New solution
en/2.7.21.md
+23 −0
| @@ -0,0 +1,23 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $2.7.21.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + |  | ||
| + | |||
| + | The ring will roll without slipping when the speed of its center of mass $v$ is equal to $\omega R$, where $\omega$ is the angular speed of rotation of the ring around its axis. This is achieved by friction $f=\mu N=\mu mg$, which increases the ring's linear momentum and at the same time decreases its angular momentum. After time interval $t$, linear impulse from friction is $\mu mgt$, and $v=\mu gt$. Similarly, angular impulse from friction is $-\mu mgRt$, and | ||
| + | |||
| + | \[\omega=\omega_0-\frac{\mu mgRt}/{mR^2}=\omega_0-\frac{\mu gt}{R}.\] | ||
| + | |||
| + | Setting $v=\omega R$ at time $t_*$, we have | ||
| + | |||
| + | \[\mu gt_*=\left(\omega_0-\frac{\mu gt_*}{R}\right)R\qquad\Rightarrow\qquad t_*=\frac{\omega_0R}{2\mu g}.\] | ||
| + | |||
| + | We also obtain the final speed of the center of mass $v_*=\omega_0R/2$ and the final angular speed of rotation of the ring around its axis $\omega_*=\omega_0/2$. Heat generated from lost kinetic energy is | ||
| + | |||
| + | \[\frac{1}{2}mR^2\omega_0^2-\frac{1}{2}mR^2\omega_*^2-\frac{1}{2}m\v_*^2=\frac{1}{2}\cdot\frac{1}{2}mR^2\omega_0^2.\] | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,23 @@ | |||
| ### Statement | |||
| $2.7.21.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| The ring will roll without slipping when the speed of its center of mass $v$ is equal to $\omega R$, where $\omega$ is the angular speed of rotation of the ring around its axis. This is achieved by friction $f=\mu N=\mu mg$, which increases the ring's linear momentum and at the same time decreases its angular momentum. After time interval $t$, linear impulse from friction is $\mu mgt$, and $v=\mu gt$. Similarly, angular impulse from friction is $-\mu mgRt$, and | |||
| \[\omega=\omega_0-\frac{\mu mgRt}/{mR^2}=\omega_0-\frac{\mu gt}{R}.\] | |||
| Setting $v=\omega R$ at time $t_*$, we have | |||
| \[\mu gt_*=\left(\omega_0-\frac{\mu gt_*}{R}\right)R\qquad\Rightarrow\qquad t_*=\frac{\omega_0R}{2\mu g}.\] | |||
| We also obtain the final speed of the center of mass $v_*=\omega_0R/2$ and the final angular speed of rotation of the ring around its axis $\omega_*=\omega_0/2$. Heat generated from lost kinetic energy is | |||
| \[\frac{1}{2}mR^2\omega_0^2-\frac{1}{2}mR^2\omega_*^2-\frac{1}{2}m\v_*^2=\frac{1}{2}\cdot\frac{1}{2}mR^2\omega_0^2.\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||