The drawing shown represents an engineer\'s concept of the vertical profile of a

Question

The drawing shown represents an engineer's concept of the vertical profile of a very tall water slide. The radii of curvature of the path in proximity to B and C are pg = 50 ft and Pc = 70 ft, respectively. The water on the slide makes the coefficient of friction negligible. A person starts from rest down the slide at point A. Determine (a) the height of point A so that when the person passes point C, the normal force the slide exerts on the person is zero and (b) the person's normal acceleration at point B. Note that the person's velocity is instantaneously horizontal at B and C.

Explanation / Answer

a) normal force at c = centripetal force - gravitational force

= mv2/R - mg = 0 where v is the velocity at c

--> mv2 = mgR

now thw change in kinetic energy equals the change in potential energy

--> 1/2(mv2 - mv02) = mg (hA-100)

since, mv2 = mgR

--> 1/2 * mv02 = mg (hA-100) - 1/2 * mgR

--> 1/2 * mv02 = mg (hA-100 - R/2)

--> 1/2 * v02 = g (hA-100 - 70/2)

v0 = 0

g = 32.174 ft/s2

--> hA-100 - 70/2 = 0

--> hA = 135 ft

b) velocity at b = Vb

1/2 * mVb2 = mg hA

--> 1/2 * Vb2 = g hA

--> Vb2 = 2g hA = 2*32.174 *135 = 8686.98

normal acceleration at b = Vb2/Rb = 8686.98/50 = 173.74 ft/s2

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