Edit to “Solution”
en/2.3.33.md
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| ### Statement | |||
| $2.3.33.$ The body slides along a flat surface that smoothly passes into another flat surface located at an angle $\alpha$ to the first one. Coefficient of friction $\mu$. Determine the kinetic energy at the end of the interface section if it is equal to $K_0$ at the beginning. | |||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | |||
| − |  | ||
| Assume that the speed $v$ of the body is much larger than $\sqrt{gr}$, where $r$ the radius of curvature of the bend, so that the gravitational field $g$ has negligible effect on centripetal acceleration. Consequently, the normal force | |||
| \[N=\frac{mv^2}{r}=\frac{2K}{r},\] | |||
| where $K$ is the kinetic energy of the body. The change in kinetic energy is due to the work done by friction, i.e. | |||
| \[dK=-\mu Nr\,d\theta=-2\mu K\,d\theta,\] | |||
| where $d\theta$ is a small angle subtended by a small arc of the bend. Integrating, we have $K(\theta)=K_0\cdot e^{-2\mu\theta}$. In the figure, the angle subtended by the bend is $\alpha$, so $K=K_0\cdot e^{-2\mu\alpha}$ at the end of the bend. | |||
| #### Answer | |||
| $K_0\cdot e^{-2\mu\alpha}$ | |||
| unchanged lines 11 | |||
| ### Statement | ### Statement | ||
| $2.3.33.$ The body slides along a flat surface that smoothly passes into another flat surface located at an angle $\alpha$ to the first one. Coefficient of friction $\mu$. Determine the kinetic energy at the end of the interface section if it is equal to $K_0$ at the beginning. | $2.3.33.$ The body slides along a flat surface that smoothly passes into another flat surface located at an angle $\alpha$ to the first one. Coefficient of friction $\mu$. Determine the kinetic energy at the end of the interface section if it is equal to $K_0$ at the beginning. | ||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | ### Solution | ||
|  | ||
| Assume that the speed $v$ of the body is much larger than $\sqrt{gr}$, where $r$ the radius of curvature of the bend, so that the gravitational field $g$ has negligible effect on centripetal acceleration. Consequently, the normal force | Assume that the speed $v$ of the body is much larger than $\sqrt{gr}$, where $r$ the radius of curvature of the bend, so that the gravitational field $g$ has negligible effect on centripetal acceleration. Consequently, the normal force | ||
| \[N=\frac{mv^2}{r}=\frac{2K}{r},\] | \[N=\frac{mv^2}{r}=\frac{2K}{r},\] | ||
| where $K$ is the kinetic energy of the body. The change in kinetic energy is due to the work done by friction, i.e. | where $K$ is the kinetic energy of the body. The change in kinetic energy is due to the work done by friction, i.e. | ||
| \[dK=-\mu Nr\,d\theta=-2\mu K\,d\theta,\] | \[dK=-\mu Nr\,d\theta=-2\mu K\,d\theta,\] | ||
| where $d\theta$ is a small angle subtended by a small arc of the bend. Integrating, we have $K(\theta)=K_0\cdot e^{-2\mu\theta}$. In the figure, the angle subtended by the bend is $\alpha$, so $K=K_0\cdot e^{-2\mu\alpha}$ at the end of the bend. | where $d\theta$ is a small angle subtended by a small arc of the bend. Integrating, we have $K(\theta)=K_0\cdot e^{-2\mu\theta}$. In the figure, the angle subtended by the bend is $\alpha$, so $K=K_0\cdot e^{-2\mu\alpha}$ at the end of the bend. | ||
| #### Answer | #### Answer | ||
| $K_0\cdot e^{-2\mu\alpha}$ | $K_0\cdot e^{-2\mu\alpha}$ | ||
| unchanged lines 11 | |||