To determine the radius of curvature $r$, we can use the equation provided in the graph:

$n = (0.400 \, \text{kg/m}) v^2 + 588 \, \text{N}$

This equation models the normal force $n$ as a function of the square of velocity $v^2$.

Step-by-Step Solution:

From Newton's second law for circular motion at the bottom of a dip, the normal force $n$ at the bottom of a circular arc is related to the gravitational force and the centripetal force:

$n = mg + \frac{mv^2}{r}$

Where:

• $n$ is the normal force,
• $m$ is the mass of the object,
• $g$ is the acceleration due to gravity (9.8 m/s²),
• $v$ is the velocity,
• $r$ is the radius of curvature we need to find.

Equation Analysis:

Comparing the equation $n = (0.400 \, \text{kg/m}) v^2 + 588 \, \text{N}$ with the theoretical form $n = mg + \frac{mv^2}{r}$, we can identify the constants:

• The term $588 \, \text{N}$ corresponds to $mg$, which means the mass $m$ is:

$mg = 588 \, \text{N}$

Solving for $m$:

$m = \frac{588 \, \text{N}}{9.8 \, \text{m/s}^2} = 60 \, \text{kg}$

• The coefficient of $v^2$ in the equation, $0.400 \, \text{kg/m}$, represents $\frac{m}{r}$. So we can solve for $r$:

$\frac{m}{r} = 0.400 \, \text{kg/m}$

Substitute $m = 60 \, \text{kg}$:

$\frac{60 \, \text{kg}}{r} = 0.400 \, \text{kg/m}$

Now, solve for $r$:

$r = \frac{60 \, \text{kg}}{0.400 \, \text{kg/m}} = 150 \, \text{m}$

Final Answer:

The radius of curvature $r$ is 150 m.

Would you like further details or have any questions?

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Here are 5 related questions:

1. What would the normal force be at a given velocity if the radius is doubled?
2. How would the normal force change if the mass were halved?
3. What happens to the radius of curvature if the coefficient of $v^2$ decreases?
4. How is the centripetal force related to the radius in circular motion?
5. Can this method be applied to non-uniform circular motion? Why or why not?

Tip: When analyzing forces in circular motion, always check for the contributions from both gravitational and centripetal forces to the normal force.