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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/ [email protected] 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 down the cone between these turning points while going around with varying angular velocity ˙ φ = p
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