A 1.0 kg ball swings in a vertical circle on the end of an 90-cm-long string. The tension in the string is 20 N. What is the ball's speed?

To find the speed of the ball in this scenario, we need to consider the forces acting on the ball and apply Newton's second law for circular motion.

### Step 1: Set up the forces
There are two forces acting on the ball:
1. **Tension in the string** (\( T \)) = 20 N
2. **Gravitational force** (\( F_g \)) = \( mg \), where \( m = 1.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).

At the bottom of the circle, the net force towards the center of the circle is the centripetal force, which is provided by the difference between the tension and the gravitational force:

\[
F_{\text{net}} = T - F_g = T - mg
\]

### Step 2: Apply the centripetal force equation
The centripetal force is also given by:

\[
F_{\text{net}} = \frac{mv^2}{r}
\]

Where:
- \( m = 1.0 \, \text{kg} \) (mass of the ball)
- \( v \) is the speed of the ball (which we need to find)
- \( r = 0.90 \, \text{m} \) (radius of the circle)

### Step 3: Set up the equation
Equating the two expressions for \( F_{\text{net}} \):

\[
T - mg = \frac{mv^2}{r}
\]

Substitute the given values:

\[
20 \, \text{N} - (1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2) = \frac{1.0 \, \text{kg} \times v^2}{0.90 \, \text{m}}
\]

\[
20 \, \text{N} - 9.8 \, \text{N} = \frac{v^2}{0.90 \, \text{m}}
\]

\[
10.2 \, \text{N} = \frac{v^2}{0.90 \, \text{m}}
\]

### Step 4: Solve for \( v \)
Multiply both sides by \( 0.90 \, \text{m} \):

\[
v^2 = 10.2 \, \text{N} \times 0.90 \, \text{m} = 9.18 \, \text{m}^2/\text{s}^2
\]

Now, take the square root of both sides:

\[
v = \sqrt{9.18} \, \text{m/s} \approx 3.03 \, \text{m/s}
\]

### Final Answer:
The ball's speed is approximately \( 3.03 \, \text{m/s} \).

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Would you like more details or have any questions? 

Here are five related questions to expand your understanding:
1. How would the speed change if the string length increased?
2. What is the role of tension in circular motion?
3. How would the result differ if the ball were at the top of the circle?
4. What is the formula for centripetal acceleration?
5. How does the mass of the object affect its speed?

**Tip:** In circular motion, tension is not always equal to the centripetal force. At different points of the motion (top vs bottom), gravity either adds to or subtracts from the tension.

Find the speed of a 1.0 kg ball swinging in a vertical circle with a tension of 20 N and a string length of 90 cm. This problem involves circular motion, centripetal force, and Newton's Second Law. Detailed steps included, suitable for students in Grades 11-12.

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