Edit to “Solution”
en/2.3.32.md
+1 −1
| ### Statement | |||
| $2.3.32.$ What is the work of the friction force for one revolution of an air sled moving along a vertical circular track? The speed of the sledge is constant and equal to $v$, the mass of the sledge is $m$, and the coefficient of friction is $\mu$. | |||
| ### Solution | |||
|  | |||
| Let $\alpha$ be the angle from the downward radius of the circle to the radius pointing to the sled measured counterclockwise. The centripetal force on the sled is | |||
| \[N-mg\cos\alpha=\frac{mv^2}{r}.\] | |||
| Consequently, the friction is | |||
| \[f=\mu N=\mu m\left(g\cos\alpha+\frac{v^2}{r}\right).\] | |||
| Thus, the work done by friction is | |||
| @@ -18,8 +18,8 @@Solution | |||
| \[\int_0^{2\pi}-fr\,d\alpha=\int_0^{2\pi}-\mu m\left(g\cos\alpha+\frac{v^2}{r}\right)r\,d\alpha.\] | |||
| − | Since the average of $\cos\alpha$ for $0\le\alpha\le2\pi$ is zero, the integral simplies to $-2\pi\mu mv^2$. | ||
| + | Since the average of $\cos\alpha$ for $0\le\alpha\le2\pi$ is zero, the integral simplifies to $-2\pi\mu mv^2$. | ||
| #### Answer | |||
| $-2\pi\mu mv^2$ | |||
| ### Statement | ### Statement | ||
| $2.3.32.$ What is the work of the friction force for one revolution of an air sled moving along a vertical circular track? The speed of the sledge is constant and equal to $v$, the mass of the sledge is $m$, and the coefficient of friction is $\mu$. | $2.3.32.$ What is the work of the friction force for one revolution of an air sled moving along a vertical circular track? The speed of the sledge is constant and equal to $v$, the mass of the sledge is $m$, and the coefficient of friction is $\mu$. | ||
| ### Solution | ### Solution | ||
|  |  | ||
| Let $\alpha$ be the angle from the downward radius of the circle to the radius pointing to the sled measured counterclockwise. The centripetal force on the sled is | Let $\alpha$ be the angle from the downward radius of the circle to the radius pointing to the sled measured counterclockwise. The centripetal force on the sled is | ||
| \[N-mg\cos\alpha=\frac{mv^2}{r}.\] | \[N-mg\cos\alpha=\frac{mv^2}{r}.\] | ||
| Consequently, the friction is | Consequently, the friction is | ||
| \[f=\mu N=\mu m\left(g\cos\alpha+\frac{v^2}{r}\right).\] | \[f=\mu N=\mu m\left(g\cos\alpha+\frac{v^2}{r}\right).\] | ||
| Thus, the work done by friction is | Thus, the work done by friction is | ||
| @@ -18,8 +18,8 @@Solution | |||
| \[\int_0^{2\pi}-fr\,d\alpha=\int_0^{2\pi}-\mu m\left(g\cos\alpha+\frac{v^2}{r}\right)r\,d\alpha.\] | \[\int_0^{2\pi}-fr\,d\alpha=\int_0^{2\pi}-\mu m\left(g\cos\alpha+\frac{v^2}{r}\right)r\,d\alpha.\] | ||
| Since the average of $\cos\alpha$ for $0\le\alpha\le2\pi$ is zero, the integral simplies to $-2\pi\mu mv^2$. | Since the average of $\cos\alpha$ for $0\le\alpha\le2\pi$ is zero, the integral simplifies to $-2\pi\mu mv^2$. | ||
| #### Answer | #### Answer | ||
| $-2\pi\mu mv^2$ | $-2\pi\mu mv^2$ | ||