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This problem involves a ball swinging counterclockwise in a vertical circle and requests three parts to be solved. Here’s a breakdown of the problem:

1. **Given Data**:
   - The rope length: \( L = 1.50 \, \text{m} \)
   - The ball's angle past the lowest point: \( 36.9^\circ \)
   - The total acceleration of the ball at that instant: 
     \[
     \vec{a} = (-22.5 \, \hat{i} + 20.2 \, \hat{j}) \, \text{m/s}^2
     \]

2. **Tasks**:
   (a) Sketch a vector diagram for the acceleration components.
   (b) Determine the magnitude of the radial (centripetal) acceleration.
   (c) Calculate the speed and velocity of the ball at that instant.

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### Solution Outline:

**Part (a) - Sketch a Vector Diagram**

To visualize the problem:
- The acceleration vector has both \( \hat{i} \) (horizontal) and \( \hat{j} \) (vertical) components.
- \( \hat{i} \) corresponds to the x-direction and \( \hat{j} \) to the y-direction.
- Draw the vector with a negative horizontal component \( -22.5 \, \text{m/s}^2 \) (pointing left) and a positive vertical component \( 20.2 \, \text{m/s}^2 \) (pointing up).

**Part (b) - Magnitude of the Radial Acceleration**

To find the radial (centripetal) acceleration, we need the total acceleration \( \vec{a} \), which has both radial and tangential components. The radial acceleration is directed towards the center of the circle, and we can calculate it by finding the magnitude of \( \vec{a} \).

1. Calculate the magnitude of the total acceleration vector:
   \[
   |\vec{a}| = \sqrt{(-22.5)^2 + (20.2)^2}
   \]
   
2. The radial acceleration is the component of the total acceleration pointing towards the center of the circular path. Since the ball is moving in a circle, the radial component corresponds to the centripetal force.

**Part (c) - Speed and Velocity**

1. The speed of the ball can be related to the radial acceleration by:
   \[
   a_{\text{radial}} = \frac{v^2}{r}
   \]
   where \( r = 1.50 \, \text{m} \) is the radius (length of the rope).
   
2. Solve for the speed \( v \) using the radial acceleration found earlier.

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I will now calculate these values.The magnitude of the total acceleration is approximately \( 30.24 \, \text{m/s}^2 \).

Next, I'll calculate the radial (centripetal) acceleration component and use it to find the ball's speed.The radial (centripetal) acceleration is approximately \( 24.18 \, \text{m/s}^2 \), and the speed of the ball at that instant is approximately \( 6.02 \, \text{m/s} \).

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Would you like further details or have any specific questions?

Here are five questions to deepen your understanding:
1. How would you calculate the tangential acceleration from the given data?
2. Why is the radial acceleration important in circular motion?
3. How does the direction of the velocity vector relate to the ball’s motion in the vertical circle?
4. What role does gravity play in the vertical circular motion of the ball?
5. How would the total energy of the system (kinetic + potential) change as the ball swings?

**Tip**: In problems involving circular motion, always separate radial and tangential components to make solving easier.

A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9° past the lowest point on its way up, its total acceleration is (-22.5 î + 20.2 ĵ) m/s². For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.

Learn how to determine the radial acceleration and speed of a ball swinging counterclockwise in a vertical circle at the end of a rope 1.50 m long. The total acceleration is given by (-22.5 î + 20.2 ĵ) m/s². Suitable for high school students in Grades 11-12, this problem covers key concepts such as vector calculations, radial acceleration, and circular motion.

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