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Consider a rod, with mass $m$ and length $l$, that is fixed about a pivot point at one of its ends.

Now imagine that we flick the rod so that the end of it attains a velocity of $v_{0}$.

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Now, if we look at the instant of impact, the rod appears to be moving towards the right with a velocity of $v_{0}$ or with a kinetic energy of

$KE=\frac{1}{2}mv_{0}^2$

(I draw upon this idea as I often use this technique to find the velocity of a rod after a collision using conservation of momentum)

However, if we calculate the kinetic energy of the rod with respect to its moment of inertia and angular velocity, we attain a different answer:

$KE = \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}\frac{1}{3}ml^2\omega^2$ (moment of inertia of a rod about a fixed end is $\frac{1}{3}ml^2$)

$KE = \frac{1}{2}\frac{1}{3}ml^2(\frac{v}{l})^2$

$KE = \frac{1}{6}mv^2$

However, these two values do not equal each other. What went wrong? Is it incorrect to imagine the rod as not being attached to the pivot for an instant? If so, why does conservation of linear momentum still hold when using this trick?

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  • $\begingroup$ Does conservation of momentum always mean that the energy is also conserved? Think about inelastic collisions $\endgroup$
    – IcyOtter
    Aug 9, 2019 at 5:46
  • $\begingroup$ I think you misread what I typed out @IcyOtter. The end of the rod simply attains a velocity of $v_{0}$, there is no collision occurring here. $\endgroup$
    – Dude156
    Aug 9, 2019 at 5:48
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    $\begingroup$ Sorry, seems so. After you flick the rod, the tip moves with $v_0$. That does not mean that the entire rod appears to move at that speed. The speed of each length element depends on its distance from the pivot. You are right that at the moment of the impact the velocity is to the right. But the magnitude is position-dependent. Specifically, $v(r) = v_0 / l * r$. Kinetic energy of that little piece is $1/2 * (v_0 / l * r)^2 * m / l * dr$. Integrate over $r$ from $0$ to $l$ to get the result. $\endgroup$
    – IcyOtter
    Aug 9, 2019 at 5:57
  • $\begingroup$ I would suggest you to go through derivation of rotational kinetic energy. $\endgroup$
    – sarthak
    Oct 21, 2020 at 13:22

2 Answers 2

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The speed of the rod varies from $0$ at the pivot to $v_0$ at the free end.

To find the total kinetic energy of the rod via the $\frac 12mv^2$ formula consider an element of the rod of length $dr$ at a distance $r$ from the pivot.

That element has a mass of $\frac m L dr$ and a speed of $\frac r L v_0$ and hence a kinetic energy of $\frac 12 \frac m L dr \left(\frac r L v_0 \right)^2$.

Integration along the length of the rod will give you the kinetic energy of the whole rod.

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You can say that the translational K.E. Is (1/2)m$v_o^2$ where $v_o$ is the speed of the center of mass, but with that you must include the rotational energy about the center of mass.

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