HW3_prob2_BeadonWireProb - v is the velocity of the bead....

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Bead on the Wire Problem From p. 29 in either edition of Goldstein, the kinetic energy T for the bead on the rotating straight wire is T = 1 2 m [ · r 2 +( ) 2 ] , (1) where ω is the uniform (i.e., constant) rotation frequency, and the equation of motion is ·· r = 2 . (2) Note that · θ should not appear in either of these expressions because θ and · θ are no longer independent dynamical variables. The total time derivative of the kinetic energy, · T ,is · T = m [ · r ·· r + r · 2 ] , (3) which reduces to · T =2 mr · 2 , (4) after eliminating the generalized acceleration, which is not an independent variable, using Eq.(2). Because r and · r are, in general, not zero, the kinetic energy is not conserved. Note that even in conservative systems, the kinetic energy is never conserved unless the particle is executing circular motion. In this problem, there is no potential energy, so the total energy E equals the kinetic, and E is not conserved. Now compare Eq.(4) with the general result from Problem 1.4(2 nd ed.) or 1.1(3 rd ed.), · T = F · v , (5) where F is the total force on the bead and
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Unformatted text preview: v is the velocity of the bead. The only real force in this problem is the force of constraint that keeps the bead on the wire. This force does not constrain the radial motion, so we deduce that it has no radial component. In polar coordinates with unit basis vectors e r and e , the velocity is v = r e r + r e , (6) and Eq.(5) yields the explicit result T = F r. (7) Comparison of Eqs.(4) and (7) gives the following result for the force of constraint F = 2 m r . (8) This is the general result for this problem ("general" means it applies to all possible cases). If you want a general form that depends only on r (and constants), you have to f nd the f rst integral of Eq.(2), r 2 = r 2 + 2 ( r 2 r 2 ) , (9) and use it to eliminate r from Eq. (8). Then you can see how the constraint force varies with the initial conditions and with r. 1...
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