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Suppose we have a cylinder with radius $r$ and mass $m_1$ rolling (without slipping and forward in the image below) on a table with a rod hanging on a point that's fixed to the periphery of our cylinder. The rod has length $l$ and mass $m_2$. How can we then find the kinetic energy of this system? I'll try to present an image so that it becomes more clear for you. enter image description here

If we let our generalized coordinates be $\alpha$ and $\beta$ respectively, we can uniqeuly define the configuration of this system since the cylinder is rolling WITHOUT slipping.

Notice that the point of contact between the cylinder and the floor is momentarily at rest, meaning we can find the energy of the cylinder as $$ \frac{1}{2}(\frac{1}{2}m_1r^2 + m_1 r^2) \dot{\alpha}^2 = \frac{3}{4}m_1 r^2 \dot{\alpha}^2$$

I'm sure of this step. Now comes the hard part, to find the kinetic energy of the rod. We know that rigid body motion can be split up into both rotational and translational motion. We see that the point that's at instaneous rest and for which we have pure rotation about is the point about the periphery. Meaning the rod has the rotational energy $$\frac{1}{2} (\frac{1}{3}m_2l^2) \dot{\beta}^2$$

However, I'm extremely unsure of whether the angular velocity is correct here. I don't really know how to think in this scenario and in problems alike these in general where there're plenty of rigid bodies are connected and whose energies have to be taken into consideration. I'd be glad if anyone could explain the details on how to think in this problem, and then how to go on in more general problems alike these.

Please, I'm not asking for a solution. Only a explanation for how to think about the rotational vector at the point of contact between the rod and cylinder.

Thanks.

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  • $\begingroup$ How is the rod attached to cylender? Is it free to rotate? (Then we need more information). Or is the joint rigid? (That would lead to two ways of solving this problem) $\endgroup$ Commented May 11, 2022 at 14:13
  • $\begingroup$ The joint is rigidly connected yes. $\endgroup$
    – Tanamas
    Commented May 11, 2022 at 14:44
  • $\begingroup$ If it rigidly connected , thus $~\beta~$ can't be generalized coordinate ? It must be a joint (hinge joint) $\endgroup$
    – Eli
    Commented May 11, 2022 at 14:57
  • $\begingroup$ @Eli Yes it's a joint, but that doesn't mean it can't oscillate around that joint. Or am I mistaken? $\endgroup$
    – Tanamas
    Commented May 11, 2022 at 14:59

1 Answer 1

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When setting up the kinetic energy in terms of generalized coordinates, it is in many instances easiest to write $$ T = T_\text{tr,cm} + T_\text{rot,cm} $$ where $T_\text{tr,cm}$ is the translational kinetic energy of the center of mass (treating it as a point mass) and $T_\text{rot,cm}$ is the rotational kinetic energy as measured about the center of mass. One can typically find $T_\text{tr,cm}$ by deriving the Cartesian coordinates $(x_{cm}, y_{cm})$ of the center of mass of the object in terms of the generalized coordinates, and then calculating $$ T_\text{tr,cm} = \frac{1}{2} m \left( \dot{x}^2_{cm} + \dot{y}^2_{cm}\right). $$

In the case you've presented, it might be helpful to find the Cartesian coordinates of the attachment point as an intermediate step.

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  • $\begingroup$ Thank you very much Michael! $\endgroup$
    – Tanamas
    Commented May 13, 2022 at 9:05

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