You are designing a slide for a water park. In a sitting position, park guests slide...

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance h down the water-slide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0∘, and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole’s moment of inertia is given by I=1/3ML^2 where L = 6.00 m is the length of the pole and M = 31.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0∘ after the collision?

Expert Solution

H= 5.8114 m. ANS.

genius_generous answered 2 years ago

#1. [Water Slide & Swing] You are designing a slide for a water park. In a...

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