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

#1. [Water Slide & Swing] 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 theta max , 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 = (1/3)ML , where L = 6.00 m is the length of the pole and M = 24.0 kg is its mass. In your design, a person of mass m = 70.0 kg is to have a maximum angle of swing of theta max = 72.0˚ after their collision with the pole.

(a) In the "collision" between the slider and the pole, why is angular momentum about the pole's pivot conserved, but linear momentum and kinetic energy are not conserved? Assume that the slider is moving horizontally when they grab the handle on the vertically hanging pole.

(b) What is the angular speed of rotation of the pole & swinger just after the swinger grabs on, in terms of the final height the swinger reaches?

(c) What is the speed of the swinger at the bottom of the slide just before reaching the pole, in terms of their speed just after grabbing the pole?

(d) For a person of mass m = 70.0 kg, what must be the starting height h in order for the pole with person to have a maximum angle of swing of theta max = 72.0˚ after the collision?

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genius_generous answered 1 year ago

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