GREPhysics.NET: GR9277 #61

GR9277 #61

Problem

GREPhysics.NET Official Solution

\prob{61}

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of comfiguration I?

$\sqrt{6/5}$
$\sqrt{3/2}$
6/5
3/2
5/3

Mechanics $\Rightarrow$ }Small Oscillations

One can derive the frequency of small oscillation for a rigid body in general by using the torque form of Newton's Laws: $\tau = I \ddot{\theta} = \vec{r} \times \vec{F}$ . (I is moment of inertia, r is moment arm)

In this case, one has a constant downwards force $F=mg$ , which acts at a moment arm angle $\theta$ . Thus, $I \ddot{\theta} = -rmg \sin\theta \approx -rmg \theta$ , where the approximation works if $\theta << 1$ .

The equation of motion for small angles is thus $\ddot{\theta}=-(mgr/I)\theta$ . This is similar in form to that of a simple harmonic oscillator with the angular frequency being $\omega=\sqrt{mgr/I}$ .

Now, the problem is to find the angular frequency for each system.

One needs not worry about the rod, since it is massless and has no moment of inertia.

The moment of inertia of system I is just $I_I = 2mr^2$ . The radius of gyration r is just $r_I = 2r$ (an r for each mass).

The moment of inertia of system II is $I_{II}=mr^2 + m (r/2)^2 = \frac{5}{4}mr^2$ . The radius of gyration r is just $r_{II} = r/2+r=3r/2$ .

Thus $\omega_{II}/\omega_{I} = \sqrt{(r_{II}I_{I})/(r_{I}I_{II})}=\sqrt{2\times 3 r/2/(2(5/4)\times r)}=\sqrt{6/5}$ , as in choice (A).

Alternate Solutions

oliTUTilo
2012-11-04 12:17:14

I suggest using the Lagrangian formalism with the angle theta as your coordinate and the small angle approximation at hand (you just need $sin(\theta) \approx \theta$ for small $\theta$ ). If you're comfortable with it, you can pretty quickly find the frequency of oscillation for the generalized pendulum of system II with pretty much no extraneous memorization. If you can't remember that the simple pendulum's frequency is $\sqrt{\frac{g}{l}}$ , you can generalize the distance of the intermediate mass to quickly find the frequencies for both systems.

Let $a$ be the distance of the intermediate mass from the pivot.

(Lagrangian) $L = T -V = \{\frac{1}{2}m\dot{\theta}^2 l^2+\frac{1}{2}m\dot{\theta}^2 a^2\}-\{-mglcos(\theta}-mgacos(\theta)\} = \frac{1}{2}m\dot{\theta}^2(l^2+a^2) + mg(l+a)cos(\theta)$

(Euler Lagrange Eq) $\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{\theta}} = \frac{\partial L}{\partial \theta}$
$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{\theta}} = \frac{\partial}{\partial t} m\dot{\theta}(l^2+a^2) = m\ddot{\theta}(l^2+a^2)$
$\frac{\partial L}{\partial \theta} = -mg(l+a)sin(\theta)$
$\Rightarrow m\ddot{\theta}(l^2+a^2)=-mg(l+a)sin(\theta) \Rightarrow \ddot{\theta} = -g\frac{l+a}{l^2+a^2}sin(\theta) \approx -g\frac{l+a}{l^2+a^2}\theta \Rightarrow \omega \approx \sqrt{g\frac{l+a}{l^2+a^2}}$

For system I, $a = l \Rightarrow \omega \approx \sqrt{g\frac{2l}{2l^2}} = \sqrt{\frac{g}{l}$ , the simple pendulum.
For system II, $a = \frac{l}{2} \Rightarrow \omega \approx \sqrt{g\frac{\frac{3}{2}l}{\frac{5}{4}l^2}} = \sqrt{\frac{6g}{5l}}$

So the ratio of frequencies of System II and I is $\sqrt{\frac{6}{5}}$

This may have looked long, but in practice you'd hopefully be better experienced with Langrangians and be able to cut corners (like knowing that out of L, only T is going to have $\ddot{\theta}$ in it and only V is going to have $\theta$ in it). Depends on how used to them you are, but you at least don't need to memorize anything you shouldn't have memorized anyways (crucial for me).

BrandeisUDR
2016-02-28 00:55:55

The discrepancy in your negative sign before square rooting comes from the fact that the potentials should be positive in the initial Lagrangian, not negative. Otherwise, great alternate solution.

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