Ping
Excellent! I totally understand everything you are saying.
Please indulge me and answer a few more questions to see if what you are stating makes sense.
Remember those old fashioned exercise bikes? The ones with a brake caliper on the flywheel on the front. And you could adjust the brake pressure by cranking down on a spring.
Lets say I had one and it was geared so that for every revolution of the pedals the flywheel went around one tenth of a revolution. And you had an identical bike set at the same brake pressure except yours was geared so that for every revolution of the pedals your flywheel went around ten times.
Our personal trainer requires us to both pedal at a constant sixty pedal revolutions a minute. After we both get out flywheels up to speed, who is doing the most work and why?
Regards,
Patrick
Hmm, I might have to crack open my old dynamics textbook for this one. It's been a while since I've done a dynamics problem. This is a pretty generic dynamics question, I just need to remember or look up the equations for torque and work.
Well it's late and I need to get to bed, but let me see if I can take a stab at it really quick. I'm thinking on the fly so if I think out of turn, please give me till the morning to revise myself.
Ok for constant torque, rotational work is defined as:
W=t*theta
W = Work
t = torque
theta = total angle covered in motion
torque = (moment arm) x Force
Ok, so let’s assume that the flywheel has a 10:1 ratio with the wheel, so that every revolution results in 1/10th revolution of the actual wheel. I’m also going to assume that these are stationary bikes so that the wheels are suspended and so I don’t need to look at any additional forces related to the ground.
Bike #1: Each pedal revolution = 1/10th flywheel revolution
Bike #2: Each pedal revolution = 10 flywheel revolutions
Now let’s translate 60 RPMs of the pedal to revolutions of the actual wheel. For calculation's sake let's assume that we are observing this for 1 minute. It is inconsequential as it just takes the time factor out (i.e. per minute) but the theory will hold true for any duration of time or even rate of time.
Bike #1: 60 pedal revolutions * 1/10th flywheel revolution *1/10th wheel revolution = 0.6 revolutions of the wheel (or 216 total degrees)
Bike #2: 60 pedal revolutions * 10 flywheel revolutions * 1/10th wheel revolution = 60 revolutions of the wheel (or 21,600 total degrees)
Ok now let’s look at the forces on the wheel. Since both wheels are identical and the brakes are set at the same pressure (assuming a dynamic coefficient of friction)
Bike #1 torque = Bike #2 torque = ConstantXvalue
This is because the friction force is the same (constant
dynamic pressure) and the moment arm (length of spoke) is the same.
Now looking at the Work of each wheel:
W = torque * theta
Bike #1 = ConstantXvalue * 216 degrees
Bike #2 = ConstantXvalue * 21,600 degrees
So by my calcs, clearly Bike #2 will have produced more work and will have “travelled” (i.e. rotated) many more times than Bike #1.
That's just off the top of my head. I think it's right but it's late and I'm not 100% sure until I re-read it in the morning.
How does the explaination sound to you? :smile: