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| ### Statement |
| ### Statement |
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| $2.7.28.$ [Insert the problem statement] |
| $2.7.28.$ A thin homogeneous stick of length $l$ and mass $m$ lies symmetrically on supports whose distance is equal to $a$. One of the supports is quickly removed. What is the reaction force of the remaining support immediately afterwards? |
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| ### Solution |
| ### Solution |
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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 |
| 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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| \[\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.\] |
| \[\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.\] |
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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 |
| 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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| \[\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^)}.\] |
| \[\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^2}.\] |
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| Equating the two expressions for the angular acceleration, we have |
| Equating the two expressions for the angular acceleration, we have |
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| \[\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.\] |
| \[\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.\] |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $N=\frac{mgl^2}{l^2+3a^2}$ |
