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Chapter 12: Problem 31

The potential energy corresponding to a certain two-dimensional force is given by \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right) .(a)\) Derive \(F_{x}\) and \(F_{y}\) and describe the vector force at each point in terms of its coordinates \(x\) and \(y .(b)\) Derive \(F_{r}\) and \(F_{\theta}\) and describe the vector force at each point in terms of the polar coordinates \(r\) and \(\theta\) of the point. ( \(c\) ) Can you think of a physical model of such a force?

Short Answer

Expert verified
The force components in Cartesian coordinates are \( F_{x} = - kx \) and \( F_{y} = - ky \). In polar coordinates, the components are \( F_{r} = -kr \) and \( F_{\theta} = 0 \). A possible physical model of this force is the force exerted by a spring attached to the origin and free to move in two dimensions.

Step by step solution

01

Derive \(F_{x}\) and \(F_{y}\)

Using the relation \( F = - \nabla U \), where \( \nabla \) is the gradient operator, it can be seen that the vector force at each point in terms of its coordinates \( x \) and \( y \) can be computed by taking the gradient of the potential energy with respect to \( x \) and \( y \) getting \(F_{x} = -\frac{\partial U}{\partial x} = - kx\) and \(F_{y} = -\frac{\partial U}{\partial y} = - ky \). The force at any point is thus a vector \( \vec{F} = -k (x \hat{i} + y \hat{j}) \)
02

Derive \(F_{r}\) and \(F_{\theta}\)

Potential Energy is a function of \( r = \sqrt{x^{2} + y^{2}} \). So, \( F_{r} = -\frac{\partial U}{\partial r} = - kr \). Using \( r = x \cos \theta = y \sin \theta \), \( F_{\theta} = - \frac{\partial U}{\partial \theta} =0 \), as \( U \) doesn't depend upon \( \theta \). So, force in polar coordinates is \( \vec{F} = -kr \hat{r} \).
03

Physical interpretation

This equation for force can be taken as the force exerted by a spring with force constant \( k \) and equilibrium position at the origin. The force is always directed towards the origin, no matter in which direction \( x \) and \( y \) are oriented.

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