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Elastic Force over Semicircle

  • Thread starter Peach
  • Start date
  • #1
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Homework Statement


A variable force F_vec is maintained tangent to a frictionless, semicircular surface View Figure . By slowly varying the force, a block with weight w is moved through the angla theta, and the spring to which it is attached is stretched from position 1 to position 2. The spring has negligible mass and force constant k. The end of the spring moves in an arc of radius a.

Calculate the work done by the force F_vec.

Homework Equations


K_1 + U_1 + Wother = K_2 + U_2


The Attempt at a Solution


I'm not sure if I'm getting this correctly. Besides the elastic force, there's also the varying force correct? So that should Wother?

This is my equations so far:

K_1 = 0
U_1 = 0

K_2 = 1/2mv^2
U_2 = 1/2k(a*theta)^2

But I'm stuck on finding the velocity for K_2 in terms of a and theta. Any help is greatly appreciated, thanks. ... Assuming I'm correct so far (which I sorta doubt).
 

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Answers and Replies

  • #2
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Actually, when the problem says it is moved very slowly, that means you can neglect the fact that it has a tiny kinetic energy. So, K_2 should be 0. The other parts of your work look correct. :)
 
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  • #3
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I entered that and it says it's incorrect. What else am I missing? What about the weight on top of the spring? Does that have something to do with it?
 
  • #4
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Ah, yes, if the block is being raised vertically you will need to include potential energy of mgh after it is raised. (Of course you'll need to express mgh in terms of w and theta.)
 
  • #5
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So I have to add the potential energy of the box to the potential energy of the spring (which is the elastic force right)? So mgh_box = (w*a*theta)?
 
  • #6
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You do have to add the potential energy of the box to that of the spring, but be careful how you do it. Remember that h is the height the object has been raised, not the distance it has traveled along the semicircle.
 
  • #7
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The height isn't theta, is it? Wait, shouldn't the height be the radius then? >< Sorry I'm so confused about this circle thing.
 
  • #8
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It's okay, to see how much the block has been raised, draw a line on the diagram to represent the height the block has moved:

http://img255.imageshack.us/img255/3154/yffigure740st3.jpg [Broken]

Look familiar?
 
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  • #9
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Got it. So the total potential energy should be:

deltaU = wasin(theta) + 1/2k(a*theta)^2

Right? Just making sure because I only have one last attempt.
 
  • #10
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That looks good to me. :)
 
  • #11
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The answer is wrong. What else could we be missing? This is so frusterating. ><
 
  • #12
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I'm not sure. I just worked the problem by integrating an infinitesimal amount of work dW = F*dx and I got the same answer.

Net force acting on the block during its movement along the semicircle (mgsin(theta) and N cancel):
[tex]
\sum F_{net} = F - kx - wcos(\theta) = 0
[/tex]

So,

[tex]
F = wcos(\theta) + kx
[/tex]

(Remember x = a*theta and dx = a*dtheta)

[tex]
dW = F dx \\
= (wcos(\theta) + kx) dx \\
= (wcos(\theta) + k(a \theta)) a d\theta \\
[/tex]

[tex]
W = \int_0^{\theta_2} w\,cos \theta \,a\,d\theta\,+\,\int_0^{\theta_2} k a \theta\,a\,d\theta \\
= w\,a\,\int_0^{\theta_2} cos \theta \,d\theta\,+\,k a^2 \int_0^{\theta_2} \theta\,d\theta \\
[/tex]
[tex]
W = w a sin(\theta_2) + \frac{1}{2}k a^2 \theta_2^2
[/tex]

where [itex]\theta_2[/itex] is the final angle (the initial angle is zero).

...which should be the correct answer...?

Is the homework an online assignment where you have to fill out a textbox? Could the error be in the format you entered?
 
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  • #13
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Yeah, it's an online hw. Erm, I just checked again the answer I entered and I didn't make a mistake entering it. I don't know what's the answer since I've maxed out my attempts already. :x I'll ask my TA if it's possible. Thanks for helping me so much already, really appreciated it.
 
  • #14
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Sure, I'd be interested in knowing the answer to this problem, so keep me posted when you find out.
 
  • #15
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Peach. The answer is correct. Check the way you entered it. Make asin(theta), apparently, is not the same as a*sin(theta). The system probably screwed you over that way.
 
  • #16
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I don't remember how I entered it, whether it was asin(theta) or a*sin(theta) but the answer that I entered is showing up as asin(theta).

I can't copy it directly but this is what it's showing up as:

(w*asin(theta)) + (1/2)*k*(a*theta)^2
 
  • #17
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is there possibly a different way to write the solution? i've thought about this problem ALOT and although the answer is right, i know i'm missing something stupid.
 
  • #18
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i had the same problem as you. i entered it with the asterisk and it was fine. i believe the system interprets "asin(theta)" as arcsin?
 

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