{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

410ahw3soln - HW#3Solutions Phys410Fall 2011...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
HW#3—Solutions —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] 7.20 ( particle on a helix ) In cylindrical coordinates we have in general T = 1 2 m ( ˙ ρ 2 + ρ 2 ˙ φ 2 + ˙ z 2 ). Here ρ = R is constrained, and φ = z/λ , so T = 1 2 m (1 + R 2 2 ) ˙ z 2 . Also the potential energy is U = mgz relative to an arbitrary z = 0 point. So L == 1 2 m (1 + R 2 2 ) ˙ z 2 - mgz . Lagrange’s equation gives ¨ z = - g/ (1 + R 2 2 ). When R 0 this becomes free-fall, - g , which makes sense because the helix becomes a straight vertical line. 7.21 ( bead on a spinning rod ) Let’s add two interesting parts (b): Suppose the rod has length L from the pivot to one end, and the bead is released with zero radial velocity at some initial radius r 0 . In the limit that r 0 goes to zero, what velocity does the bead have when it reaches the end of the rod? Give the components of the velocity along the rod and perpendicular to the rod. Compare your result to what you would expect from dimensional analysis. (c) (i) Explain why the Hamiltonian is a conserved quantity, and show that it is equal to E - Ω J , where E and J are the kinetic energy and angular momentum of the bead, and Ω is the angular velocity of the rod (called lower case omega in the book). (ii) Use this conservation law to evaluate the kinetic energy when the bead leaves the rod, and show that it agrees with what you get using your results from part (b). Solution : (a) L = 1 2 m ( ˙ r 2 + r 2 ω 2 ). The angular part of the kinetic energy looks like an upside down harmonic oscillator. The Lagrange equation is ¨ r = ω 2 r , whose solutions are r ( t ) = A cosh( ωt ) + B sinh( ωt ). Then ˙ r ( t ) = ωA sinh( ωt ) + ωB cosh( ωt ), so ˙ r (0) = B , so if the particle is released with no radial motion from r 0 the solution is r ( t ) = r 0 cosh( ωt ). For ωt 1 this becomes r ( t ) ( r 0 / 2) e ωt . The only force in this situation is the force of constraint which acts at any instant perpendicular to the rod, in the φ direction. Thus although ¨ r is nonzero, the radial component of the acceleration vector is actually zero. How can that be? It’s because it also has an equal but opposite, inward pointing radial centripetal component to its acceleration, 2 . In the rotating frame of reference (we’ll get to that later), this means there is an apparent “centrifugal force” mrω 2 , pointing outward.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}