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A uniform rod of length 1.3 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 27o above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.
HINT: A convenient reference point to choose for the torque and moment of inertia is the actual pivot point. Note that your book gives the moment of inertia in that case as 1/3 m L^2 (p. 247)

To find the initial acceleration of the center of mass (CM) of the rod, let’s go through the necessary steps with the given dataTo find the initial acceleration of the center of mass (CM) of the rod, let’s go through the necessary steps with the given 

1. **Given Data and Variables**:
   - Length of the rod, \( L = 1.3 \, \text{m} \)
   - Angle \( \theta = 27^\circ \) above the horizontal
   - Moment of inertia about the pivot, \( I = \frac{1}{3} m L^2 \)
   - Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

2. **Torque Calculation**:
   The torque (\( \tau \)) about the pivot point is generated by the weight of the rod acting at its center of mass (CM), which is located at a distance \( \frac{L}{2} \) from the pivot.

   \[
   \tau = \text{Force} \times \text{Distance from pivot} = (mg) \cdot \frac{L}{2} \cdot \sin \theta
   \]

3. **Angular Acceleration (\( \alpha \))**:
   The angular acceleration can be found using Newton's second law for rotation:
   
   \[
   \tau = I \alpha
   \]

   Substituting \( \tau \) and \( I \) from the above equations:

   \[
   mg \cdot \frac{L}{2} \cdot \sin \theta = \frac{1}{3} m L^2 \alpha
   \]

   Simplifying by canceling \( m \) and \( L \) on both sides:

   \[
   g \cdot \frac{1}{2} \cdot \sin \theta = \frac{1}{3} L \alpha
   \]

   Solving for \( \alpha \):

   \[
   \alpha = \frac{3g \sin \theta}{2L}
   \]

4. **Linear Acceleration of the CM**:
   The linear acceleration \( a_{\text{CM}} \) of the center of mass can be related to the angular acceleration \( \alpha \) as follows:

   \[
   a_{\text{CM}} = \alpha \cdot \frac{L}{2}
   \]

5. **Substitute Values**:
   Now, substitute \( g = 9.8 \, \text{m/s}^2 \), \( L = 1.3 \, \text{m} \), and \( \theta = 27^\circ \):

   \[
   \alpha = \frac{3 \cdot 9.8 \cdot \sin 27^\circ}{2 \cdot 1.3}
   \]

   First, calculate \( \sin 27^\circ \approx 0.454 \):

   \[
   \alpha = \frac{3 \cdot 9.8 \cdot 0.454}{2 \cdot 1.3} \approx \frac{13.3596}{2.6} \approx 5.14 \, \text{rad/s}^2
   \]

   Now, calculate \( a_{\text{CM}} \):

   \[
   a_{\text{CM}} = 5.14 \cdot \frac{1.3}{2} \approx 5.14 \cdot 0.65 \approx 3.34 \, \text{m/s}^2
   \]

### Final Answer:
The magnitude of the initial acceleration of the center of mass of the rod is approximately \( 3.34 \, \text{m/s}^2 \).

Would you like further details on any step or have additional questions?

Here are some questions that expand on the topic:
1. How would the initial acceleration change if the rod were released from a different angle?
2. What would be the effect on the acceleration if the rod’s length were doubled?
3. How would friction at the pivot alter the initial conditions and calculations?
4. Can you calculate the initial angular velocity of the rod if it were given an initial push?
5. How does the moment of inertia affect the overall acceleration?

**Tip:** Remember that angular and linear quantities are related in rotational dynamics, with the linear acceleration at a point a distance \( r \) from the pivot given by \( a = r \alpha \).

A uniform rod of length 1.3 m is attached to a frictionless pivot at one end. It is released from rest from an angle θ = 27° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.

Calculate the initial acceleration of a rod's center of mass, released from an angle above the horizontal. This problem involves rotational dynamics concepts such as torque, moment of inertia, and angular acceleration, ideal for undergraduate physics students.

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