t(part 3) to find Vo, which I'm pretty sure is wrong
 
  • #4
There's a simple way for getting v0

You know the time it takes for a revolution.

v=d/t

Can you figure out the distance it travels in one revolution?
 
  • #5
C=2(pie)r^2

C=50(pie) m.

V=50(pie)/32s

V=?
... this gives me velocity, a ferris wheel would have constant velocity but changing acceleration, due to the change in direction. Therefore if it has constant velocity, unless the wheel was at rest, this V value we just found is Vo?
 
  • #6
Check your circumference formula first. ;P

About your other question, though the velocity vector of the person in the ferris wheel is constantly changing direction, at the instant the coin is released it is no longer moving in a circle and will behave as a projectile.

In short, this V you obtained (once you correct your formula) is V0 for the purpose of projectile motion.
 
  • #7
Ok that makes sense, one more thing, can we assume that a clock at 10 o'clock, has an angle of 150degrees from standard position? Seeing as how from 9-12 forms a right angle with three angles in between
 
  • #8
basically now i just want to make sure i have the right angle for Vo, i think it's 60 degrees, using basic geometric properties but can anyone double check this?
 
  • #9
bump, anyone know the initial angle for Vo?
 
  • #10
60 sounds right.
 
  • #11
new question about this question

I am currently working on this exact same question. I have my Vo but am really unsure what needs to be done to get Vox and Voy.
 

1. How does the weight of the passengers affect the motion of a Ferris wheel?

The weight of the passengers does not significantly affect the overall motion of the Ferris wheel. The Ferris wheel rotates at a constant speed, and the gravitational force acting on the passengers is balanced by the centripetal force from the wheel. Therefore, the weight of the passengers does not change the motion of the wheel.

2. Why does the Ferris wheel move in a circular motion?

The Ferris wheel moves in a circular motion due to the centripetal force acting on the riders. The centripetal force, provided by the support structure of the wheel, keeps the riders moving in a circular path rather than flying off in a straight line due to inertia.

3. How does the height of the Ferris wheel affect the physics of the ride?

The height of the Ferris wheel does not directly affect the physics of the ride. However, a higher Ferris wheel may have a larger circumference, resulting in a longer ride time and potentially more force acting on the riders as they move through the higher point of the circle.

4. Does the speed of the Ferris wheel change throughout the ride?

The speed of the Ferris wheel remains constant throughout the ride, as it is powered by a motor that maintains a constant rotation. However, the apparent speed of the wheel may change for riders at different points on the wheel, as they are moving at different distances from the center of rotation.

5. Why do some Ferris wheels have enclosed cabins while others have open seats?

The design of the ride does not affect the physics of the Ferris wheel. The choice of enclosed cabins or open seats is a design preference for safety, comfort, and aesthetics.

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