Two bicycle tires are set rolling with the same initial speed of 3.50 m/s along

Question

Two bicycle tires are set rolling with the same initial speed of 3.50 m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes a distance of 17.2 m ; the other is at 105 psi and goes a distance of 92.4 m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80 m/s^2

Part A

What is the coefficient of rolling friction r for the tire under low pressure?

Explanation / Answer

coefficient of friction is mu = frictional force / normal force = m*a/mg = a/g

using kinematic equations

v^2-u^2 = 2*a*S

v = 3.5/2 = 1.75 m/sec

u = 3.5 m/sec

unser low pressure s = 17.2 m

then

1.75^2-3.5^2 = 2*a*17.2

accelaration is a = -0.269 m/s^2

mu = a/g = 0.269/9.8 =0.0274

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