hw5_solutions

hw5_solutions - Physics 325, Fall 2006 Homework Solutions...

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Physics 325, Fall 2006 Homework Solutions #5 1) a) ~ F ( x, y, z ) = α ± 9 x 2 y 2 z - 2 xyz 3 ² ˆ x + α ± 6 x 3 yz - x 2 z 3 ² ˆ y + α ± 3 x 3 y 2 - 3 x 2 z 2 y ² ˆ z ~ ∇ × ~ F = α ± 6 x 3 y - 3 x 2 z 2 - 6 x 3 y + 3 x 2 z 2 ² ˆ x + α ± 9 x 2 y 2 - 6 xyz 2 - 9 x 2 y 2 + 6 xyz 2 ² ˆ y + α ± 18 x 2 yz - 2 xz 3 - 18 x 2 yz + 2 xz 3 ² ˆ z = ~ 0 b) Using the path described in class, U ( x, y, z ) = I x + I y + I z where I x = - Z x 0 ~ F ( x 0 , 0 , 0) · ˆ xdx 0 = 0 It’s zero because z = 0 along this path. I y = - Z y 0 ~ F ( x, y 0 , 0) · ˆ ydy 0 = 0 It’s zero because z = 0 along this path. I z = - Z z 0 ~ F ( x, y, z 0 ) · ˆ zdz 0 = α ± yx 2 z 3 - 3 x 3 y 2 z ² So U ( x, y, z ) is just I z plus a constant. But this constant needs to be zero in order for U (0 , 0 , 0) = 0. U ( x, y, z ) = α ± yx 2 z 3 - 3 x 3 y 2 z ² c) The bead feels the force F as well as any normal forces from the wire. The normal forces from the wire do no work, and so the change in potential energy of the bead is minus the change in kinetic energy.
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hw5_solutions - Physics 325, Fall 2006 Homework Solutions...

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