Motorcycle horizontal acceleration-front wheel rise

AI Thread Summary
To determine the horizontal acceleration required for a motorcycle's front wheel to rise off the ground, the torque generated by the rear wheel must equal the torque due to the weight acting on the motorcycle. The center of mass is positioned 95 cm above the ground and halfway between the wheels, creating a lever arm of 85.5 cm. The equation .95ma = 1.71/2(mg) shows the balance of torques, where 'm' is the mass of the motorcycle and 'g' is gravitational acceleration. At the point of front wheel lift-off, all weight is effectively on the rear wheels, and the acceleration of the center of mass is directed forward. Understanding this balance of forces and torques is crucial for analyzing motorcycle dynamics during acceleration.
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Motorcycle horizontal acceleration--front wheel rise

The distance between the centers of the wheels of a motorcycle is 171 cm. The center of mass of the motorcycle, including the rider, is 95.0 cm above the ground and halfway between the wheels. Assume the mass of each wheel is small compared to the body of the motorcycle. The engine drives the rear wheel only. What horizontal acceleration of the motorcycle will make the front wheel rise off the ground?

2. Homework Equations

τ=rFsinϕI understand that the solution is when .95ma=1.71/2(mg), so when the torques are set equal to each other. why though? thanks!
 
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All the weight is on the rear wheels. Acceleration of the center of mass is in the forward direction. Is there any rotational acceleration about the center of mass at this point in time?
 
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