Suppose a 200 kg motorcycle is heading toward a hill at a speed of 30.0 m/s. The

Question

Suppose a 200 kg motorcycle is heading toward a hill at a speed of 30.0 m/s. The two wheels are approximately annular rings with an inner radius of 28 cm and an outer radius of 33 cm. the mass of one wheel is 12 kg.

a) what is the initial rotational kinetic energy of one of the wheels?

b) what percentage of the motorcyles total kinetic energy is translational kinetic energy?

c)what is the vertical height i.e. altitude aboce its initial height it can coast up the hill if you neglect friction.

d) how much thermal energy is lost to dissipative forces if the motorcycle only gains an alittude of 36.0 m before coming to rest?

Explanation / Answer

for wheels .

surface density of mass = 12 / ( pi * 0.33^2 - pi*0.28^2 ) = 125.23667653 kg / m^2

so.. inertia of 1 wheel = 125.23667653*pi*0.33^2 *0.33^2 / 2 - 125.23667653 * pi * 0.28^2 * 0.28^2 / 2

so.. inertia of 1 wheel = 1.1238 kg m^2

a) angular speed of 1 wheel = w = velocity / outer radius = 30 / 0.33 = 90.909 rad/sec

so.. intial rotational kinetic energy = 0.5 * inertia * w^2 = 0.5 * 1.1238* 90.909^2 = 4643.8016529 J

b) translational kinetic energy = 0.5 * mass * velocity^2 = 0.5 * 200 * 30^2 = 90000 J

total rotational kinetic energy = 2 * 4643.8016529 = 9287.6033058 J

so.. percentage = 90000 *100 / ( 90000 + 9287.6033058 ) = 90.645757379 %

c) let the vertical height be h .

so.. mass * g * h = translational enegy + rotational

so.. 200 * 9.81 * h = 90000 + 9287.6033058

so.. height h = 50.6053024 m

d) thermal energy lost = translational enegy + rotational - mass * g * 36

= 90000 + 9287.6033058 - 200*9.81*36 = 28655.6033058 J

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