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Conservative Forces

PY345

Conservative Forces

  • a force acting on a particle from one location to another is conservative if:

    1. the force depends only on the particle's position (not on velocity, time, etc.) and on properties of the particle

    2. the work done by the force is the same regardless of the path taken between two locations

  • the second condition is satisfied if ∇⃗×F⃗=0\vec{\nabla}\times \vec{F}=0

  • F⃗=−∇⃗U=−∂U∂xx^−∂U∂yy^−∂U∂zz^\vec{F}=-\vec{\nabla}U=-\frac{\partial U}{\partial x}\hat{x}-\frac{\partial U}{\partial y}\hat{y}-\frac{\partial U}{\partial z}\hat{z}

Practice

Determine the value of ∫F⃗⋅dr⃗\int\vec{F}\cdot d\vec{r} from point O to point P for each of the following paths for the vector field F⃗=yx^+2xy^\vec{F}=y\hat{x}+2x\hat{y}.

four line integrals

Solution

a) Path (a) has to be done in two pieces. For Segment OQ, dr⃗=dxx^d\vec{r}=dx\hat{x} from x=0x=0 to x=1x=1 with y=0y=0 the entire way. For Segment QP, dr⃗=dyy^d\vec{r}=dy\hat{y} from y=0y=0 to y=1y=1 with x=1x=1 the entire way.

Wa=∫aF⃗⋅dr⃗=∫01(0x^+2xy^)⋅dxx^+∫01(yx^+2(1)y^)⋅dyy^W_a=\int\limits_a\vec{F}\cdot d\vec{r}=\int\limits_0^1\left (0\hat{x}+2x\hat{y}\right )\cdot dx\hat{x}+\int\limits_0^1\left (y\hat{x}+2(1)\hat{y}\right )\cdot dy\hat{y}

Wa=0+2∫01dy=2W_a=0+2\int\limits_0^1dy=2

b) Wb=∫bF⃗⋅dr⃗=∫b(Fxdx+Fydy)=∫b(ydx+2xdy)W_b=\int\limits_b \vec{F}\cdot d\vec{r}=\int\limits_b (F_x dx+F_y dy)=\int\limits_b (ydx+2xdy)

Oftentimes when you have a dxdx and a dydy that means you are doing a double integral, but that isn't the case here (that would produce a quantity with units of [N][m][m]=[N⋅m2][N][m][m]=[N\cdot m^2], which is not a Joule). Instead, we take advantage of the fact that we know how xx and yy are related. Specifically, y=xy=x which means that dy=dxdy=dx. This enables us to change to an integral involving only one variable.

Wb=∫01(xdx+2xdx)=∫013xdx=[3x22]01=1.5W_b=\int\limits_0^1(xdx+2xdx)=\int\limits_0^1 3xdx=\left [\frac{3x^2}{2}\right ]_0^1=1.5

c) We proceed the same way as part (b), except now y=x2y=x^2 which means that dy=2xdxdy=2xdx.

Wc=∫c(ydx+2xdy)=∫01(x2dx+(2x)(2xdx))=∫015x2dx=[5x33]01W_c=\int\limits_c(ydx+2xdy)=\int\limits_0^1(x^2dx+(2x)(2xdx))=\int\limits_0^1 5x^2dx=\left [\frac{5x^3}{3}\right ]_0^1

d) This is the same as (b) and (c) except that y=sin⁡(π2x)y=\sin\left (\frac{\pi}{2}x\right ), which means that dy=π2cos⁡(π2x)dy=\frac{\pi}{2}\cos\left (\frac{\pi}{2}x\right ).

Wd=∫d(ydx+2xdy)=∫01(sin⁡(π2x)dx+2xπ2cos⁡(π2x))=∫01sin⁡(π2x)dx+π∫01xcos⁡(π2x)dxW_d=\int\limits_d (ydx+2xdy)=\int\limits_0^1\left (\sin\left (\frac{\pi}{2}x\right )dx+\cancel{2}x\frac{\pi}{\cancel{2}}\cos\left (\frac{\pi}{2}x\right )\right )=\int\limits_0^1\sin\left (\frac{\pi}{2}x\right )dx+\pi \int\limits_0^1x\cos\left (\frac{\pi}{2}x\right )dx

The first integral is easy to evaluate, but the second one is less so. Looking up the integral in a table, we see that

∫xcos⁡(ax)dx=cos⁡(ax)a2+xsin⁡(ax)a\int x\cos(ax)dx=\frac{\cos(ax)}{a^2}+\frac{x\sin(ax)}{a}.

Therefore,

Wd=[−2πcos⁡(π2x)]01+π[cos⁡(π2x)(π2)2+xsin⁡(π2x)π/2]01=2π+π[2π−(2π)2]=2π+2−4π=2−2πW_d=\left [-\frac{2}{\pi}\cos\left (\frac{\pi}{2}x\right )\right ]_0^1+\pi\left [\frac{\cos\left (\frac{\pi}{2}x\right )}{\left (\frac{\pi}{2}\right )^2}+\frac{x\sin\left (\frac{\pi}{2}x\right )}{\pi/2}\right ]_0^1=\frac{2}{\pi}+\pi\left [\frac{2}{\pi}-\left (\frac{2}{\pi}\right )^2\right ]=\frac{2}{\pi}+2-\frac{4}{\pi}=2-\frac{2}{\pi}.

