Edit to “Solution”
en/2.7.22.md
+1 −1
| ### Statement | |||
| $2.7.22.$ A homogeneous cylinder of radius $R$ and mass $m$ was pushed with an initial velocity $v_0$ without rotation along the horizontal plane. After what time will the slipping stop if the coefficient of friction of the cylinder on the plane is equal to $\mu$? How much of the initial energy will be converted to heat? | |||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | |||
| − |  | ||
| The cylinder 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 cylinder around its axis. This is achieved by friction $f=\mu N=\mu mg$, which decreases the cylinder's linear momentum and at the same time increases its angular momentum. After time interval $t$, linear impulse from friction is $-\mu mgt$, and $v=v_0-\mu gt$. Similarly, angular impulse from friction is $\mu mgRt$, and $\omega=\mu mgRt/I$, where $I$ is the moment of inertia of the cylinder around its axis. Setting $v=\omega R$, we have | |||
| \[v_0-\mu gt=\frac{\mu mgR^2t}{I}\qquad\Rightarrow\qquad t=\frac{v_0}{\mu g(1+mR^2/I)},\] | |||
| which becomes $t=v_0/(3\mu g)$ for $I=mR^2/2$. We also obtain the final speed of the center of mass $v=2v_0/3$. Heat generated from lost kinetic energy is | |||
| \[\frac{1}{2}mv_0^2-\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=\frac{1}{3}\cdot\frac{1}{2}mv_0^2.\] | |||
| #### Answer | |||
| $t=v_0/(3\mu g)$ and $1/3$ of the initial energy is converted to heat. | |||
| unchanged lines 9 | |||
| ### Statement | ### Statement | ||
| $2.7.22.$ A homogeneous cylinder of radius $R$ and mass $m$ was pushed with an initial velocity $v_0$ without rotation along the horizontal plane. After what time will the slipping stop if the coefficient of friction of the cylinder on the plane is equal to $\mu$? How much of the initial energy will be converted to heat? | $2.7.22.$ A homogeneous cylinder of radius $R$ and mass $m$ was pushed with an initial velocity $v_0$ without rotation along the horizontal plane. After what time will the slipping stop if the coefficient of friction of the cylinder on the plane is equal to $\mu$? How much of the initial energy will be converted to heat? | ||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | ### Solution | ||
|  | ||
| The cylinder 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 cylinder around its axis. This is achieved by friction $f=\mu N=\mu mg$, which decreases the cylinder's linear momentum and at the same time increases its angular momentum. After time interval $t$, linear impulse from friction is $-\mu mgt$, and $v=v_0-\mu gt$. Similarly, angular impulse from friction is $\mu mgRt$, and $\omega=\mu mgRt/I$, where $I$ is the moment of inertia of the cylinder around its axis. Setting $v=\omega R$, we have | The cylinder 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 cylinder around its axis. This is achieved by friction $f=\mu N=\mu mg$, which decreases the cylinder's linear momentum and at the same time increases its angular momentum. After time interval $t$, linear impulse from friction is $-\mu mgt$, and $v=v_0-\mu gt$. Similarly, angular impulse from friction is $\mu mgRt$, and $\omega=\mu mgRt/I$, where $I$ is the moment of inertia of the cylinder around its axis. Setting $v=\omega R$, we have | ||
| \[v_0-\mu gt=\frac{\mu mgR^2t}{I}\qquad\Rightarrow\qquad t=\frac{v_0}{\mu g(1+mR^2/I)},\] | \[v_0-\mu gt=\frac{\mu mgR^2t}{I}\qquad\Rightarrow\qquad t=\frac{v_0}{\mu g(1+mR^2/I)},\] | ||
| which becomes $t=v_0/(3\mu g)$ for $I=mR^2/2$. We also obtain the final speed of the center of mass $v=2v_0/3$. Heat generated from lost kinetic energy is | which becomes $t=v_0/(3\mu g)$ for $I=mR^2/2$. We also obtain the final speed of the center of mass $v=2v_0/3$. Heat generated from lost kinetic energy is | ||
| \[\frac{1}{2}mv_0^2-\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=\frac{1}{3}\cdot\frac{1}{2}mv_0^2.\] | \[\frac{1}{2}mv_0^2-\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=\frac{1}{3}\cdot\frac{1}{2}mv_0^2.\] | ||
| #### Answer | #### Answer | ||
| $t=v_0/(3\mu g)$ and $1/3$ of the initial energy is converted to heat. | $t=v_0/(3\mu g)$ and $1/3$ of the initial energy is converted to heat. | ||
| unchanged lines 9 | |||