A uniform horizontal rod of mass 2.8 kg and
length 1.3 m is free to pivot about one end
as shown. The moment of inertia of the rod
about an axis perpendicular to the rod and
through the center of mass is given by I =
m ℓ2/12. If a 5.1 N force at an angle of 106◦
to the
horizontal acts on the rod as shown, what is
the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9.8 m/s^2. Answer in units of rad/s.

To find the angular acceleration of the rod about the pivot point, we can use the following steps:

### Step 1: Moment of inertia about the pivot
The given moment of inertia is about the center of mass. However, we need the moment of inertia about the pivot point. To do this, we'll use the **parallel axis theorem**:

\[
I_{\text{pivot}} = I_{\text{cm}} + m d^2
\]

where:
- \( I_{\text{cm}} = \frac{m \ell^2}{12} \) is the moment of inertia about the center of mass,
- \( m = 2.8 \, \text{kg} \) is the mass of the rod,
- \( \ell = 1.3 \, \text{m} \) is the length of the rod,
- \( d = \frac{\ell}{2} = 0.65 \, \text{m} \) is the distance from the center of mass to the pivot point.

First, calculate the moment of inertia about the center of mass:
\[
I_{\text{cm}} = \frac{2.8 \times (1.3)^2}{12} = 0.395 \, \text{kg} \cdot \text{m}^2
\]

Now, using the parallel axis theorem:
\[
I_{\text{pivot}} = 0.395 + 2.8 \times (0.65)^2
\]
\[
I_{\text{pivot}} = 0.395 + 2.8 \times 0.4225 = 0.395 + 1.183 = 1.578 \, \text{kg} \cdot \text{m}^2
\]

### Step 2: Torque exerted by the force
The torque \( \tau \) due to a force \( F \) applied at a distance \( r \) from the pivot point is given by:
\[
\tau = r F \sin(\theta)
\]
where:
- \( F = 5.1 \, \text{N} \),
- \( r = 1.3 \, \text{m} \) (the length of the rod),
- \( \theta = 106^\circ \) is the angle between the force and the horizontal.

Now, calculate the torque:
\[
\tau = 1.3 \times 5.1 \times \sin(106^\circ)
\]
Using \( \sin(106^\circ) \approx 0.961 \):
\[
\tau = 1.3 \times 5.1 \times 0.961 = 6.381 \, \text{N} \cdot \text{m}
\]

### Step 3: Angular acceleration
Using Newton's second law for rotation, the angular acceleration \( \alpha \) is related to the torque by the equation:
\[
\tau = I_{\text{pivot}} \alpha
\]

Solving for \( \alpha \):
\[
\alpha = \frac{\tau}{I_{\text{pivot}}} = \frac{6.381}{1.578} \approx 4.044 \, \text{rad/s}^2
\]

### Final Answer:
The angular acceleration of the rod is approximately \( \boxed{4.04 \, \text{rad/s}^2} \).

---

Would you like more details or have any questions?

Here are some related questions:
1. How does the parallel axis theorem apply to other shapes, like disks or spheres?
2. What would happen if the force were applied at a different angle?
3. How is torque related to angular momentum?
4. What is the role of angular acceleration in rotational kinetic energy?
5. How does the distribution of mass affect the moment of inertia?

**Tip:** When calculating torques, always check the perpendicular distance and angle for accuracy!

A uniform horizontal rod of mass 2.8 kg and length 1.3 m is free to pivot about one end. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = m ℓ²/12. If a 5.1 N force at an angle of 106° to the horizontal acts on the rod, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9.8 m/s². Answer in units of rad/s².

Solve a physics problem involving the angular acceleration of a 2.8 kg rod pivoted at one end and subjected to a 5.1 N force at an angle of 106°. This problem explores key rotational dynamics concepts, including torque, moment of inertia, and the parallel axis theorem, and offers a detailed step-by-step solution suitable for advanced high school or undergraduate students.

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