Jul 9, 2008
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Fall 2008 -- Purdue University -- West Lafayette, IN
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7 comments:
The back of the book has an equation to find the radius of curvature on p.700 and try plugging that into your acceleration equation.
I'm stuck at the same point. I understand what Matt is saying about using the equation for radius of curvature on pg. 700, but I don't understand how to do the derivatives. It's probably a technique that I've forgotten. Any help would be greatly appreciated!
-John
I'm currently getting the first derivative as (2*x)/25 and the second derivative as 2/25. When I solve for rho, plugging in 0 for x, I get 12.5. Then, when I use this rho to solve for the tension, I get 0.3728. However, if I divide the 12.5 (rho) by 2 to get 6.25, my solution for tension becomes 0.7456 which is pretty close to the answer. Any ideas on what's wrong?
-John
John,
You should get a more complex value for dy/dx with a radical. Something like
(-4x/25)*(1-(x/5)^2)^(-1/2)
which for y(0) will be zero as you showed before but when you differentiate this value you'll get everything to drop out after plugging in x=0 except (-4/25)*1
That first dy/dx so you can use it to check. Just make sure you are using the chain and product rule to differentiate. Hope that helps
Nathan makes a good point that you need to take care in doing the differentiation to find the first and second derivatives of y with respect to x. The result is not as simple as you have shown in your comment.
One suggestion to simplify these calculations is differentiate the equation for the path directly rather than solving for y = y(x) first.
Differentiation of the path equation gives:
(2/25)*x + (1/8)*y*(dy/dx) = 0
Solve this equation for dy/dx ( = 0).
Differentiate the path equation again with respect to x:
(2/25) + (1/8)*[(dy/dx)^2 + y*(d^2_y/d_x^2)] = 0
Solve this equation for d^2_y/d_x^2.
Either approach works; use the one that is easiest for you.
Thanks for the help! I knew the derivative was more complicated than I was making it, I just didn't see how to solve it (forgot the chain rule). Thanks again!
-John
When I checked this thread to see if I was doing the problem correctly, I realized that there was a simple solution. This is a simplified version of how I did the problem:
Ra x m*Va = Rb x m*Vb
(x^2)/25 + (y^2)/16 = 1
d/dt of position:
(x*x-dot)*(2/25) + (y*y-dot)/16 = 0
d/dt of d/dt(pos):
(x-dot^2 + x*x-dot-dot)*(2/25) + (y-dot^2 + y*y-dot-dot)/8 = 0
Additionally, note from your FBD at point B that y-dot-dot = -Tb/m.
Solve the first equation for Vb. Vb turns out to be x-dot in the negative i-direction. In the position equation set x = 0 to get the value of y at B.
Then, all you have to do is plug the following into the (d/dt)^2 equation and solve for Tb:
y-dot-dot = -Tb/m
x-dot = -Vb
x-dot-dot = 0
y-dot = 0
x = 0
y = 4 ft
I got Tb = 0.746 lb.
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