Practice

Show that ∇⃗×F⃗≠0\vec{\nabla}\times\vec{F}\ne 0 for the vector field F⃗=yx^+2xy^\vec{F}=y\hat{x}+2x\hat{y}. What small change could you make to the vector field to make it conservative?

Solution

∇⃗×F⃗=∇⃗×(yx^+2xy^)\vec{\nabla}\times\vec{F}=\vec{\nabla}\times\left (y\hat{x}+2x\hat{y}\right )

=∣x^y^z^∂∂x∂∂y∂∂zy2x0∣=(∂∂y(0)−∂∂z(2x))x^+(∂∂zy−∂∂x(0))y^+(∂∂x(2x)−∂∂yy)z^=(2−1)z^=z^=\left |\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ y&2x&0\end{matrix}\right |=\left (\frac{\partial}{\partial y}(0)-\frac{\partial}{\partial z}(2x)\right )\hat{x}+\left (\frac{\partial}{\partial z}y-\frac{\partial}{\partial x}(0)\right )\hat{y}+\left (\frac{\partial}{\partial x}(2x)-\frac{\partial}{\partial y}y\right )\hat{z}=(2-1)\hat{z}=\hat{z}

If F⃗=2yx^+2xy^\vec{F}=2y\hat{x}+2x\hat{y}, then ∇⃗×F⃗\vec{\nabla}\times\vec{F} would have evaluated to (2−2)z^=0(2-2)\hat{z}=0, which would have made it a conservative vector field.

Practice

Prove that the gravitational force near earth is conservative. Then show that ΔU=−∫12F⃗g⋅dr⃗\Delta U=-\int\limits_1^2\vec{F}^g\cdot d\vec{r} yields the expected result.

Solution

Let's take a coordinate system in which the +y direction points upward. The gravitational force is then F⃗g=−mgy^\vec{F}^g=-mg\hat{y}.

The first condition for a force to be conservative is that the force can only depend on the particle's position and properties of the particle. Mass is a property of the particle, and the gravitational field strength is actually dependent on position in general (it just happens to be constant in this case).

The second condition is satisfied if ∇⃗×F⃗=0\vec{\nabla}\times \vec{F}=0.

∇⃗×F⃗=∣x^y^z^∂∂x∂∂y∂∂z0−mg0∣=(∂∂y(0)−∂∂z(0))x^+(∂∂z(0)−∂∂x(0))y^+(∂∂x(−mg)−∂∂y(0))z^=0\vec{\nabla}\times \vec{F}=\left | \begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z} \\ 0 & -mg & 0\end{matrix}\right |=\left (\frac{\partial}{\partial y}(0)-\frac{\partial}{\partial z}(0)\right )\hat{x}+\left (\frac{\partial}{\partial z}(0)-\frac{\partial}{\partial x}(0)\right )\hat{y}+\left (\frac{\partial }{\partial x}(-mg)-\frac{\partial}{\partial y}(0)\right )\hat{z}=0

Therefore the gravitational force is conservative.

To determine ΔU\Delta U, let's suppose that Location 1 is (x,y)(x,y) while Location 2 is (x+Δx,y+Δy)(x+\Delta x, y+\Delta y).

ΔU=−∫12(Fxgdx+Fygdy)=−∫12(0−mgdy)=−∫yy+Δy(−mg)dy=+mg[y]yy+Δy=mg(y+Δy−y)=mgΔy\Delta U=-\int\limits_1^2 (F_x^g dx+F_y^gdy)=-\int\limits_1^2(0-mgdy)=-\int\limits_y^{y+\Delta y}(-mg)dy=+mg\left [y\right ]_y^{y+\Delta y}=mg(y+\Delta y-y)=mg\Delta y.

This is the result we expect.

Practice

Prove that the spring force F⃗=−k(r⃗−r⃗0)\vec{F}=-k(\vec{r}-\vec{r}_0) is conservative by showing that ∇⃗×F⃗=0\vec{\nabla}\times\vec{F}=0, then determine an expression for ΔU\Delta U as the spring is stretched from x0x^+y0y^+z0z^x_0\hat{x}+y_0\hat{y}+z_0\hat{z} to (x0+Δx)x^+(y0+Δy)y^+(z0+Δz)z^(x_0+\Delta x)\hat{x}+(y_0+\Delta y)\hat{y}+(z_0+\Delta z)\hat{z}.

