정답: $3$

$\mathrm{A}\backslash mathrm\{A\}$, $\mathrm{B}\backslash mathrm\{B\}$의 주기를 각각 $T_{\mathrm{A}}T\_\{\backslash mathrm\{A\}\}$, $T_{\mathrm{B}}T\_\{\backslash mathrm\{B\}\}$라 하고, 실의 길이를 $LL$, 중력 가속도를 $gg$라 할 때, 단진자의 주기는 $2\pi \sqrt{{\displaystyle \frac{L}{g}}}2\backslash pi\backslash sqrt\{\backslash dfrac\{L\}\{g\}\}$이므로
$T_{\mathrm{A}}=2\pi \sqrt{{\displaystyle \frac{0.5\mathrm{m}}{10\mathrm{m}\mathrm{/}{\mathrm{s}}^2}}}T\_\{\backslash mathrm\{A\}\}\; =\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{0.5\backslash ,\backslash mathrm\{m\}\}\{10\backslash ,\backslash mathrm\{m/s^\{2\}\}\}\}$,
$T_{\mathrm{A}}=2\pi \sqrt{{\displaystyle \frac{1}{20{\mathrm{s}}^{-2}}}}T\_\{\backslash mathrm\{A\}\}\; =\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{1\}\{20\backslash ,\backslash mathrm\{s^\{-2\}\}\}\}$이다.

물체의 질량을 $mm$, 용수철 상수를 $kk$라 할 때, 용수철 진자의 주기는 $2\pi \sqrt{{\displaystyle \frac{m}{k}}}2\backslash pi\backslash sqrt\{\backslash dfrac\{m\}\{k\}\}$이므로
$T_{\mathrm{B}}=2\pi \sqrt{{\displaystyle \frac{5\mathrm{k}\mathrm{g}}{k}}}T\_\{\backslash mathrm\{B\}\}\; =\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{5\backslash ,\backslash mathrm\{kg\}\}\{k\}\}$이다.

진동 주기는 $\mathrm{B}\backslash mathrm\{B\}$가 $\mathrm{A}\backslash mathrm\{A\}$의 $22$배이므로
$2\pi \sqrt{{\displaystyle \frac{5\mathrm{k}\mathrm{g}}{k}}}=2\times 2\pi \sqrt{{\displaystyle \frac{1}{20{\mathrm{s}}^{-2}}}}2\backslash pi\backslash sqrt\{\backslash dfrac\{5\backslash ,\backslash mathrm\{kg\}\}\{k\}\}\; =\; 2\; \backslash times\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{1\}\{20\backslash ,\backslash mathrm\{s^\{-2\}\}\}\}$,
$k=25\mathrm{k}\mathrm{g}\mathrm{/}{\mathrm{s}}^{2}k\; =\; 25\backslash ,\backslash mathrm\{kg/s^\{2\}\}$,
$k=25{\displaystyle \frac{\mathrm{k}\mathrm{g}\cdot \mathrm{m}\mathrm{/}{\mathrm{s}}^2}{\mathrm{m}}}k\; =\; 25\backslash ,\backslash dfrac\{\backslash mathrm\{kg·m/s^\{2\}\}\}\{\backslash mathrm\{m\}\}$,
$k=25\mathrm{N}\mathrm{/}\mathrm{m}k\; =\; 25\backslash ,\backslash mathrm\{N/m\}$이다.

따라서 정답은 $33$번이다.

정답: $2$

강체(막대와 물체로 이루어진 계)의 질량, 관성 모멘트, 회전축으로부터 질량 중심까지의 거리를 각각 $mm$, $II$, $dd$라 하고, 중력 가속도를 $gg$라 할 때, 물리 진자의 주기는 $2\pi \sqrt{{\displaystyle \frac{I}{mgd}}}2\backslash pi\backslash sqrt\{\backslash dfrac\{I\}\{mgd\}\}$이다.

$1.1.$ $mm$ 구하기:
$m=6\mathrm{k}\mathrm{g}+3\mathrm{k}\mathrm{g}=9\mathrm{k}\mathrm{g}m\; =\; 6\backslash ,\backslash mathrm\{kg\}\; +\; 3\backslash ,\backslash mathrm\{kg\}\; =\; 9\backslash ,\backslash mathrm\{kg\}$이다.

$2.2.$ $dd$ 구하기:
막대와 물체에 각각 $11$, $22$의 번호를 매기면 $m_1=6\mathrm{k}\mathrm{g}m\_\{1\}\; =\; 6\backslash ,\backslash mathrm\{kg\}$, $m_2=3\mathrm{k}\mathrm{g}m\_\{2\}\; =\; 3\backslash ,\backslash mathrm\{kg\}$이고,
${\displaystyle \sum _{i=1}^2{m}_i=}\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}\; =\{\}$$m_1+m_2=m\_\{1\}\; +\; m\_\{2\}\; =\{\}$$6\mathrm{k}\mathrm{g}+3\mathrm{k}\mathrm{g}=6\backslash ,\backslash mathrm\{kg\}\; +\; 3\backslash ,\backslash mathrm\{kg\}\; =\{\}$$9\mathrm{k}\mathrm{g}9\backslash ,\backslash mathrm\{kg\}$이다.

