A motorcycle of mass, m = 250kk, performs a speed test, accelerating from rest t

Question

A motorcycle of mass, m = 250kk, performs a speed test, accelerating from rest to 100 km/h and then on to 200 km/h. The torque (moment) at the rear wheel is a constant 400 Nm and the wheel has a diameter, d = 0.2 m. When the motorcycle reaches 100 km/h the power being developed by the engine is 70 kk and when the motorcycle reaches 200 km/h the power being developed is Neglecting friction, determine the time taken to reach 100 km/h and 200 km/h using a momentum analysis. Also neglecting friction, determine the distance travelled when the motorcycle has reached 100 km/h and 200 km/h using a work and energy formulation. What is the efficiency of the motorcycle at the two respective speeds? If the motorcycle immediately brakes when it reaches 100 km/h and stops in the same distance it travelled to reach that speed, what is the coefficient of friction at the wheels? Unfortunately the analysis in Part A is overly simplistic. Using any means at your disposal determine the time taken to reach 100 km/h if a velocity dependant drag force exists F_f = 1.5v where v is in m/s.

Explanation / Answer

mass of bike, m = 250 kg

initial velocity u = 0 m/s

final velocity v1 = 100 km/h = 27.778 m/s

v2 = 200 km/hr = 55.558 m/s

torque, T = 400 Nm

wheel diamter,d = 0.2 m

at v1, power, P1 = 70 kJ

at v2, power, P2 = 150 kJ

A. let time taken be t

then change in momentum = T*t/d [ impule is force * time and force is torque/d]

so, for reachiong 100 km/h

250*27.778 = 400*t/0.2

t = 3.47225 s

so, for reachiong 200 km/h

250*27.778*2 = 400*t/0.2

t = 6.9445 s

B. let distance travelled be s

then from work energy theorem

0.5mv^2 = T*s/d

for reaching 100 km/h

0.5*250*27.778^2 = 400*s/0.2

s = 48.22608025 m

for reaching 200 km/h

0.5*250*27.778^2*4 = 400*s/0.2

s = 192.904321 m

C. power developed, P = Fv = Tv/d

efficiency at 100 km/h = 400*27.778/70*10^3*0.2 = 0.79 = 79 pc

efficiency at 200 km/h = 400*27.778*2/150*10^3*0.2 = 0.74 = 74 pc

D. stopping distance, d= s = 48.22608025

so for coefficient of friciton mu

friction force = mu*mg

so, frictional acceleration = mu*g

2*mu*g*d = v^2

mu = 27.778^2/2*g*48.22608025 = 0.8154

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