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The problem you've shared involves a pendulum, where a small heavy ball of mass \( m \) is attached to the end of a string of length \( l \). The string forms an angle of 30° with the horizontal when the ball is released from point \( P \).

Let’s break down the parts:

### 1. **Forces on the ball at Point \( P \) (Part a)**:
- At point \( P \), the forces acting on the ball are:
  - **Tension** \( T \), which acts along the string toward the pivot.
  - **Gravitational force** \( mg \), acting vertically downward.

The forces should be represented on the diagram at point \( P \). You've done this by drawing a vertical arrow downwards for \( mg \) and a diagonal arrow toward the pivot for \( T \).

### 2. **Determining the speed of the ball at point \( P \) (Part b)**:
To find the speed at point \( P \), energy conservation is applied. The potential energy at the highest point (where \( v = 0 \)) is converted into kinetic energy at point \( P \).

- **Potential Energy at the highest point**:
  \[
  E_{\text{initial}} = mgh = mg(l \sin 30^\circ)
  \]

- **Kinetic Energy at point \( P \)**:
  \[
  E_{\text{final}} = \frac{1}{2}mv^2
  \]

By equating the potential energy at the highest point to the kinetic energy at \( P \):
\[
mgl \sin 30^\circ = \frac{1}{2}mv^2
\]

Solve for \( v \):
\[
v^2 = 2gl \cos 30^\circ
\]
\[
v = \sqrt{2gl \cos 30^\circ}
\]
The result is the speed of the ball at point \( P \), which you've correctly calculated.

### 3. **Determining the tension in the string at point \( P \) (Part c)**:
To determine the tension, the forces acting on the ball in the radial direction need to be considered, using Newton's second law for circular motion:
\[
T - mg \cos 30^\circ = m \frac{v^2}{l}
\]

Substituting the value of \( v^2 \) from the previous step:
\[
T - mg \cos 30^\circ = m \frac{2gl \cos 30^\circ}{l}
\]
\[
T = 3mg \cos 30^\circ
\]
This gives the tension in the string at point \( P \).

### 4. **Tangential acceleration at point \( P \) (Part d)**:
The tangential acceleration is given by the tangential component of the gravitational force:
\[
a_{\text{tangential}} = g \sin 30^\circ
\]
You have calculated this as \( g \cos 30^\circ \), which corresponds to the component acting along the motion of the pendulum.

---

Would you like more details on any specific part of the solution?

Here are some related questions to deepen understanding:

1. How would the equations change if the angle were different, say 45° instead of 30°?
2. What is the significance of the tension force and how does it vary as the pendulum swings?
3. Could this problem be solved using a different method, such as angular momentum conservation?
4. How would air resistance affect the calculations for speed and tension?
5. What happens to the speed at the lowest point of the pendulum?

**Tip:** Always check your trigonometric functions when resolving forces, as mixing up sine and cosine could lead to errors!

A pendulum consisting of a small heavy ball of mass m at the end of a string of length l is released from a horizontal position. When the ball is at point P, the string forms an angle of 30° with the horizontal. The problem involves determining the forces acting on the ball, the speed of the ball at point P, the tension in the string, and the tangential acceleration of the ball.

This AP Physics problem involves calculating the forces acting on a pendulum at 30° with the horizontal, determining the speed at point P, the tension in the string, and the tangential acceleration. Learn how to apply energy conservation, Newton’s second law, and trigonometry to solve pendulum motion problems.

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