1. Conceptual Visualization & Coordinate System
We define a coordinate system to analyze the forces and geometry.
Conceptual Visualization
Vector Analysis Diagram
2. Position Vectors & String Geometry
We analyze the system at an instant where the aircraft is on the negative x-axis.
Position of the Aircraft ($A$): $$ \vec{r}_A = -R \hat{i} + h \hat{j} + 0 \hat{k} $$
Position of the Hand ($B$): $$ \vec{r}_B = -r \sin\theta \hat{i} + 0 \hat{j} + r \cos\theta \hat{k} $$
String Vector ($\vec{L}$): The vector from the aircraft to the hand is $\vec{L} = \vec{r}_B – \vec{r}_A$. $$ \vec{L} = (R – r \sin\theta) \hat{i} – h \hat{j} + r \cos\theta \hat{k} $$
The square of the string’s length is $L^2 = \vec{L} \cdot \vec{L}$: $$ L^2 = (R – r \sin\theta)^2 + (-h)^2 + (r \cos\theta)^2 $$ $$ L^2 = R^2 + r^2 \sin^2\theta – 2Rr \sin\theta + h^2 + r^2 \cos^2\theta $$ Using the identity $\sin^2\theta + \cos^2\theta = 1$:
3. Force Analysis & Equations of Motion
The forces on the aircraft are Lift ($\vec{F}$), Gravity ($m\vec{g}$), and Tension ($\vec{T}$). The tension acts along the unit vector $\hat{u}_L = \vec{L}/L$. $$ \vec{T} = T \hat{u}_L = \frac{T}{L} \left[ (R – r \sin\theta) \hat{i} – h \hat{j} + r \cos\theta \hat{k} \right] $$
Vertical Equilibrium (y-direction): The aircraft’s altitude is constant, so the net vertical force is zero. $$ \Sigma F_y = F – mg + T_y = 0 $$ $$ F – mg – T \frac{h}{L} = 0 \implies F = mg + T \frac{h}{L} \quad \text{— (Eq. 2)} $$
Horizontal Dynamics (x-direction): The aircraft at $(-R, h, 0)$ has a centripetal acceleration $\vec{a}_c = (v^2/R) \hat{i}$ towards the center $(0, h, 0)$. $$ \Sigma F_x = T_x = m a_c $$ $$ T \frac{(R – r \sin\theta)}{L} = \frac{m v^2}{R} \implies T = \frac{m v^2 L}{R (R – r \sin\theta)} \quad \text{— (Eq. 3)} $$
4. Solving for the Lift Force
We use Eq. 1 to express $(R – r \sin\theta)$ in terms of the given constants. From Eq. 1: $2Rr \sin\theta = R^2 + r^2 + h^2 – L^2 \implies r \sin\theta = \frac{R^2 + r^2 + h^2 – L^2}{2R}$.
Now, substitute this into the term $(R – r \sin\theta)$: $$ R – r \sin\theta = R – \frac{R^2 + r^2 + h^2 – L^2}{2R} = \frac{2R^2 – (R^2 + r^2 + h^2 – L^2)}{2R} $$ $$ R – r \sin\theta = \frac{R^2 + L^2 – r^2 – h^2}{2R} $$
Substitute this back into the Tension expression (Eq. 3): $$ T = \frac{m v^2 L}{R} \cdot \frac{2R}{R^2 + L^2 – r^2 – h^2} = \frac{2 m v^2 L}{R^2 + L^2 – r^2 – h^2} $$
Finally, substitute $T$ into the Lift equation (Eq. 2): $$ F = mg + \left( \frac{2 m v^2 L}{R^2 + L^2 – r^2 – h^2} \right) \frac{h}{L} $$