Does Alembert work for systems of rigid bodies? My question comes from this image. (sorry for the poor drawing!)


I'm trying to find the acceleration of the block exclusively from the sum of forces in the y component, but when I do so I get $a=g$, which is the same as an object in free fall. On the other hand, if I use the relationship between angular acceleration and linear acceleration I find the right answer. Why can I not sum the forces in the $y$ direction and take $a$, that is the only question.

Here are the calculations: e

  • $\begingroup$ Are there sliding friction? D'Alembert Principle does not hold for such forces, cf. e.g this Phys.SE post. $\endgroup$ – Qmechanic Jun 30 '17 at 12:42
  • $\begingroup$ No, completely free of sliding $\endgroup$ – George Sailor Jun 30 '17 at 13:06
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    $\begingroup$ Is the "intuitive" in the title of this question really appropriate? $\endgroup$ – joshphysics Jun 30 '17 at 15:12

This is how you can solve this problem using Lagriangian mechanics, which has D'Alemberts principle built in.

Call the height of $m$ $q$. The Langrangian becomes:

$$\mathcal{L} = T-V = \frac{1}{2}m\dot{q}^2+\frac{1}{2}I\left(\frac{\dot{q}}{R}\right)^2 - qgm$$

Therefor: $$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) = \frac{\partial \mathcal{L}}{\partial q} \implies m\ddot{q} + \frac{I}{R^2}\ddot{q}=-gm$$

Now solve for $\ddot{q}$: $$\ddot{q}=\frac{-gm}{m+\frac{I}{R^2}}$$

  • $\begingroup$ Just a question, how would it be conceptually wrong to solve the problem the way I did? The answer is wrong, I know, but I applied d'Alembert principle the way it's stated right? I mean, the sum of the forces should be mass times acceleration, which would turn out to be a=g? $\endgroup$ – George Sailor Jun 30 '17 at 12:00

You first need to construct the free body diagrams of hanging mass and rotating disc.

For the mass there are two acting forces : tension force of the rope $T$ and gravitational force $m g$. So its equation of motion will be : $m g - T = m a$

For the disc, you have reaction of the tension force on tangential axis and constraint force on the pin joint. So disc is translationally static. However, tangential force makes disc rotate which can be captured by the differential equation $T R = I \alpha$.

By kinematics, we have $\alpha = a/R$ which leads to $T = I a /R^2 $. Substituting this relation of tension force $T$ in equation for mass, you obtain $m g = a (m + I/R^2)$. Thus, $a = (m g) / (m + I/R^2)$ simply.

  • $\begingroup$ You are using $T$ in two different ways here as far as I can tell. The $T = I \alpha$ equation uses $\tau$ for torque usually, not $T$ for tension. $\endgroup$ – JMac Jun 30 '17 at 11:24
  • $\begingroup$ Corrected. There was minor error in rotational EOM. $\endgroup$ – Gorkem Secer Jun 30 '17 at 11:27
  • $\begingroup$ Alright, I understand this approach, but the theorem of equivalent forces tells us that we can consider the whole system as long as we ignore the connecting forces and only represent the external forces, as I did. I'm only worried about how to approach the next problems if I want to consider the whole system, I hope I'm being clear... $\endgroup$ – George Sailor Jun 30 '17 at 11:31
  • $\begingroup$ And one last thing, how would the answer be wrong if I would consider -W_D+R-W_A=ma, which results in a=g if I should argue with someone for that result, accounting for the fact that I only used the sum of external forces is equivallent to the effective force? $\endgroup$ – George Sailor Jun 30 '17 at 11:41
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    $\begingroup$ You are right. A shortcut method would be to consider only external forces and generalized coordinates (degrees of freedom) on the system in Lagrangian formulation. For this example, the single degree of freedom is vertical position of the mass, $z$. Total kinetic energy : $T = \dfrac{1}{2} m \dot{z}^2 + \dfrac{1}{2} I \dot{z}^2/R^2$. Total potential energy : $U = m g z$. Lagrangian $L := T - U$. The system dynamics can be found by minimizing Lagrangian as $\dfrac{d \,}{d t} \left(\dfrac{\partial L}{\partial \dot{z}}\right) - \dfrac{\partial L}{\partial z} = 0$ $\endgroup$ – Gorkem Secer Jun 30 '17 at 11:44

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