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Plate sliding on ice with friction (Physics competition question)
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[QUOTE="haruspex, post: 6850443, member: 334404"] Consider the pure rolling resistance scenario, no drag. The tyre has a greater elastic constant during deformation than during relaxation. That displaces the normal force ##N## some distance ##s## ahead of the axle. Combined with the bike's load on the axle, that produces a retrograde torque ##Ns## on the wheel. In the steady motion of the bike, the motor provides a balancing torque to the driving wheel. ##\tau_{motor}=Ns##. The reaction torque on the bike is matched by a slight shift in the distribution of weight between the two axles. At this point, no horizontal forces are involved, so there is no frictional force in the direction of travel (but some lateral to stay upright, no doubt). To overcome an external retarding force ##F## instead, the motor would need to provide axle torque ##Fr##, where r is the wheel radius. Thus the external force equivalent, as far as the motor is concerned, is ##Ns/r##. Now consider the idling wheel. With no motor to provide torque, the rolling resistance torque would tend to slow the wheel down. ##I_{wheel}\alpha=Ns##. To keep the wheel turning at the rolling speed, there will be a backwards frictional force from the ground. ##Ns=rF_{fric}##. That is transmitted through the axle to the bike and to the driving wheel. As the motor works to maintain speed, the driving wheel therefore experiences an equal forward frictional force. Adding these up, the torque the motor has to supply is ##2Ns##, whence ##2Ns/r=400N##. The frictional forces the wheels will exert on the plate are ##Ns/r## forwards at the front wheel and ##Ns/r## backwards at the rear wheel. [/QUOTE]
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Plate sliding on ice with friction (Physics competition question)
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