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 solve for the speed of the ball at an angle \(\theta = 30^\circ\), we can use the following principles:

1. **Forces in circular motion:** At any point in the circular path, the forces acting on the ball are:
   - The gravitational force (\(mg\)) acting vertically downward.
   - The tension in the string (\(T\)).
   - The centripetal force required to keep the ball moving in a circular path, provided by the net radial force (which comes from the combination of tension and the radial component of gravity).

2. **Energy conservation:** The total mechanical energy (kinetic + potential) is conserved as the ball swings.

### Given:
- Mass of the ball, \( m = 1.0 \, \text{kg} \)
- Length of the string, \( r = 90 \, \text{cm} = 0.90 \, \text{m} \)
- Tension in the string at \(\theta = 30^\circ\), \( T = 20 \, \text{N} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Analyze forces at \(\theta = 30^\circ\)

At an angle \(\theta = 30^\circ\), the forces acting radially are:

- Radial component of gravity: \( mg \cos \theta \)
- Tension in the string: \( T \)

The net radial force is the centripetal force, which is:

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

Where:
- \( v \) is the speed of the ball at \(\theta = 30^\circ\).
- \( r = 0.90 \, \text{m} \) is the radius of the circle.

### Step 2: Plug in values

Substitute the known values into the equation:

\[
20 \, \text{N} - (1.0 \, \text{kg})(9.8 \, \text{m/s}^2)\cos 30^\circ = \frac{(1.0 \, \text{kg}) v^2}{0.90 \, \text{m}}
\]

First, calculate \( mg \cos 30^\circ \):

\[
mg \cos 30^\circ = 1.0 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 8.49 \, \text{N}
\]

Now substitute:

\[
20 - 8.49 = \frac{v^2}{0.90}
\]

\[
11.51 = \frac{v^2}{0.90}
\]

Multiply both sides by 0.90:

\[
v^2 = 10.36
\]

Taking the square root:

\[
v \approx 3.22 \, \text{m/s}
\]

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

---

Would you like further details or have any other questions?

Here are 5 related questions to deepen your understanding:
1. How does the speed of the ball change at different points in the circle?
2. How can we calculate the speed of the ball at the highest point in the circle?
3. What would happen to the tension in the string if the speed of the ball increased?
4. How does the angle \(\theta\) affect the radial component of the gravitational force?
5. How would the answer change if the length of the string was increased?

**Tip:** Understanding the relationship between tension, speed, and gravitational forces is crucial when analyzing circular motion.

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∘?

Learn how to calculate the speed of a 1.0 kg ball swinging in a vertical circle with an 90-cm-long string at θ = 30°. The solution involves key physics concepts such as centripetal force and Newton's Second Law. Ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation