Physics 110A: Hints on Chap 7 probs



18. It is arbitrary whether you chose to use an angular or a linear coordinate as your Lagrangian coordinate, as they are simply related to each other. But for simplicity in grading, let's adopt the linear coordinate so we're all on the same page.

22. Form vector component first before forming the kinetic energy, because you don't want to miss cross terms.

27. For grading uniformity, let's adopt the variable x to measure down to the heaviest (4m) mass, much like on page 255. Then use y to measure from the lower pulley to the 3m mass. It is not important to carry the string length through the calculation, as these only contribute constant terms in the potential energy. As long as you have the −x and −y terms in there appropriately, you'll be good.

32. Start by writing components of the CM in the x-y frame (in terms of R, b, and θ. Oh—and use R instead of r for the radius of the cylinder to distinguish it from a variable. Keep it straight that b is the half side-length of the cube. After much simplification (best done at the stage of forming the CM position vector), the Lagrangian should have two terms multiplying θ-dot-squared. Using the small angle approximations: sinθ ≈ θ; and cosθ ≈ 1 − ½θ², the part of the Lagrangian stemming from the potential will have a single θ² term. In the end, you are looking for an equation of motion that looks like θ-double-dot = −ω²θ. In order to simplify, for small oscillations we realize that both θ and θ-dot are small. So no one will lose much sleep if squared terms in either disappear from the final statement of frequency.


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