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How long does the wheel slip before it begins to roll without slipping?

  • Thread starter PatF
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  • #1
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Homework Statement



A wheel with rotational inertia I = 0.5MR^2 about its central axle is set spinning with initial angular speed w and is then lowered onto the ground so that it touches the ground with no horizontal speed. Initially it slips, but then begins to move forward and eventually rolls without slipping.

What is the wheel's final translational speed?

Homework Equations


I know that the torque where the wheel touches the ground is MgR*mu and that the angular momentum is initially Iw. Thus the angular momentum at time t is
I*(w-2*g*mu*w*t/R)


The Attempt at a Solution


I am missing something conceptual here. What does it mean physically so that the angular momentum decreases in such a way that it translates into linear momentum?
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
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Think about the physical meaning of the phrases "rolls without slipping".

What does this imply is the relationship between rotational and translational velocity?
 
  • #3
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"Think about the physical meaning of the phrases "rolls without slipping".

What does this imply is the relationship between rotational and translational velocity?"


I am terribly sorry but I HAVE thought about this but it seems absolutely pointless. You might as well say that I should consider how the qstorb schlonges with the veeblefretz. That's how much sense it makes.

I admit freely that I am missing something. But if translational and rotational velocity are involved, I just do not see how.
 
  • #4
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OK. I think I have it now and I will write down the solution in case anyone else runs into the problem. I think the problem is lot subtler than it looks and the key is figuring out what it doesn't ask for as much as figuring out what it does.

So...

First, picture the wheel as rotating clockwise and slipping on the ground. It will eventually roll to the right.

There is a force that acts on the wheel and, at the center of the wheel we have F=M*a. That same force must oppose the slippage via torque. So, F*R=-I*alpha. alpha is the angular acceleration. (The negative sign is important, by the way.)

The above tells us that M*a*R=-.5*M*R2*alpha.

Cancelling M and R we get a=-.5*R*alpha.

Here we have two accelerations, linear and angular. Integrate them both from 0 to T where T is the time when the wheel starts rolling. Then you get:

Vf-Vi =-.5*R*(wf-wi)

where the subscripts f and i stand for initial and final. Now, Vi=0. Furthermore, "rolls without slipping" says that vf =R*wf.

Now you have
Vf=-.5*Vf+.5*R*wi

If you solve this, you find Vf=R*wi/3.

What threw me here was that the problem does not ask for or tell you about anything involving mass or time or friction and, while these do show up in the steps, they cancel each other out in the end. I was using a lot of machinery for what was, in the end, some fairly simple algebra.
 

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