Source: Principles of Physics
Problem Sets:
Problem
A particle undergoes counter-clockwise uniform circular motion. At time $t_1$, its acceleration is $\vec{a}_1 = a_{1x}\hat{i} + a_{1y}\hat{j}$. At a later time $t_2$, its acceleration is $\vec{a}_2 = a_{2x}\hat{i} + a_{2y}\hat{j}$. The time interval $\Delta t = t_2 - t_1$ is less than one period.
Let $\Delta t = t_2 - t_1$. The magnitude of the acceleration is:
$$a = \sqrt{a_{1x}^2 + a_{1y}^2}$$The angular displacement $\Delta\theta$ is found by first calculating an intermediate angle $\Delta\theta_0$:
$$\Delta\theta_0 = \text{atan2}(a_{1x}a_{2y} - a_{1y}a_{2x}, a_{1x}a_{2x} + a_{1y}a_{2y})$$Then, applying the condition for counter-clockwise motion:
$$\Delta\theta = \begin{cases} \Delta\theta_0 & \text{if } \Delta\theta_0 \ge 0 \\ \Delta\theta_0 + 2\pi & \text{if } \Delta\theta_0 < 0 \end{cases}$$The radius of the path is:
$$r = \frac{a (\Delta t)^2}{(\Delta\theta)^2}$$In uniform circular motion, the acceleration is centripetal and has a constant magnitude $a$. The acceleration vector $\vec{a}$ rotates at the same constant angular speed $\omega$ as the particle.
The magnitude of acceleration is found from the initial vector:
$$a = |\vec{a}_1| = \sqrt{a_{1x}^2 + a_{1y}^2}$$The angular displacement $\Delta\theta$ of the acceleration vector over the time interval $\Delta t$ can be determined from the dot and cross products of $\vec{a}_1$ and $\vec{a}_2$. Let $C = \vec{a}_1 \cdot \vec{a}_2 = a_{1x}a_{2x} + a_{1y}a_{2y}$ Let $S = (\vec{a}_1 \times \vec{a}_2) \cdot \hat{k} = a_{1x}a_{2y} - a_{1y}a_{2x}$ From the definitions of the dot and cross products, $C = a^2 \cos(\Delta\theta)$ and $S = a^2 \sin(\Delta\theta)$.
The unique angle of rotation $\Delta\theta_0$ in the range $(-\pi, \pi]$ is given by $\Delta\theta_0 = \text{atan2}(S, C)$. Since the motion is counter-clockwise, the angular displacement $\Delta\theta$ must be positive. As the time interval is less than a period, $0 < \Delta\theta < 2\pi$. If the calculated angle $\Delta\theta_0$ is negative, we add $2\pi$ to find the correct counter-clockwise displacement.
The constant angular speed is $\omega = \frac{\Delta\theta}{\Delta t}$.
The centripetal acceleration is $a = \omega^2 r$. Solving for the radius $r$:
$$r = \frac{a}{\omega^2} = \frac{a (\Delta t)^2}{(\Delta\theta)^2}$$