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Mar 31, 2008
6/136
When completing this problem, my final answer is off by about 40 lbs. Work energy there is no work because N is always perpendicular, translational and rotational component to kinetic energy before and after. Kinematics to set V=omega*r, where r is the radius of the disk. When solving, my r drops out anyway and so does the 200 lbs. I get an expression for v2 of the disk. Using newton, I get -m*g+N=m(vdot +v2^2/row). Vdot is a zero because there is no force component tangent to the path. I get an answer that is too large. Any suggestions?
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3 comments:
Yeah, I don't know how you got r to drop out, but everything else sounds fine. Where I got tripped up, and you may have this same problem, is remembering the potential energy is going to be related to the distance that the CENTER of the wheel travels, and also, the radius of curvature is going to be the distance from the center of the wheel to the center point of the curve... maybe that helps.
I don't quite understand your question but I'll try to help. To be honest, this problem took me forever to figure out. The hardest part is figuring the potential energy change.
To make life easier I broke it down in 4 parts:
a) First draw out the position of the circle center at the three important positions (beginning (pt. C), at the end of the first 12 in (pt. B), and at pt. A). Don't forget (the circle center) will always be 6 in above the ground at any position.
b) Second calculate the height change from C to B, and then from B to A. This is the tricky part. With your circle center positions already determined at the respective points try to connect them with lines and to the vertical line at A. You'll see some triangles there (30-60-90) and by using simple trig you should be able to calculate the numbers you want. REMEMBER you want the vertical drop (change).
c) The other numbers you should have. Do a WE analysis and you can calculate v_a.
d) Once you calculate v_a just do a Newton analysis at A using path coordinates.
That's it! If you still don't get it let me know I'll try to get a drawing done or something.
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