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The problem you've uploaded involves a **conical pendulum**, where a bob moves in a horizontal circle at constant speed. Here's the breakdown:

**Given Data:**
- Mass of the bob: \( m = 0.0420 \, \text{kg} \)
- Length of the string: \( L = 1.04 \, \text{m} \)
- Circumference of the circular path: \( C = 1.25 \, \text{m} \)

You are tasked with finding:
1. **(a)** The tension in the string.
2. **(b)** The period of the motion.

### Step 1: Calculate the Radius of the Circular Path

The radius \( r \) of the circular path can be calculated from the given circumference:
\[
C = 2\pi r \quad \Rightarrow \quad r = \frac{C}{2\pi}
\]
Substitute \( C = 1.25 \, \text{m} \):
\[
r = \frac{1.25}{2\pi} = 0.199 \, \text{m}
\]

### Step 2: Relate Geometry to Tension
The angle \( \theta \) between the string and the vertical can be found using the geometry of the conical pendulum:
\[
\cos \theta = \frac{h}{L}
\]
where \( h \) is the vertical height of the string, and \( r = L \sin \theta \). This gives:
\[
\sin \theta = \frac{r}{L} = \frac{0.199}{1.04} = 0.191
\]

Now find \( \theta \):
\[
\theta = \sin^{-1}(0.191) \approx 11.01^\circ
\]

### Step 3: Tension in the String
To calculate the tension \( T \), we consider the forces acting on the bob. The tension provides both a centripetal force \( T \sin \theta = \frac{mv^2}{r} \) and supports the weight of the bob \( T \cos \theta = mg \).

We can find \( T \) by solving the system of equations. The vertical component balances the weight:
\[
T \cos \theta = mg \quad \Rightarrow \quad T = \frac{mg}{\cos \theta}
\]
Substitute \( m = 0.0420 \, \text{kg}, \, g = 9.8 \, \text{m/s}^2, \, \cos \theta = \cos 11.01^\circ = 0.981 \):
\[
T = \frac{0.0420 \times 9.8}{0.981} \approx 0.419 \, \text{N}
\]

### Step 4: Period of the Motion
The period \( T_{\text{period}} \) can be found by using the velocity and the radius. First, calculate the velocity \( v \). Using \( T \sin \theta = \frac{mv^2}{r} \), we solve for \( v \):
\[
v = \sqrt{\frac{T \sin \theta \cdot r}{m}}
\]
Substitute \( T = 0.419 \, \text{N}, \sin \theta = 0.191, r = 0.199 \, \text{m}, m = 0.0420 \, \text{kg} \):
\[
v = \sqrt{\frac{0.419 \times 0.191 \times 0.199}{0.0420}} \approx 0.58 \, \text{m/s}
\]

The period \( T_{\text{period}} \) is:
\[
T_{\text{period}} = \frac{2\pi r}{v}
\]
Substitute \( r = 0.199 \, \text{m}, v = 0.58 \, \text{m/s} \):
\[
T_{\text{period}} = \frac{2\pi \times 0.199}{0.58} \approx 2.16 \, \text{seconds}
\]

### Final Answers:
1. **Tension in the string**: \( T \approx 0.419 \, \text{N} \)
2. **Period of motion**: \( T_{\text{period}} \approx 2.16 \, \text{seconds} \)

Would you like further clarifications or more details on any step?

Here are 5 related questions for further exploration:
1. How would the tension change if the mass of the bob were doubled?
2. How does the period of motion depend on the radius of the circular path?
3. What would happen to the period if the length of the string increased?
4. How would the tension change if the bob moved faster in the circular path?
5. What effect does gravitational acceleration have on the period?

**Tip**: Always break problems involving forces into their horizontal and vertical components for clarity in solving.

The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. The bob has a mass of 0.0420 kg, the string has length L = 1.04 m and negligible mass, and the bob follows a circular path of circumference 1.25 m. What are (a) the tension in the string and (b) the period of the motion?

Learn how to calculate the tension in the string and the period of motion for a conical pendulum where a bob of mass 0.0420 kg moves in a horizontal circular path with a string of length 1.04 m and a path circumference of 1.25 m. This problem involves key concepts like centripetal force, trigonometry, and Newton's second law.

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