Question

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of 48.6 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

Answer 1

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

$mg-N = m\frac{v^2}{r}$

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

$mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s$