Solution to Question 29
1. Calculate Force Vector
The force is given by $\vec{F} = -\nabla U$. We first find the partial derivatives of $U = kx(x^2+y^2)^{-3/2}$.
$$\frac{\partial U}{\partial x} = \frac{k(y^2-2x^2)}{(x^2+y^2)^{5/2}}, \quad \frac{\partial U}{\partial y} = \frac{-3kxy}{(x^2+y^2)^{5/2}}$$At position $\vec{r} = r_0(\hat{i} + \hat{j})$, we substitute $x=r_0, y=r_0$:
$$\vec{F} = – \left( \frac{-k r_0^2}{(2r_0^2)^{5/2}}\hat{i} + \frac{-3k r_0^2}{(2r_0^2)^{5/2}}\hat{j} \right) = \frac{k}{4\sqrt{2}r_0^3}(\hat{i} + 3\hat{j})$$2. Identify Directions
The particle moves perpendicular to $\vec{r}$.
- Normal Direction ($\hat{n}$): Along $\vec{r}$, which is $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
- Tangential Direction ($\hat{t}$): Perpendicular to $\vec{r}$, which is $\frac{-\hat{i}+\hat{j}}{\sqrt{2}}$.
3. Calculate Acceleration Components
Acceleration $\vec{a} = \vec{F}/m$. We project this onto the directions.
Tangential Acceleration ($\vec{a}_t$):
$$a_t = \frac{\vec{F} \cdot \hat{t}}{m} = \frac{k}{4\sqrt{2}mr_0^3}(1(-1) + 3(1)) \cdot \frac{1}{\sqrt{2}} = \frac{k}{4mr_0^3}$$Vector form:
$$\vec{a}_t = a_t \hat{t} = \frac{k}{4mr_0^3} \left(\frac{-\hat{i}+\hat{j}}{\sqrt{2}}\right) = \frac{k}{4\sqrt{2}mr_0^3}(-\hat{i} + \hat{j})$$Normal Acceleration ($\vec{a}_n$):
$$a_n = \frac{\vec{F} \cdot \hat{n}}{m} = \frac{k}{4\sqrt{2}mr_0^3}(1(1) + 3(1)) \cdot \frac{1}{\sqrt{2}} = \frac{k}{2mr_0^3}$$Vector form:
$$\vec{a}_n = a_n \hat{n} = \frac{k}{2mr_0^3} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) = \frac{k}{2\sqrt{2}mr_0^3}(\hat{i} + \hat{j})$$