회전축을 기준점이라고 하면 $r_1={\displaystyle \frac{1}{2}}\mathrm{m}r\_\{1\}\; =\; \backslash dfrac\{1\}\{2\}\backslash ,\backslash mathrm\{m\}$, $r_2=1\mathrm{m}r\_\{2\}\; =\; 1\backslash ,\backslash mathrm\{m\}$이므로
${\displaystyle \sum _{i=1}^2{m}_i{r}_i=m_1{r}_1+m_2{r}_2}\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}r\_\{i\}\; =\; m\_\{1\}r\_\{1\}\; +\; m\_\{2\}r\_\{2\}$,
${\displaystyle \sum _{i=1}^2{m}_i{r}_i=(6\mathrm{k}\mathrm{g}\times \frac{1}{2}\mathrm{m})+(3\mathrm{k}\mathrm{g}\times 1\mathrm{m})}\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}r\_\{i\}\; =\; \backslash left(6\backslash ,\backslash mathrm\{kg\}\; \backslash times\; \backslash dfrac\{1\}\{2\}\backslash ,\backslash mathrm\{m\}\backslash right)\; +\; (3\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 1\backslash ,\backslash mathrm\{m\})$,
${\displaystyle \sum _{i=1}^2{m}_i{r}_i=6\mathrm{k}\mathrm{g}\cdot \mathrm{m}}\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}r\_\{i\}\; =\; 6\backslash ,\backslash mathrm\{kg·m\}$이다.

${\displaystyle \frac{{\displaystyle \sum _{i=1}^2{m}_i{r}_i}}{{\displaystyle \sum _{i=1}^2{m}_i}}}={\displaystyle \frac{6\mathrm{k}\mathrm{g}\cdot \mathrm{m}}{9\mathrm{k}\mathrm{g}}}={\displaystyle \frac{2}{3}}\mathrm{m}\backslash dfrac\{\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}r\_\{i\}\}\{\backslash displaystyle\backslash sum\_\{i=1\}^\{2\}m\_\{i\}\}\; =\; \backslash dfrac\{6\backslash ,\backslash mathrm\{kg·m\}\}\{9\backslash ,\backslash mathrm\{kg\}\}\; =\; \backslash dfrac\{2\}\{3\}\backslash ,\backslash mathrm\{m\}$이므로 $d={\displaystyle \frac{2}{3}}\mathrm{m}d\; =\; \backslash dfrac\{2\}\{3\}\backslash ,\backslash mathrm\{m\}$이다.

$3.3.$ $II$ 구하기:
회전축에 대한 막대의 관성 모멘트는 $2\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}2\backslash ,\backslash mathrm\{kg·m^\{2\}\}$이므로 막대와 물체로 이루어진 계의 관성 모멘트는 $2\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}2\backslash ,\backslash mathrm\{kg·m^\{2\}\}$에 물체의 관성 모멘트를 추가한 값이다.

그러므로 $I=2\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2+m_2{r}_2^{2}I\; =\; 2\backslash ,\backslash mathrm\{kg·m^\{2\}\}\; +\; m\_\{2\}r\_\{2\}^\{\backslash ,2\}$,
$I=2\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2+3\mathrm{k}\mathrm{g}\times (1\mathrm{m}{)}^{2}I\; =\; 2\backslash ,\backslash mathrm\{kg·m^\{2\}\}\; +\; 3\backslash ,\backslash mathrm\{kg\}\; \backslash times\; (1\backslash ,\backslash mathrm\{m\})^\{2\}$,
$I=5\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}I\; =\; 5\backslash ,\backslash mathrm\{kg·m^\{2\}\}$이다.

$T=2\pi \sqrt{{\displaystyle \frac{I}{mgd}}}T\; =\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{I\}\{mgd\}\}$이므로
$T=2\pi \sqrt{{\displaystyle \frac{5\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2}{9\mathrm{k}\mathrm{g}\times 10\mathrm{m}\mathrm{/}{\mathrm{s}}^2\times {\displaystyle \frac{2}{3}}\mathrm{m}}}}T\; =\; 2\backslash pi\backslash sqrt\{\backslash dfrac\{5\backslash ,\backslash mathrm\{kg·m^\{2\}\}\}\{9\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 10\backslash ,\backslash mathrm\{m/s^\{2\}\}\; \backslash times\; \backslash dfrac\{2\}\{3\}\backslash ,\backslash mathrm\{m\}\}\}$,
$T={\displaystyle \frac{\pi }{\sqrt{3}}}\mathrm{s}T\; =\; \backslash dfrac\{\backslash pi\}\{\backslash sqrt\{3\}\}\backslash ,\backslash mathrm\{s\}$이다.

따라서 정답은 $22$번이다.