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en/2.1.63.md
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| + | ### Statement | ||
| + | |||
| + | $2.1.63.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $2.1.63.$ A horizontal disk begins to spin around its axis with an angular velocity that increases linearly with time $\omega = \varepsilon t$. At what angular velocity will a body located at a distance $r$ from the disk's axis begin to slide off it, if the coefficient of friction between them is $\mu$? | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Because the angular velocity is constantly changing, the body experiences two components of acceleration on the plane of the disk: | ||
| + | 1. Centripetal (normal) acceleration, directed towards the center: $a_n = \omega^2 r$ | ||
| + | 2. Tangential acceleration, directed perpendicular to the radius: $a_\tau = \varepsilon r$ | ||
| + | |||
| + | Since these two vectors are strictly perpendicular, we find the total magnitude of acceleration $a$ using the Pythagorean theorem: | ||
| + | $$a = \sqrt{a_n^2 + a_\tau^2} = \sqrt{(\omega^2 r)^2 + (\varepsilon r)^2} = r \sqrt{\omega^4 + \varepsilon^2}$$ | ||
| + | |||
| + | The body starts to slide at the exact moment when the required force to maintain this total acceleration exceeds the maximum possible static friction force $F_{\text{fr}} = \mu N = \mu mg$. | ||
| + | Writing Newton's second law for the threshold of slipping: | ||
| + | $$ma = \mu mg \implies a = \mu g$$ | ||
| + | |||
| + | Substitute the expression for total acceleration: | ||
| + | $$r \sqrt{\omega^4 + \varepsilon^2} = \mu g$$ | ||
| + | |||
| + | Square both sides to solve for $\omega$: | ||
| + | $$r^2 (\omega^4 + \varepsilon^2) = \mu^2 g^2$$ | ||
| + | $$\omega^4 + \varepsilon^2 = \frac{\mu^2 g^2}{r^2}$$ | ||
| + | $$\omega^4 = \frac{\mu^2 g^2}{r^2} - \varepsilon^2$$ | ||
| + | |||
| + | Taking the fourth root, we find the angular velocity $\omega$: | ||
| + | $$\omega = \left( \frac{\mu^2 g^2}{r^2} - \varepsilon^2 \right)^{1/4}$$ | ||
| + | |||
| + | <i>Note: This solution is physically meaningful only if the expression under the root is positive, i.e., $\varepsilon < \mu g / r$. If the angular acceleration $\varepsilon$ is extremely large ($\varepsilon \ge \mu g / r$), the tangential acceleration alone immediately overcomes friction, and the body slips at $t=0$ when $\omega = 0$.</i> | ||
| + | |||
| + | #### Answer | ||
| + | $\omega = 0$ for $\varepsilon \ge \mu g/r$; | ||
| + | $\omega = (\mu^2 g^2/r^2 - \varepsilon^2)^{1/4}$ for $\varepsilon < \mu g/r$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.1.63.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.1.63.$ A horizontal disk begins to spin around its axis with an angular velocity that increases linearly with time $\omega = \varepsilon t$. At what angular velocity will a body located at a distance $r$ from the disk's axis begin to slide off it, if the coefficient of friction between them is $\mu$? | |||
| ### Solution | |||
| Because the angular velocity is constantly changing, the body experiences two components of acceleration on the plane of the disk: | |||
| 1. Centripetal (normal) acceleration, directed towards the center: $a_n = \omega^2 r$ | |||
| 2. Tangential acceleration, directed perpendicular to the radius: $a_\tau = \varepsilon r$ | |||
| Since these two vectors are strictly perpendicular, we find the total magnitude of acceleration $a$ using the Pythagorean theorem: | |||
| $$a = \sqrt{a_n^2 + a_\tau^2} = \sqrt{(\omega^2 r)^2 + (\varepsilon r)^2} = r \sqrt{\omega^4 + \varepsilon^2}$$ | |||
| The body starts to slide at the exact moment when the required force to maintain this total acceleration exceeds the maximum possible static friction force $F_{\text{fr}} = \mu N = \mu mg$. | |||
| Writing Newton's second law for the threshold of slipping: | |||
| $$ma = \mu mg \implies a = \mu g$$ | |||
| Substitute the expression for total acceleration: | |||
| $$r \sqrt{\omega^4 + \varepsilon^2} = \mu g$$ | |||
| Square both sides to solve for $\omega$: | |||
| $$r^2 (\omega^4 + \varepsilon^2) = \mu^2 g^2$$ | |||
| $$\omega^4 + \varepsilon^2 = \frac{\mu^2 g^2}{r^2}$$ | |||
| $$\omega^4 = \frac{\mu^2 g^2}{r^2} - \varepsilon^2$$ | |||
| Taking the fourth root, we find the angular velocity $\omega$: | |||
| $$\omega = \left( \frac{\mu^2 g^2}{r^2} - \varepsilon^2 \right)^{1/4}$$ | |||
| <i>Note: This solution is physically meaningful only if the expression under the root is positive, i.e., $\varepsilon < \mu g / r$. If the angular acceleration $\varepsilon$ is extremely large ($\varepsilon \ge \mu g / r$), the tangential acceleration alone immediately overcomes friction, and the body slips at $t=0$ when $\omega = 0$.</i> | |||
| #### Answer | |||
| $\omega = 0$ for $\varepsilon \ge \mu g/r$; | |||
| $\omega = (\mu^2 g^2/r^2 - \varepsilon^2)^{1/4}$ for $\varepsilon < \mu g/r$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||