Question

# 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 waterslide, 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=13ML2, 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?

h =

35.835.8mm

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When Person started the slide it's energy
= Mpersongh = 70*9.81*h = 686.7h -----------(1)
When it reaches at the bottom of the pole this potential energy must be converted in to the kinetic energy
(1/2)Mperson*V2 = 686.7h (by conservation of energy)

Now this will converted into the potential energy when it reaches at the extreme position as shown in the figure.
Potential energy = MPgH + MRod*gP
Where H = L - Lcos72 = 6 - 6Cos72 = 4.146 m
P = L - (L/2)Cos72 = 5.073 m
Potential energy = (70*9.81*4.146) + (31*9.81*5.073) = 4389.808 J
Hence by energy conservation
686.7h = 4389.808
h = 6.393 m

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