I have seen many questions that use energy conservation to find the angular speed of the rod. For example, if the rod is vertical and starts to fall, the decrease in potential energy will transfer to the kinetic energy of the rotation ONLY. What I do not understand is, since the center of the rod has translational speed, wouldn't that account for some kinetic energy that is converted from potential energy?
1
-
$\begingroup$ So a continuous body, such as a rod with uniform density, will have kinetic energy due to the angular momentum of the rod and kinetic energy due to the translational momentum of the center of mass. We may write: $T = T_\omega + T_v = \frac{1}{2}I\omega^2+\frac{1}{2} M v^2$. M is the total mass of the rod, I is the moment of inertia. $\endgroup$– GeraldCommented Oct 1, 2022 at 16:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
The kinetic energy of a rigid body can be written as the sum of two contributions
$K = \dfrac{1}{2} m |\mathbf{v}_G|^2 + \dfrac{1}{2} \mathbf{\omega} \cdot \mathbb{I}_G \cdot \mathbf{\omega}$.
Assuming a planar motion of a beam hinged at one of its end, the system has only one degree of freedom. Let's choose the angle of rotation w.r.t. the vertical direction. We can write:
- angular velocity: $\mathbf{\omega} = \dot{\theta} \mathbf{\hat{k}}$
- velocity of the center of mass: $\mathbf{v}_G = \dfrac{\ell}{2} \dot{\theta} \mathbf{\hat{\theta}}$
- component of the inertia tensor in $\mathbf{\hat{k}} \otimes \mathbf{\hat{k}}$: $I_{G,kk} = \dfrac{1}{12} m \ell^2$
- vertical component of the center of mass w.r.t. the hinge $z_G = \dfrac{\ell}{2} \cos \theta$.
Thus we can write the kinetic energy as
$K = \dfrac{1}{2} m\dfrac{\ell^2}{4}\dot{\theta}^2 + \dfrac{1}{2} m\dfrac{\ell^2}{12}\dot{\theta}^2 = \dfrac{1}{6} m \ell^2 \dot{\theta}^2$,
the potential energy
$V = m g z_G = m g \dfrac{\ell}{2} \cos \theta$
and the principle of conservation of energy as
$E = \dfrac{1}{6} m \ell^2 \dot{\theta}^2 + m g \dfrac{\ell}{2} \cos \theta = \text{const.}$,
and taking the time derivative you can get the equation of motion
$\left( \dfrac{1}{3} m \ell^2 \ddot{\theta} + m g \dfrac{\ell}{2} \sin \theta \right) \dot{\theta} = 0$
-
$\begingroup$ But in this example toppr.com/ask/question/… The answer only has KE = Iw^2/2, ignoring translational kinetic energy $\endgroup$ Commented Oct 1, 2022 at 16:55
-
$\begingroup$ They just write the kinetic energy using the velocity of the hinge $H$ (that is zero, so you don't see any contribution of the translation) and the rotational part uses the inertia w.r.t. to $H$, that is $I_H = \frac{1}{3} m \ell^2$. You got the same result, since you can write the kinetic energy as $K = \frac{1}{6} m \ell^2 \dot{\theta}^2 = \frac{1}{2} I_H \dot{\theta}^2$ $\endgroup$– basicsCommented Oct 1, 2022 at 17:06
-
$\begingroup$ Let's obtain the result of the problem you mentioned. At the beginning $\theta_0 = 0$, $\dot\theta_0 = 0$, at the end you know that $\theta_1 = \pi / 2$ and $\dot \theta_1$ is the unknown. You can write $E_0 = E_1$ and thus $m g \dfrac{\ell}{2} = \dfrac{1}{6} m \ell^2 \dot \theta_1^2$ and thus $\dot \theta_1 = \sqrt{\dfrac{3 g }{\ell}}$ $\endgroup$– basicsCommented Oct 1, 2022 at 17:07
-
$\begingroup$ I understand all your comments, except that "the velocity of the hinge H (that is zero)". Did not the hinge move and have a translational velocity (at least on one end)? That was exact my original confusion. $\endgroup$ Commented Oct 1, 2022 at 17:14
-
$\begingroup$ The point of the rod that is hinged to the ground is fixed in space (and thus has zero velocity), while the rod can rotates around it. When you open/close a door, the points on the hinges don't move (if the hinge is infinitely small, a point hinge), while all the other points of the doors describe a rotational motion around the axis of the hinge $\endgroup$– basicsCommented Oct 1, 2022 at 17:18