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Wheel Climbing a Step

  • Thread starter e(ho0n3
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  • #1
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Just need to know if I'm on the right track (again):

A wheel of mass M has radius R. It is standing vertically on the floor, and we want to exert a horizontal force F at its axle so that it will climb a step against which it rests. The step has height h, where h < R. What minimum force F is needed?

This problem was whispering to me: "Use energy methods...". The wheel moves about the corner of the step so I could take that as my axis of rotation and calculate the torque about it. The I proceeded to calculate the work done by this torque, i.e. [itex] W = \int{\tau} d\theta[/itex], to get the wheel up the step and equate this with the negative change of potential energy (since [itex]-W = \Delta{U}[/itex]) of the center of mass so that I could solve for F. This gave me the following (assuming I didn't screw up somewhere):
[tex]
F = \frac{mgh(1-R)}{\sqrt{2hR - h^2}}
[/tex]​
For F to be 'valid' I must have [itex]2hR - h^2 > 0 \Rightarrow R > h/2[/itex]. This is what worries me since I was explicitly given that R > h. I know what I did is wrong then, but I don't know what to do now. Any tips?
 

Answers and Replies

  • #2
Doc Al
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Torque!

Consider that the wheel pivots about the corner of the step. So take moments about that point. The horizontal force F must exert enough torque to overcome the torque exerted by the wheel's weight.
 
  • #3
Gza
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Your method looks fine to me, so I'm assuming that you probably did make an error in the algebra. I remember having this problem on a final a couple quarters back and the best advice I could give would be to draw a good picture, with all the angles in place so you can see the relationships clearly. And just a little tip; as soon as problems begin whispering to you, please take a break, because you're studying too hard. :smile:
 
  • #4
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Doc Al said:
Consider that the wheel pivots about the corner of the step. So take moments about that point. The horizontal force F must exert enough torque to overcome the torque exerted by the wheel's weight.
Let [itex]\tau_F[/itex] and [itex]\tau_w[/itex] be the torques produced by F and the weight respectively. My understanding of your response is telling me that [itex]\tau_F > \tau_w[/itex], but then this doesn't make sense so maybe I don't understand (nothing new there).

I did this problem again by setting [itex]\tau_F - \tau_w = I\alpha[/itex] and then figuring out what [itex]\alpha[/itex] is using [itex]\alpha = a/R[/itex] and of course I can find [itex]a[/itex] using Netwon's 2. law. This gives me
[tex]
F = \frac{\sqrt{2hR - h^2}}{R-h}
[/tex]​
but of course this implies that [itex]R \geq h/2[/itex].
 
Last edited:
  • #5
Doc Al
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e(ho0n3 said:
Let [itex]\tau_F[/itex] and [itex]\tau_w[/itex] be the torques produced by F and the weight respectively. My understanding of your response is telling me that [itex]\tau_F > \tau_w[/itex], but then this doesn't make sense so maybe I don't understand (nothing new there).
What doesn't make sense? The minimum horizontal force needed will be just enough so that [itex]\tau_F = \tau_w[/itex]
I did this problem again by setting [itex]\tau_F - \tau_w = I\alpha[/itex] and then figuring out what [itex]\alpha[/itex] is using [itex]\alpha = a/R[/itex] and of course I can find [itex]a[/itex] using Netwon's 2. law. This gives me
[tex]
F = \frac{\sqrt{2hR - h^2}}{R-h}
[/tex]​
but of course this implies that [itex]R \geq h/2[/itex].
No need to calculate α; just set the torques equal. You left out the weight of the wheel:
[tex]
F = mg \frac{\sqrt{2hR - h^2}}{R-h}
[/tex]​
And don't forget that the denominator implies [itex]R > h[/itex]
 
Last edited:
  • #6
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Doc Al said:
What doesn't make sense? The minimum horizontal force needed will be just enough so that [itex]\tau_F = \tau_w[/itex]
Come to think of it, I don't remember why I said that. If the wheel rotates about the corner of the step, then the angular acceleartion is greater than 0 so [itex]\tau_F - \tau_w > 0[/itex].
No need to calculate α; just set the torques equal. You left out the weight of the wheel:
[tex]
F = mg \frac{\sqrt{2hR - h^2}}{R-h}
[/tex]​
And don't forget that the denominator implies [itex]R > h[/itex]
I've scribbled so much that the mg term 'magically' disappeared. I trust this is the answer then. But why did my original analysis not yield the same answer? I think everything I did was legal wasn't it?
 
  • #7
Doc Al
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e(ho0n3 said:
But why did my original analysis not yield the same answer? I think everything I did was legal wasn't it?
Since the minimum force is not constant as the wheel rolls up the step, it's not obvious to me how to apply energy methods to this problem in a simple manner. We only need to find the initial force to get the wheel up the step.
 
  • #8
Gokul43201
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e(ho0n3 said:
But why did my original analysis not yield the same answer? I think everything I did was legal wasn't it?
No, unfortunately, everything was not legal. For instance, if you are assuming a constant force F, then the wheel WILL have a KE after it has climbed the step. So it's not right to say that all the work done went into raising the PE.

Also, there's clearly some algebraic error in your original approach since the answer is dimensionally wonky ! You cannot have a (1-R) term - what dimension does that have ?
 
  • #9
Doc Al
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Gokul43201 said:
No, unfortunately, everything was not legal. For instance, if you are assuming a constant force F, then the wheel WILL have a KE after it has climbed the step. So it's not right to say that all the work done went into raising the PE.
Exactly right. The horizontal force needed to just raise the wheel with no added KE is not constant. (That was the point I was trying to make in my last post.)
 

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