First see the picture below for the problem.
My attempt is below
Part (1)
For the moment of inertia i calculated that $$I=\frac{1}{3}ml^2$$
this is mainly from using resources online since i have no idea how to calculate it it was confusing me because in some texts they've written it as an intergral but i couldn't find any examples for systems similar to this, i know that the center of mass for the rod $cm$ is when $$cm=\frac{l}{2}l$$ ?
For part (2) i'm struggling on choosing an appropriate coordinate system to use? I have been trying to use polar coordinates but it is too hard, because i don't know how to include the rotational inertia in the formula for finding the kinetic energy, i know that $$T=\frac{1}{2}mv^2$$ where t is the kinetic energy,but how can one calculate v in this question?
for part (3) i found that the potential energy is $$mg(1-cos\theta)$$ but i dont think this is correct either because i havent been able to figure out what coordinate system to use for the problem.
i know that the lagrangian is given by $$L=T-U$$
i havent been able to attempt part 5 or 6 as of yet so i dont really need any information regarding them yet, i just want to know and understand how to attempt the first three since it would help with the revision that i am doing, none of the examples we have done in class have been similar, so i'm quite lost.
i have been using the following textbooks to help but to no prevail
(1) classical mechanics by john r taylor.
(2) schaums outline series lagrangian dynamics (the problems in this book are a lot more advanced than the ones given in lectures)
any help or direction would be extremely appreciated as i'm quite stuck.
Best Answer
The distance between the center of the rod and the point of suspension is $$d = l \sqrt{\frac{5}{4} + \cos(\theta_2 - \theta_1)}$$ Use parallel axis theorem to find $$I = \dfrac{1}{12}ml^2 + md^2 = \left(\frac{4}{3} +\cos(\theta_2 - \theta_1) \right)ml^2$$