Motorcycle horizontal acceleration-front wheel rise

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SUMMARY

The discussion focuses on calculating the horizontal acceleration required for a motorcycle to lift its front wheel off the ground. Given the motorcycle's wheelbase of 171 cm and a center of mass located 95.0 cm above the ground, the relationship between torque and acceleration is established using the equation τ = rFsinϕ. The critical condition for the front wheel to rise occurs when the torque due to the weight acting on the rear wheels equals the torque generated by the horizontal acceleration, leading to the equation 0.95ma = 1.71/2(mg).

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  • Understanding of torque and its calculation (τ = rFsinϕ)
  • Basic principles of rotational dynamics
  • Knowledge of center of mass concepts
  • Familiarity with Newton's second law of motion
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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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