oblem 
1
2
.
7
6
 ﻿The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let 
L
=
5
.
3
f
t
 ﻿and assume that the pendulum is placed in motion at 
θ
=
0
°
.
 ﻿What is the minimum angular velocity at this position for the pendulum to swing through a full circle?

To solve the problem of finding the minimum angular velocity for a pendulum to swing through a full circle, we need to analyze the pendulum's motion under the influence of gravity.

### Step-by-step solution:

1. **Modeling the Pendulum System:**
   - The pendulum consists of a point mass \( m \) at the end of a rigid, massless bar of length \( L = 5.3 \, \text{ft} \).
   - The pendulum is released from \( \theta = 0^\circ \) (the lowest point) and we want to find the minimum angular velocity \( \omega_{\text{min}} \) at this position for the pendulum to complete a full circle.

2. **Energy Consideration:**
   To determine the minimum angular velocity, we can use **conservation of mechanical energy**. For the pendulum to swing through a full circle, the tension in the string or rod must remain positive, and the total energy at the bottom must be enough to reach the top of the swing.

   - At the bottom (when \( \theta = 0^\circ \)), the pendulum has kinetic energy and no potential energy (relative to the lowest point).
   - At the top of the swing (when \( \theta = 180^\circ \)), the pendulum has potential energy and zero kinetic energy (at the minimum velocity).

3. **Mechanical Energy Conservation:**
   The total mechanical energy at the lowest point is all kinetic energy, and at the highest point (top of the swing), it is all potential energy.

   - At \( \theta = 0^\circ \) (the lowest point):
     \[
     E_{\text{bottom}} = \frac{1}{2} m v^2
     \]
     where \( v = \omega L \) is the linear velocity.

   - At \( \theta = 180^\circ \) (the highest point):
     \[
     E_{\text{top}} = m g (2L)
     \]
     The potential energy is \( mgh \), where \( h = 2L \) is the height difference from the bottom to the top.

4. **Setting the Energies Equal:**
   Since the total mechanical energy is conserved:
   \[
   \frac{1}{2} m v^2 = m g (2L)
   \]
   Simplifying, we cancel out \( m \):
   \[
   \frac{1}{2} v^2 = g (2L)
   \]
   Substituting \( v = \omega L \), we get:
   \[
   \frac{1}{2} (\omega L)^2 = g (2L)
   \]
   Simplifying for \( \omega \):
   \[
   \frac{1}{2} \omega^2 L^2 = 2gL
   \]
   \[
   \omega^2 = \frac{4g}{L}
   \]
   Taking the square root:
   \[
   \omega = \sqrt{\frac{4g}{L}}
   \]

5. **Substitute Values:**
   - Gravitational acceleration \( g = 32.2 \, \text{ft/s}^2 \).
   - Length of the pendulum \( L = 5.3 \, \text{ft} \).

   Substituting into the formula:
   \[
   \omega = \sqrt{\frac{4 \times 32.2}{5.3}}
   \]
   Let's calculate the value of \( \omega \).The minimum angular velocity \( \omega_{\text{min}} \) for the pendulum to swing through a full circle is approximately \( 4.93 \, \text{rad/s} \).

Would you like more details on the calculation or any related concepts?

Here are five related questions to explore further:
1. How would the result change if the length of the pendulum were increased?
2. What is the tension in the pendulum rod at the lowest point of the swing?
3. How is the minimum angular velocity related to the energy of the system?
4. Can the pendulum complete a full circle with a shorter length and higher velocity?
5. How does replacing the cord with a rigid bar affect the pendulum's dynamics?

**Tip:** Remember, angular velocity is different from linear velocity, but they are connected through the radius (or length) of the pendulum.

The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let L = 5.3 ft and assume that the pendulum is placed in motion at θ = 0°. What is the minimum angular velocity at this position for the pendulum to swing through a full circle?

Find out how to calculate the minimum angular velocity required for a pendulum with a length of 5.3 ft to complete a full circle. This problem explores key physics concepts, including conservation of mechanical energy and circular motion, suitable for Grades 11-12.

Math Problem Statement

oblem 1 2 . 7 6 The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let L

5 . 3 f t and assume that the pendulum is placed in motion at θ

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