DouglasReay/PhysicsProblem

The Problem

Opinions / Solutions

The question is, how far do the feet travel?  Consider a foot moving from the 10:30 o clock position to the 1:30 o clock position (a quarter of a turn).  During this period, the bike moves 2.5 meters, as does the foot attached to the pedal that is attached to the bike.  And since during this quarter turn of the pedal, that direction of motion has a component of at least 1/sqrt(2) in the direction that the force is being applied, doesn't it need to be taken into account? --DR

Not sure that's right.  It's two different frames of reference.  The feet are travelling in perfectly circular arcs around the pivot point which is stationary with respect to the bicycle frame and rider.  If the whole assembly was running on a treadmill the equations would be functionally identical, the important point is the distance from the pivot, which is constant.  One further point: do not forget that there is an equal and opposite foot moving from 3 to 9 o'clock on the other side providing a balance in vectors within the plane of motion of the crank disc (and isn't from 9 o'clock to 3 o'clock a half turn?). --Jumlian

Yep, you're right.  I knew there was a reason I prefer digital watches.  :-)  Times corrected.

On your further point: How would the other foot cancel out the effect?  In the same time that the uppermost foot is travelling 2.5 meters rather than 0.16 meters, the bottom most foot can, at best, be doing zero work rather than negative work.

On reference to frames of reference: Yes, I suspect this is right.  But what then is the true definition of work done?  What does it mean for a force to be applied in a particular frame of reference?  And how to I explain it to A-level maths students?

Hmmmm., Thinks.  The linear distance travelled by the foot is essentially irrelevant.  The foot is doing work in an angular frame of reference.  In terms of explaining this, the cyclist and the bike can be regarded as one unit.  The cyclist does work on the (cyclist+bike) and the (cyclist+bike) (through the wheels) does work on the road, and moves linearly as a result.  The linear motion of the (cyclist+bike) is a consequence of the work done, and occurs in a separate frame.  The work done is NOT dependent on the linear motion of the foot, not for classical, A-level mechanics.  If you wanted to get into serious detail, then the angular movement of the feet is essentially constant, but the component resolved to the horizontal plane varies as a sine wave.  The velocity of the feet are opposing sine waves, relative to the pivot point, giving net zero momentum and force transfer, and net zero energy since we assume that the feet have zero mass.  In real life, there is some mass movement here and hence some force applied, some energy needed, since human cyclists are not massless inelastic engines.  But we assume they are or we'd have to consider things like how long their hair is for air resistance purposes, or chop off their feet and weigh them.  The (cyclist+bike) is assumed to be one single united mass moving on a road, relative movement of its component parts is not considered.  The crank + pedal + foot definitions in the question describe the levers through which the force is applied, the further conditions of "pushing in a circle with no wasted force" show that the problem is one of an angular frame of reference.  Force exerted to the crank, from each foot, at all times, is therefore constant.  Think of it this way:  You exert force x on each pedal and are travelling at velocity v, no acceleration.  Thus net force on the rider+bike is zero (F=ma).  So the resistances equal the effort.  The work you are doing is spent overcoming the resistances, and is equivalent to the moment exerted on each pedal (x), multiplied by the length of the lever (10cm), multiplied by how far you push that lever (0.2 pi) multiplied by the number of revolutions of the crank (10).  If you suddenly decide to don your massless scuba gear and cycle in a big vat of syrup, the resistive force increases markedly as a function of your velocity.  So, in order to balance forces your speed drops, as you're not working any harder.  Now for each rotation of the crank, you still do exactly the same work but are travelling at a lower speed.  The confusion appears to be located in thinking that the velocity of the component parts of the bicycle + rider is directly related to the work done.  Not true.  Only true through the medium of air resistance where F is proportional to V (and viscosity, and cross sectional area).  A simpler case: consider work against gravity.  Say you have an apple, handily weighing about 100g in a world where g=10 rather than 9.81.  It therefore exerts a force of 1N.  Performing 1J of work in lifting it upwards lifts it one metre, as 1J=1Nm.  There is no implication about speed here, no units of time, the joule could be exerted instantly, over a second, a few days or weeks.  That is the work done against a resistive force.  The same argument applies if it was air resistance not gravity, and we come back to the bicycle problem...  --Jumlian  Ye deities it's been ages since I had to think about stuff like this...

B. By definition (code for: I'm lazy and can't be bothered to do this right from Newton II), Work = integral (Force) wrt (Distance) = integral (Couple) wrt (Angle passed through), either of which gets you answer B. That's how much work was done. Everything else is just question fluff. It's easier to see the answer if you talk about the integral of the couple with respect to the angle (i.e. couple * angle as your force is constant), as the couple is the product of the magnitude of the two opposed forces (i.e. F) and the perpendicular distance between them (20cm) and the angle is just the number of times the pedals went round times 2 pi. For the advanced student, you can of course take off the toe clips so your force is some other function (a pair of step functions, say) and assume the resistive force is to a first approximation constant - do you need any more energy to do the same distance at the same speed? The answer's quite easy - of course you don't, because you're pushing the bike the same distance against the same force - but if you start by calculating the force required on the pedals (2F) or by integrating that step function (nontrivial unless you think rather than applying menial arithmetic, because the step function is discontinuous) then you can tie yourself up in knots trying to prove it. --Requiem

Of course, the second interesting comment is what do the other answers read? A reads that you have pushed with both feet for 100 metres - i.e. you are riding a bicycle with no gearing and pedals as far apart as the wheels! C reads that not only have you cycled the entire distance, you have then sat down and pushed the bicycle all the way back with your feet. I like exploring what false answers mean, rather than just discarding them as false. --Requiem