• Support PF! Buy your school textbooks, materials and every day products Here!

Water Slide Problem

  • Thread starter solarcat
  • Start date
  • #1
22
3

Homework Statement:

Imagine a water slide where the rider starts at a height h above the ground before sliding down the slide. After reaching the bottom of the slide, the rider grabs a pole hanging above which is pivoted about a stationary, frictionless axle. The pole and rider swing up 72 degrees and the person then lets go. The person can be modeled as a point mass. The mass of the pole is 24 kg and it is 6 meters long. The person's mass is 70 kg. What is the initial height h of the water slide?

Relevant Equations:

L = Iw = mvr sin (theta)
Moment of inertia = m(person)r^2 + 1/3 m(rod)r^2
Rotational energy = 1/2 Iw^2
Kinetic energy = 1/2 mv^2
Potential energy = mgh
v = wr
First I found the moment of inertia of the rod + rider to be I = (70 kg * 62 m2+ 1/3 * 24 kg * 62 m2)= 2808

Then I found initial angular momentum of the pole and rider just as the rider grabs the pole:
L = Iw = 2808 v0/ 6 m = 468 v0
Then I found the final angular momentum of the pole and rider when the pole has rotated 72 degrees:
L = (70+24 kg)(vf)(6 m) sin 72 deg = 539.256 vf
Using conservation of momentum, you get v0 = 1.152 vf.
Since v0 = √2gh = 4.427√h, vf = 3.842√h

Then I thought to use conservation of energy:
Initial energy as the rider grabs pole = Linear kinetic + Rotational kinetic + Potential = 1/2 (M+m)v02 + 1/2 I(v/r)2 + 0
= 1/2 (70+24)(v02) + 1/2 (2808)(v0/6)2 = 86 v02 = 86*4.427√h = 380.722 √h
Final energy after pole rotates 72 degrees= Linear kinetic + Rotational kinetic + Potential = 1/2 (M+m)vf2 + 1/2 I(vf/r)2 + (M+m)g*hfinal = 1/2(70+24 kg)(3.842√h)2 + 1/2 (2808) (3.842√h/6)2 + (70+24)(9.8)(hfinal) = 693.765 h + 575.678 h + 921.2 hfinal = 1269.443 h + 921.2 hfinal

1269.443 h + 921.2 hfinal = 380.722 √h
This is where I got stuck. I thought maybe you could use geometry and the Law of Sines to find h final, but even that wouldn't give you an equation you could solve. That's why I'm convinced I've done something wrong
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,550
5,460
You need to break it into three stages:
  1. Descent of slide
  2. Seizing the pole
  3. Upswing
(It doesn’t say so, but you need to assume there is no remaining angular velocity at 72 degrees.)
For each stage, what conservation laws can be used?
 
  • #3
22
3
1. Descent of slide - Use conservation of energy to find speed at bottom of slide
mpersongh = 1/2 mpersonv2
gh = 1/2 v2
2gh = v2
v = √2gh

2. Seizing the pole - Use conservation of angular momentum
Moment of inertia of person = (70 kg) (6 m)2 = 2520
Moment of inertia of rod + person = 2808, as calculated above

w0*r = v = √2gh
w0 = (√2gh)/6 = 0.738 √h

2520 (0.738 √h) = 2808 wf
wf = 0.662 √h

3. Upswing - Conservation of energy
1/2 (70+24 kg) (0.662 √h rad/s * 6 m)2+ 1/2 (2808)( 0.662 √h)2 = (70+24)(9.8) (h final)
When I used the law of sines to find h, I got h = 7.053 sin 36 deg = 4.146 m
741.519 h + 615.295 h = 3818.972
h = 2.81
that can't be right...
 
Last edited:
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,550
5,460
There are many benefits in keeping everything symbolic, not plugging in any numbers until the very end.
1/2 (70+24 kg) (0.662 √h rad/s * 6 m)2
That's treating the whole mass of the rod as at 6m from the axle.
Anyway, you don't need that term. The entire KE is expressed in the next term:
1/2 (2808)( 0.662 √h)2
(70+24)(9.8) (h final)
How far does the mass centre of the rod rise?
 
  • #5
22
3
How far does the mass centre of the rod rise?
If you calculate the potential energy based on the mass center, you'd have to have some initial potential energy, right?

Initial Energy = 1/2 (2808)( 0.662 √h)2 + (94 kg) (9.8 m/s/s) (3 m) = 615.295 h + 2763.6

I'm having trouble calculating the final height. I drew this diagram, and I used 6 m/sin 54 degrees = x/sin 72 degrees, so x = 7.053. Then h1 = 7.053 sin 36 degrees = 4.146 and h2 = 3 cos 72 degrees = 0.927 total height = 5.072 m?
 

Attachments

  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,550
5,460
(94 kg) (9.8 m/s/s) (3 m)
The rider does not start at a height of 3m before the upswing.
= 615.295 h
Why do you have a reference to h on the right of the equation?
2763.6
How is this calculated? I am not saying it is wrong, but hard to follow when you just post numbers.
 
  • #7
22
3
The rider does not start at a height of 3m before the upswing.
No, but you asked "How far does the mass centre of the rod rise?" If you consider the change in height of the mass center of the rod, instead of the end of the rod as I did previously, you have to account for the fact that the mass center is initially at height R/2 = 3 m. Anyway, the problem gives you the moment of inertia of the rod as being I = 1/3 MR^2, and there is a hint given that says to first calculate the angular speed of rotation of the rider and rod just after the rider grabs on, in terms of the final height the *rider* reaches.

The 615.295 h comes from the initial kinetic energy just after the rider grabs on:
W = 0.662 √h, as calculated in my second post
W^2 = 0.438244 h
I = 2808, as calculated in my first post
Kinetic energy = 1/2 IW^2 = 1/2 (2808) (0.438224 h ) = 615.226 h

2763.6 is equal to the initial potential energy of the rod and rider, (94 kg) (9.8 m/s/s) (3 m. However, I don't think this is correct based on what I stated above.
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,550
5,460
you have to account for the fact that the mass center is initially at height R/2 = 3 m
Sure, but you have 94kg in that term, as though the rider also starts at 3m.
The 615.295 h comes from the initial kinetic energy just after the rider grabs on
I had misunderstood your equation. I thought the right hand side was intended to be the final energy, so involving the 72 degrees. If you had kept everything symbolic, as I requested, it wouid have been obvious. Please try to get into that habit; there are lots of benefits.
 

Related Threads on Water Slide Problem

  • Last Post
Replies
9
Views
7K
Replies
5
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
6
Views
463
Replies
3
Views
14K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
Top