A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string. If the angular speed at the bottom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?

To find the tension in the string when the object is at the bottom of the circle, we need to consider both the gravitational force and the centripetal force acting on the object at that point.

### Step 1: Identify the forces acting on the object
At the bottom of the circle, the forces acting on the object are:
1. The gravitational force (\( F_g \)) pulling the object downward.
2. The tension in the string (\( T \)), which acts upward to provide the necessary centripetal force and to counter the gravitational force.

### Step 2: Calculate the gravitational force (\( F_g \))
The gravitational force is given by:
\[
F_g = mg
\]
where:
- \( m = 0.300 \, \text{kg} \) (mass of the object)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[
F_g = 0.300 \times 9.8 = 2.94 \, \text{N}
\]

### Step 3: Calculate the centripetal force required (\( F_c \))
The centripetal force required to keep the object moving in a circular path at the bottom of the circle is:
\[
F_c = m \omega^2 r
\]
where:
- \( \omega = 8.00 \, \text{rad/s} \) (angular speed at the bottom)
- \( r = 0.500 \, \text{m} \) (radius of the circular path, which is the length of the string)

Substituting the values:
\[
F_c = 0.300 \times (8.00)^2 \times 0.500
\]
\[
F_c = 0.300 \times 64 \times 0.500
\]
\[
F_c = 9.6 \, \text{N}
\]

### Step 4: Calculate the tension in the string (\( T \))
At the bottom of the circle, the tension in the string must provide enough force to both support the weight of the object and provide the centripetal force. Therefore, the total tension is the sum of the gravitational force and the centripetal force:
\[
T = F_g + F_c
\]
Substituting the values we calculated:
\[
T = 2.94 + 9.6
\]
\[
T = 12.54 \, \text{N}
\]

### Final Answer:
The tension in the string when the object is at the bottom of the circle is:
\[
\boxed{12.54 \, \text{N}}
\]

Would you like a further breakdown or have any other questions?

---

Here are five related questions you might find helpful:
1. What would the tension be if the object were at the top of the circle?
2. How would the tension change if the length of the string were increased?
3. How does the tension vary with the angular speed at different points in the circle?
4. How would the tension be affected if the object's mass were doubled?
5. What happens to the tension if gravity is ignored (e.g., in a space setting)?

**Tip:** Remember that in circular motion, the centripetal force always points toward the center and can result from different combinations of tension, gravity, or normal force, depending on the situation and position in the circle.

A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string. If the angular speed at the bottom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?

Learn how to calculate the tension in a string for a 0.300-kg object moving in vertical circular motion with an angular speed of 8.00 rad/s at the bottom of the circle. This problem covers concepts of centripetal force, gravitational force, and tension forces, ideal for physics students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation