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| ### Statement |
| ### Statement |
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| $2.1.43.$ [Insert the problem statement] |
| $2.1.43.$ The horizontal axis of radius $R$ , which rotates at an angular velocity $\omega$ , is compressed by a sleeve equipped with a counterweight so that it does not rotate when moving along the axis. Determine the steady-state velocity of the bushing under the action of a force $F$ applied to it along the axis. Maximum friction force of the axle against the bushing $ F_{tr} > F$. |
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| ### Solution |
| ### Solution |
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| The shaft ( radius R) spins with angular velocity $ \omega$ , but the sleeve (bushing) is prevented from rotating by the counterweight. So at the contact surface between shaft and sleeve there is relative sliding made of two perpendicular components: |
| The shaft ( radius R) spins with angular velocity $ \omega$ , but the sleeve (bushing) is prevented from rotating by the counterweight. So at the contact surface between shaft and sleeve there is relative sliding made of two perpendicular components: |
| 1) a circumferential component, from the shaft's rotation: $ u = \omega R$ |
| 1) a circumferential component, from the shaft's rotation: $ u = \omega R$ |
| 2) an axial component, from the sleeve's motion along the shaft: $v$ ( the steady-state velocity we want) |
| 2) an axial component, from the sleeve's motion along the shaft: $v$ ( the steady-state velocity we want ) |
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| Since these |
| Since these two velocity components are mutually perpendicular ( one is along the surface's circumference, the othe along the axis), the resultant relative sliding speed is: |
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| $$ u_{rel} = \sqrt{ v^2 + ( \omega R)^2 } $$ |
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| Kinetic friction always acts opposite to the relative sliding velocity at the contact, and its magnitutde equals the maximum friction force $F_{tr}$ (given). So the friction force vector has magnitude $F_{tr}$, directed opposite to $ \vec{u}_{rel}$ . |
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| The axial component of this friction force (the part that resists the applied force $F$) is the projection of $F_{tr}$ along the axis: |
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| $$ F_{tr, axial} = F_{tr} \ \cdot \ \frac{v}{u_{rel}} = F_{tr} \ \cdot \ \frac{v}{\sqrt{ v^2 + (\omega R)^2}} $$ |
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| (this follows just from similar triangles: the axial component of friction is to $F_{tr}$ as $v$ to $u_{rel}$) |
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| "Steady-state" means the bushing moves with constant velocity, so the net exial force is zero: the applied force $F$ is exactly balanced by the avial component of friction: |
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| $$ F = F_{tr} \ \cdot \ \frac{v}{\sqrt{ v^2 + (\omega R)^2}}$$ |
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| Now solve for $v$: |
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| $$ F \sqrt{v^2 + \omega^2 R^2} = F_{tr} \ v$$ |
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| Square both sides: |
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| $$ F^2 (v^2 + \omega^2 R^2 ) = F_{tr}^2 v^2$$ |
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| $$F^2 \omega^2 R^2 = v^2 (F_{tr}^2 - F^2)$$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $$ v = \dfrac{F \omega R}{ \sqrt{F_{tr}^2 - F^2}}$$ |
