本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
If the rotational inertia + /l2dg*giwyjtkl it )kup- r953d)bh.pzm:+sbsxj 8 of a sphere about an axis through the center of the sphexhk lbt t. mzd93 j)ps+br: )pi8ks5-ure is $I$, what is the rotational inertia of another sphere that has the same density, but has twice the radius?
A. $2I$
B. $4I$
C. $8I$
D. $16I$
E. $32I$
参考答案: E
本题详细解析:
The rotational inertia of a solid sphere;4 ;te-gua mqx is $I = \frac{2}{5}MR^2$.
The mass $M$ is related to density $\rho$ and volume $V = \frac{4}{3}\pi R^3$ by $M = \rho V = \rho (\frac{4}{3}\pi R^3)$.
Substitute this expression for $M$ into the inertia equation:
$I = \frac{2}{5} \left( \rho \frac{4}{3}\pi R^3 \right) R^2 = \left( \frac{8}{15}\rho\pi \right) R^5$.
This shows that for a constant density $\rho$, the rotational inertia $I$ is proportional to the radius to the fifth power: $I \propto R^5$.
If the new sphere has twice the radius ($R' = 2R$), its new inertia $I'$ will be:
$I' \propto (R')^5 = (2R)^5 = 32 R^5$.
Therefore, $I' = 32I$.
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