410ahw10soln - HW#10Solutions Phys410Fall 2011 Prof. Ted...

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HW#10—Solutions —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ jacobson@umd.edu 13.12 ( bead on spinning rod ) L = 1 2 m r 2 + r 2 ω 2 ), so p = ∂L/∂ ˙ r = m ˙ r , and H = p ˙ r - L = p 2 / (2 m ) - 1 2 2 r 2 . There is no potential energy, and the kinetic energy is T = p 2 / (2 m ) + 1 2 2 r 2 , so H 6 = T + U . 13.13 ( particle on cylinder with restoring force ) The force arises from the potential U = 1 2 kr 2 , so L = 1 2 m ( ˙ z 2 + R 2 ˙ φ 2 ) - 1 2 k ( R 2 + z 2 ), and H = p 2 z / (2 m )+ p 2 φ / (2 mR 2 )+ 1 2 k ( R 2 + z 2 ). Hamilton’s equations read ˙ φ = p φ / ( mR 2 ), ˙ p φ = 0, ˙ z = p z , ˙ p z = - kz . This is simple harmonic motion in the z direction with angular frequency p k/m , while circulating with constant angular velocity. 13.14 ( particle on cone ) From Example 13.4, ˙ z = 0 ⇐⇒ p z = 0. At these turning points z tp in the z motion, the total energy is given by E = H = U eff ( z tp ; p φ ), where the effective potential is given by U eff = [ p 2 φ / (2 m ( c 2 + 1))] /z 2 tp + mgz tp , where p φ is the conserved angular momentum, and c is the pitch of the cone, ρ = cz . For a given p φ , U eff has a minimum at some value of z . The energy must be greater than or equal to the value at the minimum. When E is equal to this minimum, the orbit is circular with constant angular velocity on a horizontal cross section of the cone. When E is greater than that, there are two solutions for the turning point, and the mass oscillates up and
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This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.

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410ahw10soln - HW#10Solutions Phys410Fall 2011 Prof. Ted...

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