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The problem described in the image involves a merry-go-round with vertical rods inclined at an angle of \( 37^\circ \) from the vertical due to rotation. We are asked to calculate the frequency of revolution of the merry-go-round.

### Given Data:
- Angle with vertical: \( \theta = 37^\circ \)
- Length of the horizontal rod: \( L_h = 2.1 \, \text{m} \)
- Length of the vertical rod (during rotation): \( L_v = 1.5 \, \text{m} \)
- Gravitational acceleration: \( g = \pi^2 \, \text{m/s}^2 \)
- \( \sin 37^\circ = 0.6 \)

### Steps to Solve:

#### 1. Horizontal and Vertical Forces
During the circular motion, the vertical rod makes an angle \( \theta \) with the vertical due to the centrifugal force. The forces acting on the rider (or the object at the end of the vertical rod) include:
- Centripetal force, \( F_c \), which acts horizontally.
- Gravitational force, \( F_g = mg \), acting vertically downward.

The tension in the rod will have two components:
- Horizontal component: \( T \sin \theta \), which provides the centripetal force.
- Vertical component: \( T \cos \theta = mg \), balancing the weight.

#### 2. Centripetal Force
From the vertical equilibrium:
\[
T \cos \theta = mg
\]
From the horizontal component:
\[
T \sin \theta = m \cdot \frac{v^2}{r}
\]
where \( v \) is the tangential velocity and \( r \) is the horizontal distance from the center of the rotation to the rider.

The total distance \( r \) is:
\[
r = L_h + L_v \sin \theta = 2.1 + 1.5 \times 0.6 = 2.1 + 0.9 = 3.0 \, \text{m}
\]

#### 3. Relationship Between Forces
Dividing the two force equations:
\[
\frac{T \sin \theta}{T \cos \theta} = \frac{\frac{v^2}{r}}{mg} 
\]
\[
\tan \theta = \frac{v^2}{rg}
\]
\[
v^2 = rg \tan \theta
\]
Substituting \( r = 3.0 \, \text{m} \), \( g = \pi^2 \, \text{m/s}^2 \), and \( \tan 37^\circ = \frac{\sin 37^\circ}{\cos 37^\circ} \approx \frac{0.6}{0.8} = 0.75 \):
\[
v^2 = 3.0 \times \pi^2 \times 0.75 = 7.065 \, \text{m}^2/\text{s}^2
\]
\[
v = \sqrt{7.065} \approx 2.66 \, \text{m/s}
\]

#### 4. Frequency of Revolution
The frequency \( f \) is related to the velocity by:
\[
v = 2 \pi r f
\]
Solving for \( f \):
\[
f = \frac{v}{2 \pi r} = \frac{2.66}{2 \pi \times 3.0} \approx \frac{2.66}{18.8496} \approx 0.141 \, \text{Hz}
\]

Thus, the frequency of revolution of the merry-go-round is approximately \( 0.141 \, \text{Hz} \).

Would you like further clarification on any part of this process? Here are some related questions to explore the topic further:

1. How does the angle of inclination affect the frequency of rotation?
2. What would happen if the length of the rods were changed?
3. How is the centripetal force related to the tangential velocity in circular motion?
4. Why is the tangent of the angle used in the force equation?
5. Can this method be applied to other rotating systems, like a conical pendulum?

**Tip:** Remember that the force responsible for circular motion is always directed toward the center of the circle, known as the centripetal force.

A merry-go-round has vertical rods inclined at an angle of 37° due to rotation. Given the horizontal rod length (2.1 m), vertical rod length (1.5 m), gravitational acceleration (g = π^2 m/s²), and sin 37° = 0.6, calculate the frequency of revolution.

Solve the problem of finding the frequency of revolution for a merry-go-round where vertical rods are inclined at 37° from the vertical. This physics problem involves circular motion, centripetal force, and trigonometric concepts. Suitable for Grades 11-12.

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