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 this problem, we can apply the concepts of circular motion and Newton's second law.

### Step-by-step solution:

1. **Understanding the forces at play**:
   At any point in the circular motion, there are two main forces acting on the ball:
   - The tension \( T \) in the string (given as 20 N when \( \theta = 30^\circ \)).
   - The gravitational force \( mg \), where \( m \) is the mass of the ball (1.0 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²).

2. **Breaking down the forces**:
   The tension and the component of the gravitational force both act towards the center of the circular path. We need to consider the net radial force, which provides the centripetal force \( F_{\text{centripetal}} = m\frac{v^2}{r} \) that keeps the ball moving in a circular path.

   - The tension \( T \) provides part of the centripetal force.
   - The component of the gravitational force along the radial direction is \( mg \cos \theta \).

   So, the total centripetal force is the sum of these forces:
   \[
   T - mg \cos \theta = m \frac{v^2}{r}
   \]
   where \( v \) is the speed of the ball at \( \theta = 30^\circ \), and \( r = 0.90 \) m is the length of the string.

3. **Substitute the known values**:
   \[
   20 \, \text{N} - (1.0 \, \text{kg})(9.8 \, \text{m/s}^2)\cos 30^\circ = (1.0 \, \text{kg}) \frac{v^2}{0.90 \, \text{m}}
   \]

   First, calculate \( mg \cos 30^\circ \):
   \[
   mg \cos 30^\circ = (1.0 \, \text{kg})(9.8 \, \text{m/s}^2)(0.866) = 8.487 \, \text{N}
   \]

   Now, substitute this into the equation:
   \[
   20 \, \text{N} - 8.487 \, \text{N} = \frac{v^2}{0.90 \, \text{m}}
   \]
   \[
   11.513 \, \text{N} = \frac{v^2}{0.90 \, \text{m}}
   \]

4. **Solve for \( v^2 \)**:
   \[
   v^2 = 11.513 \, \text{N} \times 0.90 \, \text{m} = 10.3617 \, \text{m}^2/\text{s}^2
   \]

5. **Solve for \( v \)**:
   \[
   v = \sqrt{10.3617} \approx 3.22 \, \text{m/s}
   \]

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

Would you like more details on any part of the solution, or do you have any further questions?

Here are 5 related questions to expand on this concept:
1. What would the speed of the ball be at the top of the circle?
2. How does the tension in the string change as the ball moves through different points in the circle?
3. What is the minimum speed required to keep the ball moving in a vertical circle without the string going slack?
4. How would the problem change if the length of the string were doubled?
5. What would happen to the tension if the mass of the ball were increased?

**Tip**: Always check if the forces acting on an object include components in different directions. Decomposing forces into radial and tangential components is key in circular motion problems.

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 in a vertical circular motion when the tension is 20 N at an angle of θ = 30°. This problem involves circular motion, forces, and Newton's Second Law. Suitable for students in Grades 11-12.

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