Today, we will be doing one of my favorite problems of all time: USAPhO 2011 A2!.

Solution

First, we find the moment of inertia, using the parallel axis theorem. Since the stick is suspended a distance $R$ away from the center, we can write the moment of intertia as

\[I = \frac{1}{12} ML^2 + MR^2\]

Next, we can compute the torque about this axis.

\[\tau = -mgR \sin{\theta} \approx -mgR \theta\]

Using Newton’s Second Law for rotational motion, we can write the equation

\[\left (\frac{1}{12} ML^2 + MR^2 \right) \ddot{\theta} = - mgR \theta \Rightarrow \ddot{\theta} = - \frac{12gR}{L^2 + 12R^2} \theta\]

This equation of motion is in the form of a simple harmonic oscillator, so we can simply write down that its period is

\[T = 2 \pi \sqrt{\frac{L^2 + 12 R^2}{12gR}}\]

To use the data we are provided, we must turn this into a linear equation that can be plotted. Doing some algebraic manipulation on this equation, we get

\[T^2 R = \frac{12 \pi^2 R^2}{3g} + \frac{\pi^2 L^2}{3g}\]

This indicates that we can plot $R^2$ on the x-axis and $T^2 R$ on the y-axis. If we do so, we will find that the slope will be

\[m = \frac{12 \pi^2}{3g}\]

Similarly, the y-intercept of the line will be

\[b = \frac{\pi^2 L^2}{3g}\]

We first compute $R^2$ and $T^2 R$ for each data point given and report the results below, using an appropriate number of significant figures.

$R^2$ (m^2) $T^2 R$ (m/s^2)
0.0025 0.74
0.0056 0.75
0.0104 0.770
0.0243 0.826
0.0392 0.886
0.0445 0.908
0.0912 1.10
0.150 1.33
0.203 1.55
0.346 2.12

To graph this, we use Asymptote, because it looks cool and because we can. I’ve also included the code below, in case you wanted to tinker with the display.

[asy]

unitsize(150);

draw((0,0) -- (2,0), arrow=Arrow(HookHead), blue);
draw((0,0) -- (0, 2), arrow = Arrow(HookHead), blue);

label("$0$", (0, 0), SW, olive);
label("$R^2$", (2, 0), E, green);
label("$T^2R$", (0, 2), N, green);

real[] X = {0.0025, 0.005625, 0.010404, 0.024336, 0.039204, 0.044521, 0.091204, 0.149769, 0.203401, 0.345744};
real[] Y = {0.7380482, 0.7508172, 0.769692918, 0.825957756, 0.88569855, 0.907611436, 1.09596555, 1.331676675, 1.548557659, 2.12268};

for(int i = 0; i < 9; ++i)
{
    dot((X[i], Y[i]), red);
}

real sumProducts = 0;
real sumX = 0;
real sumY = 0;
real sumXSquare = 0;
int N = 10;

for(int i = 0; i < 9; ++i)
{
    sumProducts += X[i] * Y[i];
    sumX += X[i];
    sumY += Y[i];
    sumXSquare += X[i]^2;
}

real m = (N * sumProducts - sumX * sumY)/(N * sumXSquare - sumX^2);
real b = (sumY - m * sumX)/(N);

real f(real x)
{
return m * x + b;
}

path g = graph(f, 0, 0.27);
draw(g, arrow = Arrow(HookHead));


[/asy]

Diagram 6

We see that $m = 4.09$ and $b = 0.713$. Solving the equations above, we see that

\[g = \frac{12 \pi^2}{3m} = \frac{12 \pi^2}{3 (4.09)} = 9.65\ \frac{\text{m}}{\text{s}^2}\] \[L = \sqrt{\frac{12b}{m}} = \sqrt{\frac{12 (0.713)}{(4.09)}} = 1.45\ \text{m}\]