Pendulums — Question 4

Question 4

Two different pivots can give the same physical-pendulum period

A body has radius of gyration $k_g=.20\,\mathrm{m}$. It can hang from pivots on the same line through its center of mass, each above the center in its own hanging configuration. One pivot distance is $d_1=.10\,\mathrm{m}$. Use the effective length $\ell_{\mathrm{eff}}=d+k_g^2/d$.

1. Find a distinct pivot distance $d_2$ giving the same small-angle period.

2. Find the common effective length and period for $g=10\,\mathrm{m/s^2}$. Derive the product relation between distinct equal-period pivot distances.

Question 4 - Solution

Two different pivots can give the same physical-pendulum period

(a) Derivation and calculation

Equal periods require equal effective lengths:$$L_{\mathrm{eff}}=d+\frac{k_g^2}{d}.$$$$k_g=.200,\ d_1=.100:\quad L_{\mathrm{eff}}=.100+\frac{.0400}{.100}=.500\,\mathrm m.$$$$d^2-.500d+.0400=0\Rightarrow d=.100\ \text{or}\ .400\,\mathrm m.$$

(b) Result and interpretation

$$\boxed{d_2=.400\,\mathrm m},\qquad\boxed{T=2\pi\sqrt{.500/10}=1.40496\,\mathrm s}.$$$$d_1+\frac{k_g^2}{d_1}=d_2+\frac{k_g^2}{d_2}\Rightarrow(d_1-d_2)\left(1-\frac{k_g^2}{d_1d_2}\right)=0.$$$$d_1\ne d_2\Rightarrow\boxed{d_1d_2=k_g^2}.$$

Physical check

The two distances lie on opposite sides of the minimum-period distance $d=k_g$.