We are tasked with finding the ratio of the radial accelerations at the bottom and top of a vertical circular motion, given the speeds $v_1$ at the top and $v_2$ at the bottom of the circle.

Key points:

• Radial acceleration is given by the equation: $a_{\text{rad}} = \frac{v^2}{R}$ where $v$ is the speed of the object and $R$ is the radius of the circular path.

Radial acceleration at the top of the circle:

• At the top of the circle, the car's speed is $v_1$, so the radial acceleration is: $a_{\text{rad,top}} = \frac{v_1^2}{R}$

Radial acceleration at the bottom of the circle:

• At the bottom of the circle, the car's speed is $v_2$, so the radial acceleration is: $a_{\text{rad,bottom}} = \frac{v_2^2}{R}$

Ratio of radial accelerations:

Now, we find the ratio of the radial acceleration at the bottom to that at the top: $\frac{a_{\text{rad,bottom}}}{a_{\text{rad,top}}} = \frac{\frac{v_2^2}{R}}{\frac{v_1^2}{R}} = \frac{v_2^2}{v_1^2}$

Thus, the ratio of the radial accelerations is: $\boxed{\frac{v_2^2}{v_1^2}}$

Summary:

The ratio of the radial acceleration at the bottom of the circle to the top is $\frac{v_2^2}{v_1^2}$.

---

Would you like more details or clarification on any part of this solution?

Here are five related questions to explore this concept further:

1. How would the tension in the string (or normal force on the car) change between the top and bottom of the circle?
2. How does gravity affect the car’s speed at different points in the circle?
3. What would happen to the car's speed if friction were considered?
4. How can energy conservation help us relate $v_1$ and $v_2$?
5. How would the centripetal force differ at the top and bottom?

Tip: In circular motion, radial acceleration is always directed towards the center, and its magnitude depends on both the speed and the radius of the path.