2. (a) A bicyclist is coasting down a straight road that is inclined to the hori

Question

2. (a) A bicyclist is coasting down a straight road that is inclined to the

horizontal at an angle ? as shown. Find an expression for the maximum

coasting speed (no pedaling, no braking, assume no friction). The total mass

(of the cyclist and the bike) is m, the drag coecient is cD and the e?ective

frontal area is A. Assume sea-level air for which ? = 1.2 kg/m3

(b) For an upright bike, White (Table 7.3) gives cDA = 0.51 m2

. For m = 65

kg and ? = 3 degrees, ?nd the maximum coasting speed in m/sec and also

in mi/hr.(c) For a racer on a road bike, White gives cDA = 0.30 m2

. Repeat the

calculation of part (b) for the racer.

(d) Now include mechanical friction that is proportional to speed. The term

to be added into the force balance is fV where the friction coe?cient for an

average bike is f = 1.0 (N s)/m. Recalculate the maximum coasting speed

for cases (b) and (c) with mechanical friction.

A bicyclist is coasting down a straight road that is inclined to the horizontal at an angle a as shown. Find an expression for the maximum coasting speed (no pedaling, no braking, assume no friction). The total mass (of the cyclist and the bike) is m, the drag coecient is cD and the effective frontal area is A. Assume sea-level air for which rho = 1.2 kg/m3 For an upright bike, White (Table 7.3) gives cDA = 0.51 m2. For m = 65 kg and Alpha = 3 degrees, find the maximum coasting speed in m/sec and also in mi/hr. For a racer on a road bike, White gives cDA = 0.30 m2. Repeat the calculation of part for the racer. Now include mechanical friction that is proportional to speed. The term to be added into the force balance is fV where the friction coefficient for an average bike is / = 1.0 (N s)/m. Recalculate the maximum coasting speed for cases (b) and (c) with mechanical friction.

Explanation / Answer

a) ma=mg*sin(alpha) - 1/2*rho*V^2*C(d)*A equation ofsystem. At maximum velocity value i.e,terminal velocity,acceleration will be zero (will reach state of equilibrium)
hence simply equating mg*sin(alpha) =1/2*rho*V^2*C(d)*A and solving for v we get max velocity of bycyle
V^2=2mgsin(alpha) / rho*C(d)*A

b)substituting all the values
we will get terminal velocity as 10.43 m/s i.e,23.331245526 miles/hr

c)13.60 m/s i.e,30.422333572 miles per hour

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