Finding Kinetic Energy on a Vertical Circle with Friction

[jhae2.718]

It seems that I used the wrong value for C to get positive KE. I get -0.045 J as well after using the value for C I have above...

So, it seems that the problem's answer was incorrect, as I find that the KE becomes zero at around 53.4 degrees...

 

[turin]

So, it seems that the problem's answer was incorrect, ...

Do you have an official answer that we can compare to?

 

[jhae2.718]

0.71 J is the book's answer; unfortunately there isn't anything given to support that answer.

 

[turin]

Sorry, I see that you included the official answer at the bottom of your first post.

I "solved" it, and I also get a negative value for the final kinetic energy at φ=+π/3. I will have to think about this some more.

 

[jhae2.718]

No problem.

I think the answer given is probably incorrect; there have been other wrong answers given.

Personally, it seems counter-intuitive to only lose approx. 0.27 J due to friction on an arc like that, especially since for φ>0, the component of the weight tangent to the circle is also pulling it down. (Of course, regarding the weight, for φ<0 the weight is accelerating the block, but by the time it gets to P₂, there's been more energy being lost, but maybe I'm going off in a wrong direction here...)

 

[turin]

Just as I suspected: you are sharper than the author of this particular physics problem (and sharper than me). K_f = 0.71 J is the result that I get if I ignore the centripetal acceleration (as I would have done if you did not mention it)! In other words, the official answer is incorrect, and you have identified a critical flaw in this problem.

This is actually quite interesting: it demonstrates the profound importance that the centripetal acceleration plays in a problem where the author probably just assumed (without verification) that it could be neglected. As it turns out, not only is the correct answer (which includes the effect of centripetal acceleration) quite different, it is in fact utter nonsense. (Or, you could classify this as a trick question, because it asks for a quantity that is not allowed to exist according to the problem.) This shows the danger of making "simplifying assumptions", which is the phrase that was most likely in the subconscious of the problem's author. Let this be a good lesson for all of us.

 

[jhae2.718]

It's actually quite fascinating, how the inclusion of centripetal acceleration changes the answer by so much.

I suppose it could be a trick question, but only if the answer said as much...otherwise, I'll chalk that up to an error in the problem (or, rather, an error in the attempt made at a solution by the problem's author).

Of course, the part I found most beneficial was learning how to set up the problem as an ODE, and learning how to use integrating factors...

Thank you very much for your help!

 

[turin]

I suppose it could be a trick question, but only if the answer said as much ...

That would be quite a nasty trick, especially since the official answer is wrong.

Of course, the part I found most beneficial was learning how to set up the problem as an ODE, and learning how to use integrating factors...

I would hate to beat this problem to death, especially since it has no solution (as stated). But, for the sake of learning and practicing diff. eq., you may benefit from actually approaching this problem without setting up the integral equation. That is, try to solve the problem directly in terms of force rather than energy. With some trickery, you can derive the same diff. eq., and it will demonstrate another common "trick" in solving diff. eq.s: nontrivial usage of the chain rule, both implicit and explicit.

So, in case you find some spare time and decide to do this, here is your first hint:

You still want the independent variable to be position, but the force equation will give you a diff. eq. with time as the independent variable. What physical quantity relates position and time? Use the chain rule (implicitly) to convert.

Thank you very much for your help!

The pleasure was mine!