- #1
blueberrypies
- 6
- 0
\underbrace{d}_{dt} (\dot{x})^{2} = 2\ddot{x} ? or is it just 2\dot{x} ?
Understanding the Derivative of Angular Velocity in Lagrangian Mechanics
- Thread starter blueberrypies
- Start date
Click For Summary
The discussion focuses on the derivative of angular velocity in Lagrangian mechanics, specifically questioning whether the derivative of the squared velocity should be 2\ddot{x} or 2\dot{x}. Participants emphasize the importance of using the chain rule for differentiation. Clarifications are made regarding the notation used in the Lagrangian, particularly the interpretation of θ' as a variable that can be replaced by x. The conversation reflects a collaborative effort to understand the underlying principles of Lagrangian mechanics. Overall, the thread highlights the complexities of applying calculus in this context.
- #2
tiny-tim
Science Advisor
Homework Helper
- 25,837
- 258
welcome to pf!
hi blueberrypies! welcome to pf!
neither
…
use the chain rule
hi blueberrypies! welcome to pf!
neither
…use the chain rule
- #3
blueberrypies
- 6
- 0
nevermind. I'm going to work on this more.
- #4
- #5
tiny-tim
Science Advisor
Homework Helper
- 25,837
- 258
no, that would be ∂L/∂t
for ∂L/∂θ', remember that θ' is just a symbol, you can replace it by x (or anything), and forget the '
for ∂L/∂θ', remember that θ' is just a symbol, you can replace it by x (or anything), and forget the '
- #6
blueberrypies
- 6
- 0
tiny-tim said:no, that would be ∂L/∂t
for ∂L/∂θ', remember that θ' is just a symbol, you can replace it by x (or anything), and forget the '
Ohhh!
I see what you are saying :) Thanks!
Similar threads
- · Replies 8 ·
- · Replies 65 ·
- · Replies 2 ·
- · Replies 4 ·
- · Replies 16 ·
Undergrad
Question on analytic mechanics
- · Replies 3 ·
Undergrad
Lagrangian for pendulum
- · Replies 11 ·