 #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 * 6^{2} m^{2}+ 1/3 * 24 kg * 6^{2} m^{2})= 2808
Then I found initial angular momentum of the pole and rider just as the rider grabs the pole:
L = Iw = 2808 v_{0}/ 6 m = 468 v_{0}
Then I found the final angular momentum of the pole and rider when the pole has rotated 72 degrees:
L = (70+24 kg)(v_{f})(6 m) sin 72 deg = 539.256 v_{f}
Using conservation of momentum, you get v_{0} = 1.152 v_{f}.
Since v_{0} = √2gh = 4.427√h, v_{f} = 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)v_{0}^{2} + 1/2 I(v/r)^{2} + 0
= 1/2 (70+24)(v_{0}^{2}) + 1/2 (2808)(v_{0}/6)^{2} = 86 v_{0}^{2} = 86*4.427√h = 380.722 √h
Final energy after pole rotates 72 degrees= Linear kinetic + Rotational kinetic + Potential = 1/2 (M+m)v_{f}^{2} + 1/2 I(v_{f}/r)^{2} + (M+m)g*h_{final} = 1/2(70+24 kg)(3.842√h)^{2} + 1/2 (2808) (3.842√h/6)^{2} + (70+24)(9.8)(h_{final}) = 693.765 h + 575.678 h + 921.2 h_{final} = 1269.443 h + 921.2 h_{final}
1269.443 h + 921.2 h_{final} = 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
Then I found initial angular momentum of the pole and rider just as the rider grabs the pole:
L = Iw = 2808 v_{0}/ 6 m = 468 v_{0}
Then I found the final angular momentum of the pole and rider when the pole has rotated 72 degrees:
L = (70+24 kg)(v_{f})(6 m) sin 72 deg = 539.256 v_{f}
Using conservation of momentum, you get v_{0} = 1.152 v_{f}.
Since v_{0} = √2gh = 4.427√h, v_{f} = 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)v_{0}^{2} + 1/2 I(v/r)^{2} + 0
= 1/2 (70+24)(v_{0}^{2}) + 1/2 (2808)(v_{0}/6)^{2} = 86 v_{0}^{2} = 86*4.427√h = 380.722 √h
Final energy after pole rotates 72 degrees= Linear kinetic + Rotational kinetic + Potential = 1/2 (M+m)v_{f}^{2} + 1/2 I(v_{f}/r)^{2} + (M+m)g*h_{final} = 1/2(70+24 kg)(3.842√h)^{2} + 1/2 (2808) (3.842√h/6)^{2} + (70+24)(9.8)(h_{final}) = 693.765 h + 575.678 h + 921.2 h_{final} = 1269.443 h + 921.2 h_{final}
1269.443 h + 921.2 h_{final} = 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