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jbriggs444
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Any other forces?UMath1 said:Rolling resistance
Any other forces?UMath1 said:Rolling resistance
Any other forces?UMath1 said:static friction of front wheel
I am considering the bike and the rider one system
Any other forces?UMath1 said:Air resistance
Gravity? The normal force of the pavement on the front tire? The normal force of the pavement on the rear tire?UMath1 said:I can't think of any
For some possible calculations, the force from gravity and the normal force on the two tires will cancel out, yes. But if we are trying to be careful, we should list them first and argue that they cancel for the particular calculation we are trying to perform afterward. For instance, we could do a torque balance on the bicycle as a whole to see how large static friction could be before the bicycle pulls a "wheelie". Gravity and the normal forces would be crucial for that.UMath1 said:Right. Yes, gravity and normal force, but they cancel out, right?
UMath1 said:Earlier my calculation for the force of static friction was: (Tu-Tl)Rg/Rw. But this cannot be the case, because it were you'd have break the maximum force of static friction to get the bike to accelerate.
So you now agree that the force of static friction does not need to exceed the maximum force from static friction in order for the bicycle to accelerate?UMath1 said:I think it could be neglected for the sake of simplicity. Even if it were a unicycle, you still wouldn't have to break the maximum force of static friction.
You still have not told us what makes you think that ##(T_u - T_l)\frac{R_g}{R_w}## has to be incorrect.UMath1 said:I agree because otherwise you'd run into the original issue of sliding before rolling. However I don't understand why. If the equation is not (Tu-Tl)Rg/Rw, what is it?
And what makes you think that the torque from the chain grows and grows without bound?UMath1 said:It has to be incorrect because that would mean that static friction keeps growing as the torque from the chain grows. It would keep growing until it reaches the maximum value. But I have been told repeatedly that reaching the maximum value is unnecessary and that would be "burning rubber".
If you recall, we assumed that the wheel has negligible moment of inertia in order to simplify the calculations. Accordingly, it takes negligible net torque for it to accelerate.UMath1 said:In order for the wheel to rotationally accelerate the torque of the chain MUST be greater than the torque of static friction.
If the rotational inertia of the rear wheel is neglected, that equation tells you that the chain torque is equal to the static friction torque. If the rotational inertial of the rear wheel is not neglected, then the chain torque has to be a little higher than the static friction torque, but part of the chain torque is consumed in accelerating the rear wheel, so the static friction torque is the same.UMath1 said:In order for the wheel to rotationally accelerate the torque of the chain MUST be greater than the torque of static friction. Based on that equation the only way that can happen is if maximum static friction value is attained. Otherwise, static frictions torque would continue to match the torque of the chain, meaning no angular acceleration.
We've already done that in previous posts.jbriggs444 said:If you recall, we assumed that the wheel has negligible moment of inertia in order to simplify the calculations. Accordingly, it takes negligible net torque for it to accelerate.
If you want to re-do the calculations with a non-negligible moment of inertia in the rear wheel then we can do so.
Huh?UMath1 said:Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached
I want to echo Chet's puzzlement. The torques will be negligibly different until maximum static friction is reached. [At which time the angular acceleration of the wheel becomes significant in spite of the wheel's neglibible moment of inertia]UMath1 said:Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached
Your own first equation in post #52.UMath1 said:The equation is based on neglecting moment of inertia and taking the torques to be equal, but they are not. They are negligibly different as you say...but they are still different.
So what equation indicates that the torque of the chain is little higher than the torque of static friction.
It leads to a set of simultaneous equations involving the mass of rider and bicycle and the moment of inertia of the front and rear wheels which can be solved to obtain a formula for the force of static friction in terms of the force of the chain, the two radii, the mass of the rider and bicycle and the moment of inertia of the front and rear wheels.UMath1 said:Right but that doesn't express static friction solely in terms of the force of the chain and the two radii. It only arbitrarily shows that there is net torque but it doesn't show how a bigger gear would cause more static friction.
Can you write an equation for the angular velocity of the wheel in terms of the linear velocity of the bicycle?UMath1 said:How can I eliminate acceleration?
With some type setting, that's ##\omega = \frac{v}{r_{wheel}}## where ##\omega## is the angular velocity of the wheel, ##v## is the velocity of the bicycle and ##r_{wheel}## is the radius of the rear wheel.UMath1 said:W=v/r
I am trying to lead you there. We'll get rid of the angular acceleration and the linear acceleration in due course. First I want you to write down the equation.UMath1 said:Right..but I don't want angular acceleration or acceleration in the equation. It would alpha=a/r.