In first gear,a motorcycle engine produces 902 Nm of torque on the rear wheel.If the rear wheel is lifted off the road and free to spin without friction,how long will it take the engine to spin the wheel from rest to 200 revolutions per minute (corresponding to the outer part of the wheel moving at approximately 24 km/h)if the wheel were on the road? Treat the rear wheel as a hollow cylinder with inner diameter 50 cm and outer diameter 64 cm,with a mass of 13 kg.
A)0.025 s
B)0.015 s
C)0.031 s
D)0.013 s
E)0.040 s

Question

Quizplus · Accepted Answer

The correct answer is A because the torque is 902 Nm, the wheel's moment of inertia is 0.32 kg m^2, and the final angular velocity is 200 rpm. Using the equation:torque = moment of inertia * angular accelerationwe can find the angular acceleration:angular acceleration = torque / moment of inertia = 902 Nm / 0.32 kg m^2 = 2818.75 rad/s^2Then, using the equation:final angular velocity = initial angular velocity + angular acceleration * timewe can find the time it takes to reach 200 rpm:time = (final angular velocity - initial angular velocity) / angular acceleration = (200 rpm * 2π rad/min / 60 s/min - 0 rad/s) / 2818.75 rad/s^2 = 0.025 s

In First Gear,a Motorcycle Engine Produces 902 Nm of Torque