Новое решение
en/2.7.28.md
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| + | ### Statement | ||
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| + | $2.7.28.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + |  | ||
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| + | Since the moment of inertia and the clockwise moment with respect to the center of mass are $ml^2/12$ and $Na/2$, respectivly, the angular acceleration is | ||
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| + | \[\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.\] | ||
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| + | Similarly, since the moment of inertia and the clockwise moment with respect to the remaining support are $ml^2/12+m(a/2)^2$ and $mga/2$, respectively, the angular acceleration is | ||
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| + | \[\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^)}.\] | ||
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| + | Equating the two expressions for the angular acceleration, we have | ||
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| + | \[\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.\] | ||
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| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.7.28.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| Since the moment of inertia and the clockwise moment with respect to the center of mass are $ml^2/12$ and $Na/2$, respectivly, the angular acceleration is | |||
| \[\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.\] | |||
| Similarly, since the moment of inertia and the clockwise moment with respect to the remaining support are $ml^2/12+m(a/2)^2$ and $mga/2$, respectively, the angular acceleration is | |||
| \[\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^)}.\] | |||
| Equating the two expressions for the angular acceleration, we have | |||
| \[\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||