How to calculate torque required to accelerate

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gearhead
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I have a bicycle that has a mass of 100kilograms with the rider on it. The wheels have a diameter of 26" or 0.722meters. How do i calculate how much torque is required to: 1.) break the static coefficient of friction and cause the wheel to start turning, and 2.) determine how much torque is required to accelerate at a rate of 1m/s^2?
 
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Assuming no slipping, the torque, [itex]\tau[/itex], delivered to the wheels of radius, R, is related to the force, F, that pushes the vehicle by

[tex]\tau=RF[/itex][/tex]
 
so then i must calculate the force of static friction on the bike first to determine the amount of torque required to overcome this. F=ma, f=100(9.8) = 980N*0.5(the static coefficient of friction)=Force of friction=490N. So then required torque is then, t=0.361m*490=176.86N*m of torque to counteract friction and cause the wheel to start turning.
 
gearhead said:
I have a bicycle … How do i calculate how much torque is required to: 1.) break the static coefficient of friction and cause the wheel to start turning

Hi gearhead! :smile:

The wheel does not move relative to the ground, so the static friction to be overcome is only between the axle and the bearings.

For deformable wheels (eg rubber), there is also rolling resistance (loss of energy through deformation … it's what slows the Moon down!): see http://en.wikipedia.org/wiki/Rolling_resistance :wink:
 
tiny-tim said:
... the static friction to be overcome is only between the axle and the bearings.
Ah, yes, that is a better interpretation of the question than what I was thinking of. Then gearhead will also need the radius of the axle.
 
oh allright, thanks guys, so then theoretically any amount of torque you generate will accelerate you forward right?
 
so then, i'll try to calculate the amount of torque required to accelerate the bike at 1m/s^2.
using
F=ma, F=(100kg)*(1m/s^2)=100Newtons.
T=R*F. T=(0.361m)*(100N)=36.1N*m of torque, right?