(zero at axle): (U = m_1 g (-x) + m_2 g (x - l) ) — careful: Let’s set (U=0) at axle, then (U_1 = -m_1 g x) (if (x) positive down, (m_1) below axle, height negative), (U_2 = m_2 g (l - x))? Wait, if (x) is distance below axle for (m_1), then (m_2) is above axle by (x)? Actually, in standard Atwood: when (m_1) goes down by (x), (m_2) goes up by (x). Let (y) = downward displacement of (m_1) from fixed pulley center. Then height of (m_1) = (-y), height of (m_2) = (-(L-y))? Better: Let the pulley center be (y=0). String length (L) fixed: (y_1 + y_2 = \textconst). Let (q) = (y_1), then (y_2 = c - q). (T = \frac12 m_1 \dotq^2 + \frac12 m_2 \dotq^2). (U = m_1 g y_1 + m_2 g y_2 = m_1 g q + m_2 g (c - q) = (m_1-m_2)g q + \textconst).
While I cannot directly generate a downloadable , you can easily save this response as one by pressing Ctrl+P (or Cmd+P) on your keyboard and selecting "Save as PDF." Lagrangian Mechanics: Core Problems and Solutions
Ideal for quick review sessions before physics midterms or graduate-level qualifiers.
Visuals showing how the generalized coordinates are defined. lagrangian mechanics problems and solutions pdf
You don’t need to calculate the tension in a string or the normal force of a surface.
This is an advanced problem where the potential is velocity-dependent, requiring the use of the generalized potential , leading to the Lorentz force equation.
for a specific problem (e.g., "particle on a rotating hoop") (zero at axle): (U = m_1 g (-x)
(\sin\theta \approx \theta) → (\ddot\theta + \fracgL\theta = 0) → period (T = 2\pi\sqrtL/g).
For small oscillations, ( \sin\theta \approx \theta ), the equation becomes ( \ddot\theta + \fracgl \theta = 0 ), describing simple harmonic motion with frequency ( \omega = \sqrtg/l ).
𝜕L𝜕θ=mR2ω2sinθcosθ−mgRsinθthe fraction with numerator partial cap L and denominator partial theta end-fraction equals m cap R squared omega squared sine theta cosine theta minus m g cap R sine theta Setting up the equation of motion: Let (y) = downward displacement of (m_1) from
Write Lagrangian. (b) Find acceleration.
vm2=(Ẋ+ẋcosα)2+(−ẋsinα)2=Ẋ2+ẋ2+2Ẋẋcosαv sub m squared equals open paren cap X dot plus x dot cosine alpha close paren squared plus open paren negative x dot sine alpha close paren squared equals cap X dot squared plus x dot squared plus 2 cap X dot x dot cosine alpha