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Tadhg
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Hi folks, this is a problem I am having in a collage project I am doing. I'm not too sure about my results and keep chasing my tail using the same formulas, e.g. F=ma etc. I think I need to allow for the fact that the force is applied as a collision but I am not sure how or if I need to do it.
So the big question, is the force that is applied to the ground to accelerate the bike the same as the force that is applied to the object that it impacts? I have tried to provide all the relevant data, but please ask if you need more.
Thank you.
1. Homework Statement
A motorcycles engine provides a torque of 46.91 Nm the transmission multiples this torque to 1053.87 Nm. This torque applied to the ground through a wheel with an outer diameter of 0.72 m. The motorcycles front wheel hits an object at a distance of 1.3 m. Bike and rider mass=194.37 kg.Force = torque/distance
Force = 1053.87/(0.72/2)
Force applied to the ground to accelerate the motorcycle = 2924.97 N
Force=mass x acceleration
F=ma
a = 2924.97/194.37
a=15.05 m/s^2
Final Velocity^2 = Initial Velocity^2 + 2 x acceleration x distance
V^2=U^2+2as
V = squrt(0+2x15.05x1.3)
velocity = 6.14 m/s
Kinetic Energy = 1/2 x mass x velocity^2
Ek = 0.5 x 194.37 x 6.14^2
Kinetic Energy = 3664.67 Joules
Assuming no losses and that there is an instantaneous transfer of torque to the ground etc.
So the big question, is the force that is applied to the ground to accelerate the bike the same as the force that is applied to the object that it impacts? I have tried to provide all the relevant data, but please ask if you need more.
Thank you.
1. Homework Statement
A motorcycles engine provides a torque of 46.91 Nm the transmission multiples this torque to 1053.87 Nm. This torque applied to the ground through a wheel with an outer diameter of 0.72 m. The motorcycles front wheel hits an object at a distance of 1.3 m. Bike and rider mass=194.37 kg.Force = torque/distance
Force = 1053.87/(0.72/2)
Force applied to the ground to accelerate the motorcycle = 2924.97 N
Force=mass x acceleration
F=ma
a = 2924.97/194.37
a=15.05 m/s^2
Final Velocity^2 = Initial Velocity^2 + 2 x acceleration x distance
V^2=U^2+2as
V = squrt(0+2x15.05x1.3)
velocity = 6.14 m/s
Kinetic Energy = 1/2 x mass x velocity^2
Ek = 0.5 x 194.37 x 6.14^2
Kinetic Energy = 3664.67 Joules
Assuming no losses and that there is an instantaneous transfer of torque to the ground etc.