Physics Help Forum D'Alembert and moment inertia of wheel

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 Aug 20th 2017, 10:25 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 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. Attached Thumbnails     Last edited by kiwiheretic; Aug 20th 2017 at 10:37 PM.
 Aug 20th 2017, 11:34 PM #2 Junior Member   Join Date: Jul 2017 Posts: 6 It seems to be that you have forgotten the inertia of the mass $m$ and accounted only the inertia of pulley. In such, you have to use $F_{inertial}=\frac{I\ddot{s}}{R^2} + m\ddot{s}$. Adopting this yields the desired result. __________________ . .. ... Be patient, TEX will load, eventually...
 Aug 20th 2017, 11:38 PM #3 Junior Member   Join Date: Jul 2017 Posts: 6 To collaborate more to your question, I checked the definition of D'Alembert principle from wiki. https://en.wikipedia.org/wiki/D%27Al...%27s_principle It clearly suggests that the sum needs to be taken over each particle in the system. __________________ . .. ... Be patient, TEX will load, eventually...
Aug 21st 2017, 12:35 AM   #4
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 Originally Posted by zemozamster To collaborate more to your question, I checked the definition of D'Alembert principle from wiki. https://en.wikipedia.org/wiki/D%27Al...%27s_principle It clearly suggests that the sum needs to be taken over each particle in the system.
That is an interesting page since it uses the term "inertial force" when it should read "force of inertia." Inertial force is describe at

https://en.wikipedia.org/wiki/Fictitious_force

which also confuses inertial force with force of inertia.

Aug 21st 2017, 12:13 PM   #5
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 Originally Posted by zemozamster To collaborate more to your question, I checked the definition of D'Alembert principle from wiki. https://en.wikipedia.org/wiki/D%27Al...%27s_principle It clearly suggests that the sum needs to be taken over each particle in the system.
I presume that means every particle that is somehow connected with bits of string and pulleys, etc

Thanks for that. I will try some other problems and see if that completes my understanding on it.

Aug 21st 2017, 12:36 PM   #6
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 Originally Posted by kiwiheretic I presume that means every particle that is somehow connected with bits of string and pulleys, etc Thanks for that. I will try some other problems and see if that completes my understanding on it.
As I recall, D'Alembert's principle is derived for a discrete system of particles whereas Lagrangian dynamics is far more general than that by which I mean that it can be used for systems which are composed, at least in part, of continuous media.

 Tags dalembert, inertia, moment, wheel