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| | | ### Statement |
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| | − | $2.1.43.$ 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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| | + | $2.1.43.$ 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 is $F_{\text{fr}} > F$. |
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| | | ### 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 the shaft and the sleeve, there is relative sliding made of two perpendicular components: |
| | + | 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) |
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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: |
| | − | 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 ) |
| | + | Since these two velocity components are mutually perpendicular (one is along the surface's circumference, the other along the axis), the resultant relative sliding speed is: |
| | + | $$u_{\text{rel}} = \sqrt{v^2 + (\omega R)^2}$$ |
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| | − | 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: |
| | − | $$ u_{rel} = \sqrt{ v^2 + ( \omega R)^2 } $$ |
| | + | Kinetic friction always acts opposite to the relative sliding velocity at the contact, and its magnitude equals the maximum friction force $F_{\text{fr}}$ (given). So the friction force vector has magnitude $F_{\text{fr}}$, directed opposite to $\vec{u}_{\text{rel}}$. |
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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}$ . |
| | + | The axial component of this friction force (the part that resists the applied force $F$) is the projection of $F_{\text{fr}}$ along the axis: |
| | + | $$F_{\text{fr, axial}} = F_{\text{fr}} \cdot \frac{v}{u_{\text{rel}}} = F_{\text{fr}} \cdot \frac{v}{\sqrt{v^2 + (\omega R)^2}}$$ |
| | + | (This follows just from similar triangles: the axial component of friction is to $F_{\text{fr}}$ as $v$ is to $u_{\text{rel}}$) |
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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: |
| | + | "Steady-state" means the bushing moves with constant velocity, so the net axial force is zero: the applied force $F$ is exactly balanced by the axial component of friction: |
| | + | $$F = F_{\text{fr}} \cdot \frac{v}{\sqrt{v^2 + (\omega R)^2}}$$ |
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| | − | $$ F_{tr, axial} = F_{tr} \ \cdot \ \frac{v}{u_{rel}} = F_{tr} \ \cdot \ \frac{v}{\sqrt{ v^2 + (\omega R)^2}} $$ |
| | − | (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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| | + | $$F \sqrt{v^2 + \omega^2 R^2} = F_{\text{fr}} v$$ |
| | | Square both sides: |
| | + | $$F^2 (v^2 + \omega^2 R^2) = F_{\text{fr}}^2 v^2$$ |
| | + | $$F^2 \omega^2 R^2 = v^2 (F_{\text{fr}}^2 - F^2)$$ |
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| | − | $$ F^2 (v^2 + \omega^2 R^2 ) = F_{tr}^2 v^2$$ |
| | − | $$F^2 \omega^2 R^2 = v^2 (F_{tr}^2 - F^2)$$ |
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| | | #### Answer |
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| | − | $$ v = \dfrac{F \omega R}{ \sqrt{F_{tr}^2 - F^2}}$$ |
| | + | $$v = \frac{F \omega R}{\sqrt{F_{\text{fr}}^2 - F^2}}$$ |
