Source: Principles of Physics
Problem Sets:
Problem
A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance $x$ between the pivot point and the rod's center. The rod's length is $L = 2.20$ m and its mass is $m = 22.1$ g.
- What is the minimum period?
- If $x$ is chosen to minimize the period and then $L$ is increased, does the period increase, decrease, or remain the same?
- If, instead, $m$ is increased without $L$ increasing, does the period increase, decrease, or remain the same?
[Q1] 2.26 s [Q2] Increase [Q3] Remain the same
The period of a physical pendulum is $T = 2\pi \sqrt{I / (mgd)}$. For a thin rod pivoted at distance $x$ from its center, the distance to the center of mass is $d=x$. By the parallel axis theorem, the moment of inertia about the pivot is $I = I_{cm} + mx^2 = \frac{1}{12}mL^2 + mx^2$. The period as a function of $x$ is:
$$T(x) = 2\pi \sqrt{\frac{m(L^2/12 + x^2)}{mgx}} = 2\pi \sqrt{\frac{L^2/12 + x^2}{gx}}$$To find the minimum period, we minimize the term inside the square root by taking its derivative with respect to $x$ and setting it to zero. This gives the optimal distance $x = L/\sqrt{12}$. Substituting this value of $x$ into the period equation gives the minimum period:
$$T_{min} = 2\pi \sqrt{\frac{2(L^2/12)}{g(L/\sqrt{12})}} = 2\pi \sqrt{\frac{L}{g\sqrt{3}}}$$From this expression, $T_{min}$ is proportional to $\sqrt{L}$ and is independent of mass $m$.