- #1
solarcat
- 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
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