Answer
$$\frac{L_1}{L_2}=\frac{\sqrt3}{3}$$
Work Step by Step
- The radius of the rotation circle at point A is $r_A=\sqrt{L_1^2+L_2^2}$
- The radius of the rotation circle at point B is $r_B=L_1$
According to the given information, $a_{c, A}=2a_{c, B}$. Therefore, $$\frac{r_A\omega_A^2}{r_B\omega_B^2}=2$$
The angular speed $\omega$ describes the entire rotation and does not change for any specific points inside that rotation. Therefore, $\omega_A=\omega_B$
$$\frac{r_A}{r_B}=2$$ $$\frac{\sqrt{L_1^2+L_2^2}}{L_1}=2$$
Bring the equation to the power of 2: $$\frac{L_1^2+L_2^2}{L_1^2}=1+\frac{L_2^2}{L_1^2}=4$$ $$\Big(\frac{L_2}{L_1}\Big)^2=3$$ $$\frac{L_1}{L_2}=\frac{\sqrt3}{3}$$
