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2007美国US F=MA物理竞赛 (id: 8af1298c9)

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admin 发表于 2025-12-20 23:13:49 | 显示全部楼层 |阅读模式
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 indrf- c4h .9g9imm1)acj. hp zqo2+, ha eojsjk -(an;aertia of a sphere about an axis through the center of the sphere oj-ae sn;,h2ja+oak(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 spher c8dq6q )r gm:0l0u9+mjz stfme 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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