Solution

F⃗=−k(r⃗−r⃗0)=−k((x−x0)x^+(y−y0)y^+(z−z0)z^)\vec{F}=-k(\vec{r}-\vec{r}_0)=-k\left ((x-x_0)\hat{x}+(y-y_0)\hat{y}+(z-z_0)\hat{z}\right )

∇⃗×F⃗=∣x^y^z^∂∂x∂∂y∂∂z−k(x−x0)−k(y−y0)−k(z−z0)∣\vec{\nabla}\times\vec{F}=\left |\begin{matrix}\hat{x}&\hat{y}&\hat{z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ -k(x-x_0)& -k(y-y_0)& -k(z-z_0)\end{matrix}\right |

When performing the cross product, each differential operator operates on an element that is diagonal to it. By inspection, we can see that all of the derivatives are going to be zero, so that means that ∇⃗×F⃗\vec{\nabla}\times\vec{F} is zero.

Now we calculate ΔU\Delta U.

ΔU=−∫12(−k)((x−x0)dx+(y−y0)dy+(z−z0)dz)=k(∫x0x0+Δx(x−x0)dx+∫y0y0+Δy(y−y0)dy+∫z0z0+Δz(z−z0)dz)\Delta U=-\int\limits_1^2(-k)\left ((x-x_0)dx+(y-y_0)dy+(z-z_0)dz\right )=k\left (\int\limits_{x_0}^{x_0+\Delta x}(x-x_0)dx+\int\limits_{y_0}^{y_0+\Delta y}(y-y_0)dy+\int\limits_{z_0}^{z_0+\Delta z}(z-z_0)dz\right )

All three integrals are of the same form, so let's just focus on one.

∫x0x0+Δx(x−x0)dx=[x22−x0x]x0x0+Δx=(x0+Δx)22−x0(x0+Δx)−x022+x02=x022+x0Δx+Δx22−x02−x0Δx−x022+x02=12Δx2\int\limits_{x_0}^{x_0+\Delta x}(x-x_0)dx=\left [\frac{x^2}{2}-x_0x\right ]_{x_0}^{x_0+\Delta x}=\frac{(x_0+\Delta x)^2}{2}-x_0(x_0+\Delta x)-\frac{x_0^2}{2}+x_0^2=\cancel{\frac{x_0^2}{2}}+\cancel{x_0\Delta x}+\frac{\Delta x^2}{2}-\cancel{x_0^2}-\cancel{x_0\Delta x}-\cancel{\frac{x_0^2}{2}}+\cancel{x_0^2}=\frac{1}{2}\Delta x^2

Therefore we can say that

ΔU=12k(Δx2+Δy2+Δz2)\Delta U=\frac{1}{2}k\left(\Delta x^2+\Delta y^2+\Delta z^2\right ).

Practice

The gravitational potential energy of a system of two arbitrary masses (that is, not restricted to an object near the surface of earth), is U=−Gm1m2r.U=-G\frac{m_1m_2}{r}. Use F⃗−∇⃗U\vec{F}-\vec{\nabla}U to find the corresponding gravitational force.

Solution

We first recognize that r=x2+y2+z2r=\sqrt{x^2+y^2+z^2}.

F⃗−∇⃗U=∂∂xGm1m2x2+y2+z2x^+∂∂yGm1m2x2+y2+z2y^+∂∂zGm1m2x2+y2+z2z^\vec{F}-\vec{\nabla} U=\frac{\partial}{\partial x}\frac{Gm_1m_2}{\sqrt{x^2+y^2+z^2}}\hat{x}+\frac{\partial}{\partial y}\frac{Gm_1m_2}{\sqrt{x^2+y^2+z^2}}\hat{y}+\frac{\partial}{\partial z}\frac{Gm_1m_2}{\sqrt{x^2+y^2+z^2}}\hat{z}

Since the three derivates are of a similar form, let's focus on just one.

∂∂xGm1m2x2+y2+z2=Gm1m2∂∂x(x2+y2+z2)−1/2=−12Gm1m2(x2+y2+z2)−3/2(2x)\frac{\partial}{\partial x}\frac{Gm_1m_2}{\sqrt{x^2+y^2+z^2}}=Gm_1m_2\frac{\partial}{\partial x}\left (x^2+y^2+z^2\right )^{-1/2}=-\frac{1}{\cancel{2}}Gm_1m_2\left (x^2+y^2+z^2\right )^{-3/2}(\cancel{2}x)

Now that we have take the derivative, we can revert this to being back in terms of rr again. Putting it all together, we get

F⃗=−Gm1m2(xr3x^+yr3y^+zr3z^)=−Gm1m2r3(xx^+yy^+zz^)=−Gm1m2r3r⃗=−Gm1m2r2r⃗r=−Gm1m2r2r^\vec{F}=-Gm_1m_2\left (\frac{x}{r^3}\hat{x}+\frac{y}{r^3}\hat{y}+\frac{z}{r^3}\hat{z}\right )=-G\frac{m_1m_2}{r^3}(x\hat{x}+y\hat{y}+z\hat{z})=-G\frac{m_1m_2}{r^3}\vec{r}=-G\frac{m_1m_2}{r^2}\frac{\vec{r}}{r}=-G\frac{m_1m_2}{r^2}\hat{r}.