Well maybe if you aren't an instructor you shouldn't have a solutions manual

>>7652346
Still, it would be incredibly helpful whenever I get stuck.

>>7652349
well maybe if you got stuck, you'd actually learn the material.

>>7652357
>well maybe if you got stuck, you'd actually learn the material

what a fucking retarded comment
he, by his own admission >>7652349, is already getting stuck

>>7652357
>>7652364
Let me put it this way: the exercsise in the OP post is an incredibly easy pendulum exercise, or so it seems. The exercise asks to express [math]\ddot φ[/math] and [math]\dot φ^2[/math] as functions of φ, using the law of angular momentum (in polar corrdinates of course since it's easier). This is the fifth exercise on the first chapter and I'm already stumped, even though I've solved lots of rigid-body exercises and physical pendulum exercises before. I'm trying to get "into the exercise's head" and figure out the thought process for expressing [math]\ddot φ[/math] and [math]\dot φ^2[/math] as functions of φ, but I can't think of anything. I dropped the book for a couple of months waiting for the solutions manual (which, admittedly, they're not obligated to send me) but since this book is highly recommended, I'd like to pick it up again. I've looked through other physical pendulum exercises in other books but they all use the classical Newtonian force diagram approach and not the angular momentum approach.

>>7652375
1.) Write down the expresions for the kinetic energy and the potential energy.
2.) Plug it in the E-L equation to get the equations of motion.
3.) Solve them.

[math]T = \frac{m_1}{2} b^2 \dot \phi^2 + \frac{m_2}{2} (2b)^2 \dot \phi^2 = \frac{1}{2} \left( m_1 + 4 m_2\right) b^2 \dot \phi^2[/math]
[math]V = m_1 g b \cos(\phi) + m_2 g (2b) \cos(\phi) = \left(m_1 + 2 m_2 \right) g b \cos(\phi)[/math]
[math]L = T - V = \frac{1}{2} \left( m_1 + 4 m_2\right) b^2 \dot \phi^2 - \left(m_1 + 2 m_2 \right) g b \cos(\phi)[/math]

[math]\left( m_1 + 4 m_2\right) b^2 \ddot \phi + \left(m_1 + 2 m_2 \right) g b \sin(\phi) = 0[/math]
[math]\ddot \phi + \left( \frac{m_1 + 2 m_2 }{ m_1 + 4 m_2}\right) \frac{g}{b} \sin(\phi) = 0[/math]

>>7652409
[math]m_1[/math] should be [math]m_2[/math] and [math]m_2[/math] should be [math]m_3[/math].

>>7652409
>>7652410
Thanks, I did that too, but these are energy considerations and not angular momentum consideration.

Guess I'll drop the book and use the less-recommended Fowles and Cassidy Analytical Mechanics.