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∘ ?

To find the ball's speed at the moment when the angle \(\theta = 30^\circ\), we need to consider the forces acting on the ball at that point, particularly the tension in the string and the gravitational force.

Here are the key points in solving the problem:

### Step 1: Draw a Free Body Diagram (FBD)
At the point where \(\theta = 30^\circ\), the forces acting on the ball are:
- The tension in the string (\(T = 20 \, \text{N}\)) acting radially inward.
- The gravitational force (\(mg\)) acting vertically downward.

We will break down the forces into radial (along the string) and tangential components, but first, let's calculate the necessary physical quantities.

### Step 2: Apply Newton’s Second Law in Radial Direction
In the radial direction, the forces providing the centripetal acceleration are:
- The component of the gravitational force that acts along the radial direction (\(mg\cos\theta\)).
- The tension force (\(T\)).

Using Newton's second law in the radial direction, we get:
\[
T - mg\cos\theta = m \frac{v^2}{r}
\]
Where:
- \(T = 20 \, \text{N}\) (tension),
- \(m = 1.0 \, \text{kg}\) (mass of the ball),
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity),
- \(r = 0.90 \, \text{m}\) (radius of the circle),
- \(\theta = 30^\circ\),
- \(v\) is the speed of the ball at this point.

First, calculate the component of the gravitational force in the radial direction:
\[
mg\cos\theta = (1.0)(9.8)\cos(30^\circ) = 9.8 \times 0.866 = 8.49 \, \text{N}
\]

Now plug this into the equation:
\[
20 \, \text{N} - 8.49 \, \text{N} = 1.0 \cdot \frac{v^2}{0.90}
\]
\[
11.51 = \frac{v^2}{0.90}
\]

### Step 3: Solve for the Speed \(v\)
Now solve for \(v\):
\[
v^2 = 11.51 \times 0.90 = 10.36
\]
\[
v = \sqrt{10.36} = 3.22 \, \text{m/s}
\]

### Final Answer:
The ball's speed when \(\theta = 30^\circ\) is approximately \(3.22 \, \text{m/s}\).

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Would you like further details or have any questions on this solution?

Here are 5 related questions for expanding the concept:
1. How would the speed change if the angle \(\theta\) were different, say \(\theta = 45^\circ\)?
2. What would happen if the tension in the string increased to 25 N?
3. How can we determine the total mechanical energy of the system at this point?
4. How does the tension in the string vary at different points of the swing?
5. What would the speed be at the highest point of the swing?

**Tip**: For circular motion problems, it's crucial to separate the forces into radial and tangential components when analyzing them with Newton’s laws.

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∘?

Find the speed of a 1.0 kg ball swinging in a vertical circle on a 90-cm string when the tension is 20 N and the angle is θ=30°. This solution explores circular motion, Newton's second law, and forces. Suitable for Grades 10-12 physics students.

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