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We are tasked with finding the period of the circular motion for a small rock swinging on a rope at an angle of 36.9° from the vertical. Here's how we approach the problem:

### Step 1: Analyze the Forces
The tension in the rope has two components:
1. **Vertical component**: Balances the weight of the rock.
   \[
   T \cos(\theta) = mg
   \]
2. **Horizontal component**: Provides the centripetal force that keeps the rock moving in a circle.
   \[
   T \sin(\theta) = \frac{mv^2}{r}
   \]

Where:
- \( T \) is the tension in the rope,
- \( m = 0.500 \, \text{kg} \) (mass of the rock),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( r \) is the radius of the circular motion,
- \( v \) is the velocity of the rock,
- \( \theta = 36.9^\circ \).

### Step 2: Solve for Tension \( T \)
Using the vertical component equation:
\[
T = \frac{mg}{\cos(\theta)}
\]
Substitute \( m = 0.500 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( \theta = 36.9^\circ \):
\[
T = \frac{0.500 \times 9.8}{\cos(36.9^\circ)} = \frac{4.9}{0.8} \approx 6.125 \, \text{N}
\]

### Step 3: Determine the Radius \( r \)
The radius \( r \) of the circular motion is the horizontal distance from the center of the circle to the rock, which is:
\[
r = L \sin(\theta)
\]
where \( L = 1.50 \, \text{m} \) is the length of the rope.

Substituting the values:
\[
r = 1.50 \times \sin(36.9^\circ) \approx 1.50 \times 0.6 = 0.90 \, \text{m}
\]

### Step 4: Find the Velocity \( v \)
Now, use the horizontal component equation for centripetal force:
\[
T \sin(\theta) = \frac{mv^2}{r}
\]
Substituting for \( T = 6.125 \, \text{N} \), \( \sin(36.9^\circ) \approx 0.6 \), \( m = 0.500 \, \text{kg} \), and \( r = 0.90 \, \text{m} \):
\[
6.125 \times 0.6 = \frac{0.500 \times v^2}{0.90}
\]
\[
3.675 = \frac{0.500 \times v^2}{0.90}
\]
\[
v^2 = \frac{3.675 \times 0.90}{0.500} = 6.615
\]
\[
v \approx \sqrt{6.615} \approx 2.57 \, \text{m/s}
\]

### Step 5: Find the Period \( T_{\text{period}} \)
The period of the circular motion is given by:
\[
T_{\text{period}} = \frac{2 \pi r}{v}
\]
Substituting \( r = 0.90 \, \text{m} \) and \( v = 2.57 \, \text{m/s} \):
\[
T_{\text{period}} = \frac{2 \pi \times 0.90}{2.57} \approx \frac{5.65}{2.57} \approx 2.2 \, \text{seconds}
\]

### Conclusion:
The period of the motion is approximately **2.2 seconds**, which corresponds to answer **(b)**.

Would you like more details on any step, or further clarification?

Here are 5 related questions:
1. What are the conditions for uniform circular motion?
2. How does the angle of the rope affect the period?
3. What would happen to the period if the mass of the rock were doubled?
4. How would the period change if the rope were longer?
5. Why does the tension in the rope have both vertical and horizontal components?

**Tip:** In circular motion problems, always separate the forces into components and carefully analyze both the radial and tangential directions.

A small rock with mass m = 0.500 kg is swinging in a horizontal circle on the lower end of a 1.50 m long light rope. The rope makes a constant angle of θ = 36.9° with respect to the vertical. Find the period of the motion, the time it takes the rock to complete one revolution around its circular path.

Learn how to determine the period of a rock swinging in a circular motion on a rope with a 36.9° angle from the vertical. This problem uses key concepts of circular motion and trigonometry and includes step-by-step calculations with Newton's second law and centripetal force.

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