Edits to “Statement”, “Solution”, “Answer”
en/2.1.59.md
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| @@ -1,15 +1,11 @@ | |||
| ### Statement | |||
| − | $2.1.59.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $2.1.59.$ A weight of mass $m$, attached by a spring of stiffness $k$ to an axis, moves around this axis in a circle of radius $R$ with an angular velocity $\omega$. What is the length of the undeformed spring? | |||
| ### Solution | |||
| + |  | ||
| + | |||
| We write Newton's second law for the weight, considering the elastic force of the spring $F_{\text{el}}$: | |||
| $$m a_{\text{cp}} = F_{\text{el}} = k\Delta x$$ | |||
| Let's express $\Delta x$ through the length of the undeformed spring $l$: | |||
| $$m a_{\text{cp}} = F_{\text{el}} = k(R - l) \quad (1)$$ | |||
| Let's find the centripetal acceleration through the angular velocity of rotation $\omega$: | |||
| $$a_{\text{cp}} = \omega^2 R$$ | |||
| Substitute this into (1): | |||
| $$m \omega^2 R = k(R - l)$$ | |||
| $$\frac{k}{m}(R - l) = \omega^2 R$$ | |||
| From which we find $l$: | |||
| $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | |||
| @@ -28,7 +24,3 @@Solution | |||
| #### Answer | |||
| $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,15 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $2.1.59.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.1.59.$ A weight of mass $m$, attached by a spring of stiffness $k$ to an axis, moves around this axis in a circle of radius $R$ with an angular velocity $\omega$. What is the length of the undeformed spring? | $2.1.59.$ A weight of mass $m$, attached by a spring of stiffness $k$ to an axis, moves around this axis in a circle of radius $R$ with an angular velocity $\omega$. What is the length of the undeformed spring? | ||
| ### Solution | ### Solution | ||
|  | |||
| We write Newton's second law for the weight, considering the elastic force of the spring $F_{\text{el}}$: | We write Newton's second law for the weight, considering the elastic force of the spring $F_{\text{el}}$: | ||
| $$m a_{\text{cp}} = F_{\text{el}} = k\Delta x$$ | $$m a_{\text{cp}} = F_{\text{el}} = k\Delta x$$ | ||
| Let's express $\Delta x$ through the length of the undeformed spring $l$: | Let's express $\Delta x$ through the length of the undeformed spring $l$: | ||
| $$m a_{\text{cp}} = F_{\text{el}} = k(R - l) \quad (1)$$ | $$m a_{\text{cp}} = F_{\text{el}} = k(R - l) \quad (1)$$ | ||
| Let's find the centripetal acceleration through the angular velocity of rotation $\omega$: | Let's find the centripetal acceleration through the angular velocity of rotation $\omega$: | ||
| $$a_{\text{cp}} = \omega^2 R$$ | $$a_{\text{cp}} = \omega^2 R$$ | ||
| Substitute this into (1): | Substitute this into (1): | ||
| $$m \omega^2 R = k(R - l)$$ | $$m \omega^2 R = k(R - l)$$ | ||
| $$\frac{k}{m}(R - l) = \omega^2 R$$ | $$\frac{k}{m}(R - l) = \omega^2 R$$ | ||
| From which we find $l$: | From which we find $l$: | ||
| $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | ||
| @@ -28,7 +24,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | $$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||