\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]
This question aims to find an expression for the Force f which is expressed in terms of the unit vectors i^ and j^.
The concepts needed for this question include potential energy function, conservative forces, and unit vectors. Potential Energy Function is a function that is defined as the position of the object only for the conservative forces like gravity. Conservative forces are those forces that do not depend on the path but only on the initial and final positions of the object.
Expert Answer
The given potential energy function is given as:
\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]
The conservative force of motion in two dimensions is the negative partial derivative of its potential energy function multiplied by its respective unit vector. The formula for conservative force in terms of its potential energy function is given as:
\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat{i} + \dfrac { dU }{ dy } \hat{j} \Big) \]
Substituting the value of U in the above equation to get the expression for Force f.
\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{i} + \dfrac { d }{ dy } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{j} \Big) \]
\[ \overrightarrow{F} = – \Big( a \dfrac { d }{ dx } \Big( \dfrac{1} {x^2} \Big) \hat{i} + a \dfrac { d }{ dy } \Big( \dfrac{1} {y^2} \Big) \hat{j} \Big) \]
\[ \overrightarrow{F} = 2a \dfrac{ 1 }{ x^3 } \hat{i} + 2a \dfrac{ 1 }{ y^3 } \hat{j} \]
\[ \overrightarrow{F} = 2a \Big( \dfrac{ 1 }{ x^3 } \hat{i} + \dfrac{ 1 }{ y^3 } \hat{j} \Big) \]
Numerical Result
The expression for the force $\overrightarrow {f}$ is expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}$ is calculated to be:
\[ \overrightarrow{F} = \Big( \dfrac{ 2a }{ x^3 } \hat{i} + \dfrac{ 2a }{ y^3 } \hat{j} \Big) \]
Example
Potential energy function is given for an object moving in XY-plane. Derive an expression for the force f expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}.
\[ U(x, y) = \big( 3x^2 + y^2 \big) \]
We can derive an expression for force by taking the negative of the partial derivative of the potential energy function and multiplying it by respective unit vectors. The formula is given as:
\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat {i} + \dfrac { dU }{ dy } \hat {j} \Big) \]
\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } \big( 3x^2 + y^2 \big) \hat {i} + \dfrac { d }{ dy } \big( 3x^2 + y^2 \big) \hat {j} \Big) \]
\[ \overrightarrow{F} = – \big( 6x \hat {i} + 2y \hat {j} \big) \]
\[ \overrightarrow{F} = – 6x \hat {i}\ -\ 2y \hat {j} \]
The expression of force f is calculated to be $- 6x \hat {i}\ -\ 2y \hat {j}$