A 3.0 kg stone at the end of a 0.50 m long string is whirled in a vertical circle with constant velocity. If the tension in the string is 10 N at the top of the circle the tension at the bottom is?

To solve this, we need to analyze the forces acting on the stone at both the top and the bottom of the circular path.

### 1. **At the top of the circle**:
The forces acting on the stone are:
- The tension in the string, \( T_{\text{top}} = 10 \, \text{N} \), directed downward.
- The weight of the stone, \( mg = 3.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \), also directed downward.

Since the stone is moving in a circular path with constant velocity, the net force is the centripetal force, which is the sum of the tension and the weight at the top:
\[
F_{\text{net}} = T_{\text{top}} + mg = m \frac{v^2}{r}
\]
Substituting the known values:
\[
10 \, \text{N} + 29.4 \, \text{N} = 3.0 \, \text{kg} \times \frac{v^2}{0.50 \, \text{m}}
\]
Solving for \( v^2 \):
\[
39.4 = 6v^2
\]
\[
v^2 = \frac{39.4}{6} = 6.5667 \, \text{m}^2/\text{s}^2
\]
\[
v = \sqrt{6.5667} \approx 2.56 \, \text{m/s}
\]

### 2. **At the bottom of the circle**:
The forces acting on the stone at the bottom are:
- The tension in the string, \( T_{\text{bottom}} \), directed upward.
- The weight of the stone, \( mg = 29.4 \, \text{N} \), directed downward.

At the bottom, the tension must overcome both the centripetal force and the weight of the stone:
\[
T_{\text{bottom}} - mg = m \frac{v^2}{r}
\]
Substituting the known values:
\[
T_{\text{bottom}} - 29.4 \, \text{N} = 3.0 \, \text{kg} \times \frac{6.5667}{0.50 \, \text{m}}
\]
\[
T_{\text{bottom}} - 29.4 = 39.4 \, \text{N}
\]
\[
T_{\text{bottom}} = 39.4 \, \text{N} + 29.4 \, \text{N} = 68.8 \, \text{N}
\]

### Final Answer:
The tension in the string at the bottom of the circle is **68.8 N**.

Do you want more details or have any questions?

### Related Questions:
1. What would be the tension at the midpoint of the vertical circle?
2. How would increasing the velocity affect the tension at the bottom of the circle?
3. What is the minimum speed needed for the stone to maintain circular motion?
4. How does changing the length of the string affect the centripetal force?
5. What would the tension be if the stone were whirled in a horizontal circle?

### Tip:
Centripetal force always acts toward the center of the circle, but its magnitude depends on the velocity and radius of the motion.

A 3.0 kg stone at the end of a 0.50 m long string is whirled in a vertical circle with constant velocity. If the tension in the string is 10 N at the top of the circle, the tension at the bottom is?

Find the tension at the bottom of a vertical circular path for a 3 kg stone on a 0.50 m string. This problem involves forces, centripetal force, and Newton's Second Law. Learn the detailed step-by-step solution with the necessary formulas. Suitable for students in Grades 10-12.

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