Athena is tracking the motion of a proton undergoing uniform circular motion in a cyclotron. At time 𝑡1 = 3.25 ns she reported the electron to be at ⟨50.0 cm, 50.0 cm⟩ with velocity −(4.29 × 10^7 m/s)î and acceleration directed along the +𝑦-axis. Then at time 𝑡2 = 58.2 ns it had velocity −(4.29 × 10^7 m/s)î and acceleration along the −𝑥-axis. The time 𝑡2 − 𝑡1 is less than one period.

Evaluate this first then answer the following:
A. What is the period of revolution of the electron?
B. Where is the center of the circular path located?
C. What is the centripetal acceleration of the electron?

To solve this problem, let's break down the information provided:

1. **Uniform Circular Motion**: The proton is undergoing uniform circular motion, meaning it travels in a circular path at constant speed, and the acceleration is always directed toward the center of the circle (centripetal acceleration).

2. **Given Data**:
   - Time \( t_1 = 3.25 \) ns (nanoseconds)
   - Position at \( t_1 = \langle 50.0 \, \text{cm}, 50.0 \, \text{cm} \rangle \)
   - Velocity at \( t_1 = -(4.29 \times 10^7) \, \text{m/s} \, \hat{i} \)
   - Acceleration at \( t_1 \) is along the \( +y \)-axis.
   - Time \( t_2 = 58.2 \) ns
   - Velocity at \( t_2 = -(4.29 \times 10^7) \, \text{m/s} \, \hat{i} \)
   - Acceleration at \( t_2 \) is along the \( -x \)-axis.
   - The time difference \( \Delta t = t_2 - t_1 = 58.2 \, \text{ns} - 3.25 \, \text{ns} = 54.95 \, \text{ns} \), which is less than one period of revolution.

Let's proceed step by step to answer each part:

### A. Period of Revolution of the Electron

1. **Calculate the Fraction of the Circle Traveled**:

   Since the velocity of the electron at \( t_1 \) and \( t_2 \) is in the \( -\hat{i} \) direction but the acceleration changes from \( +y \) to \( -x \), this suggests that the electron travels a quarter (1/4) of a full circle between \( t_1 \) and \( t_2 \).

2. **Determine the Period of the Circular Motion**:

   The time to complete a quarter-circle is \( \Delta t = 54.95 \, \text{ns} \).

   Therefore, the full period \( T \) (time to complete a full circle) is:
   \[
   T = 4 \times \Delta t = 4 \times 54.95 \, \text{ns} = 219.8 \, \text{ns}
   \]

### B. Center of the Circular Path

3. **Finding the Center**:

   At \( t_1 \), the position of the electron is \( (50.0, 50.0) \) cm. The acceleration is directed along the \( +y \)-axis, which means the center of the circular path must be to the left (in the negative \( x \)-direction) of this point because the centripetal acceleration always points toward the center.

   At \( t_2 \), the acceleration is along the \( -x \)-axis, meaning the center is below this point (negative \( y \)-direction).

   Since the path forms a circle and we know the acceleration directions at two points, the center of the circular path must be at the same distance from both positions but offset by the radius.

   Given that at time \( t_1 \), the acceleration is along the \( +y \)-axis, the center must lie on the line where \( x = 50.0 \, \text{cm} \). Similarly, since at time \( t_2 \) the acceleration is along the \( -x \)-axis, the center must lie on the line where \( y = 50.0 \, \text{cm} \). Therefore, the center of the circle is at:
   \[
   \text{Center} = (0, 0) \, \text{cm}
   \]

### C. Centripetal Acceleration of the

Athena is tracking the motion of a proton undergoing uniform circular motion in a cyclotron. At time t1 = 3.25 ns she reported the electron to be at ⟨50.0 cm, 50.0 cm⟩ with velocity −(4.29 × 10^7 m/s)î and acceleration directed along the +𝑦-axis. Then at time t2 = 58.2 ns it had velocity −(4.29 × 10^7 m/s)î and acceleration along the −𝑥-axis. The time t2 − t1 is less than one period. Evaluate this first then answer the following: A. What is the period of revolution of the electron? B. Where is the center of the circular path located? C. What is the centripetal acceleration of the electron?

Explore how to determine the period, center of the circular path, and centripetal acceleration of an electron undergoing uniform circular motion in a cyclotron. Learn step-by-step calculations involving kinematic concepts and centripetal force theorems. Suitable for undergraduate physics students.

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