4.23

4.23 Which of the followings are conservative force.

a) $\mathbf{F} = k(x,2y,3z)$
$\begin{array}{rcl} (\nabla \times \mathbf{F})_x &=& \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \\ &=& \dfrac{\partial}{\partial y}(3kz) - \dfrac{\partial}{\partial z}(2ky) \\ &=& 0\end{array}$

In this case, the other 2 components will give the same results.
So, we can conclude that $\nabla \times \mathbf{F} = 0$ The given force is conservative force.

$\begin{array}{rcl} U(\mathbf{r}) &=& - \int_0^r{\mathbf{F}(r)}d\mathbf{r} \\ &=& - \int_0^r{[F_xdx + F_ydy + F_zdz]} \\ &=& -k{\int_{0}^{x}xdx +\int_{0}^{y}2ydy + \int_{0}^{z}3zdz} \\ &=& -k(\dfrac{1}{2}x^2 + y^2 + \dfrac{3}{2}z^2) \end{array}$

$U(\mathbf{r}) = -\dfrac{1}{2}k(x^2, 2y^2 , 3z^2)$

ิb) $\mathbf{F} = k(y,x,0)$
$\begin{array}{rcl} (\nabla \times \mathbf{F})_x &=& \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \\ &=& \dfrac{\partial}{\partial y}(0) - \dfrac{\partial}{\partial z}(x) \\ &=& 0\end{array}$

$\begin{array}{rcl} (\nabla \times \mathbf{F})_y &=& \dfrac{\partial F_z}{\partial x} - \dfrac{\partial F_x}{\partial z} \\ &=& \dfrac{\partial}{\partial x}(0) - \dfrac{\partial}{\partial z}(y) \\ &=& 0\end{array}$

$\begin{array}{rcl} (\nabla \times \mathbf{F})_z &=& \dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \\ &=& \dfrac{\partial}{\partial x}(x) - \dfrac{\partial}{\partial y}(y) \\ &=& 1-1 \\ &=& 0 \end{array}$

So, we can obviously see that $\nabla \times \mathbf{F} = 0$ The given force is conservative force.

$\begin{array}{rcl} U(\mathbf{r}) &=& - \int_0^r{\mathbf{F}(r)}d\mathbf{r} \\ &=& - \int_0^r{[F_xdx + F_ydy + F_zdz]} \\ &=& -k{\int_{0}^{x}xdx +\int_{0}^{y}ydy} \\ &=& -k(\dfrac{1}{2}x^2 + \dfrac{1}{2}y^2 ) \end{array}$

$U(\mathbf{r}) = -\dfrac{1}{2}k(x^2, y^2 , 0)$

c) $\mathbf{F} = k(-y,x,0)$
$\begin{array}{rcl} (\nabla \times \mathbf{F})_x &=& \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \\ &=& \dfrac{\partial}{\partial y}(0) - \dfrac{\partial}{\partial z}(x) \\ &=& 0\end{array}$

$\begin{array}{rcl} (\nabla \times \mathbf{F})_y &=& \dfrac{\partial F_z}{\partial x} - \dfrac{\partial F_x}{\partial z} \\ &=& \dfrac{\partial}{\partial x}(0) - \dfrac{\partial}{\partial z}(-y) \\ &=& 0\end{array}$

$\begin{array}{rcl} (\nabla \times \mathbf{F})_z &=& \dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \\ &=& \dfrac{\partial}{\partial x}(x) - \dfrac{\partial}{\partial y}(-y) \\ &=& 1-(-1) \\ &=& 2 \end{array}$

So, As wee see that $\nabla \times \mathbf{F} \neq 0$ The given force is non-conservative force.
We can not write the formula for U(r).

P.Buphamalai 5205086

คุณสมบัติของเด็กดี