Bicycle Normal Force in Valley

## Question:

A boy on a bicycle has a combined weight of 125 pounds. He coasts
the bicycle without pedaling down a hill with a vertical drop of
30 feet. At the top of the hill the speed of the bicycle was 10
feet per second. At the bottom of the hill the road curves upward
with a radius of curvature of 50 feet. Determine the normal force
on both bicycle wheels when he arrives at the bottom of the hill.
Neglect friction and the mass of the wheels.
## Answer:

The 125 pound boy/bicycle combination has a mass of 3.88 slugs.
At their 10 f/s initial velocity the kinetic energy is 1/2*m*v^{2} = 1/2*3.88*100 = 194 foot
pounds. The potential energy is m*g*h =
3.88*32.2*30 foot pounds = 3750 foot pounds. The combined
energy is 3944 foot pounds which will be entirely kinetic at the
foot of the hill. So at the foot of the hill
1/2*3.88*v^{2}=3944, or
v^{2}=2033 f^{2}/s^{2}. So v=45 f/s. The total normal force on the bicycle
wheels must provide the centripetal acceleration to follow the 50
foot radius curved path and support the weight of the boy/bicycle
combination. The centripetal acceleration is v^{2}/r or
2033/50 = 40.66 ft/s^{2}. This
times the mass of 3.88 slugs gives the force due to the curved
path. That force is 157.8 pounds. To this we add the weight of
the apparatus, 125 pounds to get a total of
282.7 pounds.
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