Understanding Peel-Out From a Torque/Rotational Dynamics Perspective

[RobL14]

Understanding "Peel-Out" From a Torque/Rotational Dynamics Perspective

I've been having a lot of trouble understanding rotational dynamics involving wheels. I understand the simple stuff, but I don't understand the concepts behind the more difficult problems.

Let me make this clear: this problem is not for credit -- I know the answer and the path to the solution, but I do not know why. I'm becoming extremely frustrated, because the book doesn't make this situation clear.

Homework Statement

If the coefficient of static friction between a wheel and the ground is u, calculate the minimum applied torque required to "lay rubber." The wheel has radius R and mass M.

Homework Equations

net torque = I * alpha

The Attempt at a Solution

In the solution, they simply state that the applied torque must be equal to the torque due to friction.

frictional torque = applied torque < uMgR

I can understand that reasoning intuitively, but if you write the net torque expression, then you get that alpha is zero, which contradicts that the wheel is spinning at all. Clearly I'm missing something, and I'd appreciate an in-depth response, because I need to understand the concept.

 

[ashishsinghal]

if you write the net torque expression, then you get that alpha is zero, which contradicts that the wheel is spinning at all. Clearly I'm missing something, and I'd appreciate an in-depth response, because I need to understand the concept.

If your acceleration is zero, does that mean that you are not moving. You may be moving with constant velocity. The same situation is occurring here. By the way, what does lay rubber mean?

 

[RobL14]

Oh, I think I understand what's happening here. It was staring me in the face. "Lay rubber" means to "peel-out" or essentially have your tires move without rolling, i.e. the angular velocity times the radius is faster than the velocity of the center of mass.

If the torques add to zero, then the wheel will start spinning with constant velocity, and the wheel will move forward due to the force of friction. However, once the applied torque is greater than the friction torque, the wheels will spin faster than the body translates, i.e. "lay rubber." But I am correct in thinking that this is only about to happen when the angular acceleration is zero, right?

 

[eczeno]

If the torques add to zero, then the wheel will start spinning with constant velocity

this is correct. when the torques are ballanced, the wheel roles with constant speed.

and the wheel will move forward due to the force of friction.

also correct.

However, once the applied torque is greater than the friction torque, the wheels will spin faster than the body translates, i.e. "lay rubber." But I am correct in thinking that this is only about to happen when the angular acceleration is zero, right?

it is correct that \omega > v/r when the there is slipping (burning rubber). but comparing \omega to v directly does not make sense (but i know what you are trying to say).

when the applied torque exceeds that which static friction can apply,spinning will happen. so yes, it 'starts' to happen when \alpha is still zero, but really, spinning means that \alpha > 0. the reason is, once it starts spinning, we are dealing with kinetic friction instead of static, so the torque from friction could drop by a significant amount as soon as the slipping begins, depending on the ratio of coefficients of friction.

so, in order to start the wheel spinning, one must supply a torque > the maximum torque that static friction can supply. but once, that is done, the wheel will continue to accelerate rotationally due to the decrease in the frictional torque.

hope this helps

 

[RobL14]

Thank you for helping. It's still weird to me, but let me put it another way, and you can tell me if I'm thinking about it correctly.

When the net torque is zero, the wheel will spin with some constant velocity. At the same time, the wheel will translate with a linear acceleration proportional to the force of friction. Therefore, from this perspective, the wheel will have zero angular acceleration, but non-zero translational acceleration. Once the torque is non-zero, it will now have an angular acceleration, but that angular acceleration \alpha < a/R.

There are a few things that confuse me about this problem. First, how can we say that the friction force causing the torque is the same friction force causing the forward linear motion? I thought that the friction causing the torque was static.

Second, why can't I solve this problem by finding the torque necessary for the wheel's linear acceleration to be unrelated to its angular acceleration? For example, start by assuming that the wheel is rolling without slipping, then:

<br /> (\vec{F}_{net})_x = F_{fr} = m(\vec{a}_{cm})<br /> <br />

<br /> (\vec{a}_{cm}) = F_{fr}/m<br />

<br /> (\vec{F}_{net})_y = F_G - N = 0<br />

<br /> (\tau_{net}) = (\tau_{applied})-(\tau_{fr}) = I(\alpha) = I(\vec{a}_{cm})/R<br />

<br /> (\tau_{applied})-(\tau_{fr}) = I(\alpha) = I*(F_{fr}/m)/R<br />

<br /> (\tau_{applied}) = I(\alpha) = I*F_{fr}/(mR) + (\tau_{fr}) = I*(F_{fr})/(mR) + R(F_{fr})<br />

<br /> (\tau_{applied}) = (F_{fr})(I/(mR) + R)<br />

Because the maximum force of static friction is given by F_{static} &lt; (\mu)Mg, then the maximum torque that can be applied to the system, and it still be true that \alpha = a/R, must be given by:

<br /> (\tau_{applied}) = (\mu)Mg(I/(mR) + R)<br /> <br />

Clearly, this is wrong, so I have a flaw in my reasoning. Can anyone point it out?

 

[eczeno]

your basic idea, to find the maximum applied torque where the equation \\alpha[/tex = a/R sitll holds is correct. i think what is confusing you is that when the wheel is rolling at constant velocity this is trivially true because alpha and a are both zero. the problem really boils down to:<br /> <br /> R * max static friction = max applied torque = answer. <br /> <br /> now what happens after this torque is exceeded needs a whole new analysis, because the friction switches from static to kinetic.<br /> <br /> as for why static friction makes us move, think about how you walk. you push backward on the floor with your foot and the floor pushes back. this is static friction because your foot and the floor are not moving relative to each other (that is, your foot is not sliding along the floor, it grips) when they are in contact. <br /> <br /> using kinetic friction to get yourself going is not as effective. this is the case when we try to start running on ice. our feet slide backward along the ice and it pushes us forward a little bit, but not very much.<br /> <br /> cheers.

 

[RobL14]

Thanks, I now understand how friction is linearly accelerating the wheel. But what's wrong with my math? Shouldn't I just reach the same conclusion?

 

[eczeno]

oh, right, i forgot about that.

you are mixing systems.

your first two equations are Newtons second law applied to the car as a whole. equation 4, the first torque equation, is correct if you are only considering one wheel. the wheel feels a friction force from the road (1/4 that from equation 1), but it is also pushing the whole car forward, and so feels a reaction force from the axle in the opposite direction. so f_friction/4 - f_axel = m_wheel * a would be the corresponding force equation.

but, equation 4 is the only one you need, if you apply it correctly. set alpha to zero, and let the frictional torque = R * max friction force and you have it.

cheers.