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 when its angle from the highest point on the circle is θ=30∘. What is the ball's speed when  θ=30∘ ? Remember, this is a vertical circle and the angle is measured from the highest point of that vertical circle

We can approach this problem using the principles of forces in circular motion and energy conservation. Here's the step-by-step solution:

### Step 1: Identify the forces acting on the ball
At an angle \( \theta = 30^\circ \), there are two forces acting on the ball:
- The gravitational force (\( F_g = mg \))
- The tension in the string (\( T = 20 \, \text{N} \))

The components of these forces will be analyzed along the radial direction to find the centripetal force, which is responsible for the circular motion.

### Step 2: Radial force equation
The net force acting towards the center of the circle provides the centripetal force. The equation for the net force in the radial direction is:

\[
T - mg \cos(\theta) = \frac{mv^2}{r}
\]

Where:
- \( T \) is the tension in the string (20 N),
- \( m \) is the mass of the ball (1.0 kg),
- \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)),
- \( \theta = 30^\circ \),
- \( v \) is the speed of the ball at \( \theta = 30^\circ \),
- \( r \) is the length of the string (0.90 m).

### Step 3: Plug in the known values
First, calculate the component of gravitational force acting along the radial direction:

\[
mg \cos(\theta) = (1.0 \, \text{kg})(9.8 \, \text{m/s}^2) \cos(30^\circ)
\]
\[
mg \cos(30^\circ) = 9.8 \times 0.866 = 8.49 \, \text{N}
\]

Now, substitute these values into the radial force equation:

\[
T - mg \cos(30^\circ) = \frac{mv^2}{r}
\]
\[
20 \, \text{N} - 8.49 \, \text{N} = \frac{(1.0 \, \text{kg}) v^2}{0.90 \, \text{m}}
\]
\[
11.51 \, \text{N} = \frac{v^2}{0.90 \, \text{m}}
\]
\[
v^2 = 11.51 \times 0.90 = 10.36 \, \text{m}^2/\text{s}^2
\]
\[
v = \sqrt{10.36} = 3.22 \, \text{m/s}
\]

### Final Answer:
The speed of the ball when \( \theta = 30^\circ \) is approximately **3.22 m/s**.

---

Would you like more details or have any further questions? Here are some related questions to expand on the topic:

1. How would the speed change if the angle \( \theta \) were smaller, say \( 15^\circ \)?
2. What would happen to the tension if the string were longer or shorter?
3. How would the speed at the bottom of the circle (at \( \theta = 90^\circ \)) compare to the speed at \( \theta = 30^\circ \)?
4. How does the gravitational force affect the motion of the ball at different points in the vertical circle?
5. What role does centripetal force play in maintaining the ball's circular motion?

**Tip**: In circular motion, always consider both forces: tension and gravity. They combine to create the centripetal force at different points on the circle.

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 when its angle from the highest point on the circle is θ=30∘. What is the ball's speed when θ=30∘? Remember, this is a vertical circle and the angle is measured from the highest point of that vertical circle.

Find the speed of a 1.0 kg ball swinging on a 90-cm string in a vertical circle. Given the tension in the string is 20 N at θ=30°, this solution explains the use of radial forces and circular motion principles to calculate the ball's speed, suitable for Grades 10-12.

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