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I am looking at problem 26 of the physics 2001 GRE and am faced with the following problem:

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown above, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

I understand that the total potential energy of the rod is $MgL/2$, and that this quantity must be conserved. I then think to myself, the sum of all of the energy in the rod right when it strikes the ground must be entirely translational, since the potential energy is 0 at the ground. This gives $$E_{translational} = \sum{}\frac{1}{2}m_iv_i^2 = \int{\frac{1}{2}dmv(r)^2} = \int{\frac{1}{2}dmr^2\omega^2} = \frac{1}{2}I\omega^2 = \frac{mL^2}{6}\omega^2$$ Where v(r) is the velocity of any mass dm a distance r from the pivot.

Solving for $\omega$ and using this to find $v(L)$ yields $v(L) = \sqrt{3gL}$ which is the correct answer. However, this derivation seems like a bad approach on the PGRE in that I haven't really used any center of mass ideas except for the fact that the potential energy of the rod can be treated as if the rod were a point mass at height $L/2$, that is, $U = U_{com}$. What other approaches are there to solving this problem?

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  • $\begingroup$ I think you have solved it in the most efficient way I can think of. $\endgroup$ – mwengler Sep 6 '17 at 20:28
  • $\begingroup$ This looks like a list question... $\endgroup$ – DanielSank Sep 7 '17 at 1:26
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Another approach: the moment of inertia of a rod of mass $m$ and length $\ell$, pivoted from one end, is $I=\frac13 m \ell^2$. The rotational kinetic energy is $E = \frac12 I \omega^2$, which must be equal to the potential energy at the start, which is $\frac12 m g \ell$.

Also, we know that the velocity at the tip of the rod $v=\omega \ell$

So

$$\frac12 \left(\frac13 m \ell^2 \right)\omega^2 = \frac12 m g \ell\\ v^2 = 3g\ell$$

Which is the same result by a (slightly) different method.

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