**D'Alembert and moment inertia of wheel**
Yet another D'Alembert problem. This time a wheel (pulley?) with only one mass but this time I have to account for the moment of inertia.
When I tried to frame it mathematically I wrote it down as $\displaystyle (F_{applied}-F_{inertial}) \delta s = 0$ I ended up with the following.
$\displaystyle F_{applied} = mg$
$\displaystyle F_{inertial} = \frac{I \ddot{s}}{R^2}$
so my final solution ended up as $\displaystyle \ddot{s} = \frac{mgR^2}{I}$ which was wrong. Looking at the book solution they added an $\displaystyle m \ddot{s}$ term but am not sure why. Why does this need to be included as well when I thought the inertial term was just the rotating wheel or do I need to include the acceleration of each moving element independently, that is the wheel and the block?
I am sure I still don't know what I'm doing. This is still harder for me to wrap my head around than using vectors and Newton's second law. I thought all I had to do was to put all the explicit forces (applied forces) like gravity, etc, on the left, adjust for the direction of the virtual displacement, and then account for all the accelerations of the individual elements on the right and then multiply by their respective masses (inertial forces). Thus I thought I had an epiphany when I started thinking of D'Alembert's principle as an analog of F=ma along generalised co-ordinates but I'm not sure if that understanding is valid.
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Last edited by kiwiheretic; Aug 20th 2017 at 10:37 PM.
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