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### Given Data:
- Mass of the rod \( m = 1.7 \, \text{kg} \)
- Length of the rod \( \ell = 0.98 \, \text{m} \)
- Force \( F = 4.2 \, \text{N} \)
- Angle of force application \( \theta = 74^\circ \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
  
### Step 1: **Moment of Inertia about the Pivot Point**
The moment of inertia \( I \) for a rod pivoted at one end is different from the center of mass. The formula for the moment of inertia for a rod pivoted about one end is:

\[
I_{\text{pivot}} = \frac{1}{3} m \ell^2
\]

Substitute the values:
\[
I_{\text{pivot}} = \frac{1}{3} \times 1.7 \, \text{kg} \times (0.98 \, \text{m})^2 = 0.5447 \, \text{kg} \cdot \text{m}^2
\]

### Step 2: **Torque Produced by the Force**
The torque \( \tau \) generated by a force applied at an angle is given by:

\[
\tau = F \ell \sin(\theta)
\]

Substitute the values:
\[
\tau = 4.2 \, \text{N} \times 0.98 \, \text{m} \times \sin(74^\circ)
\]
First, calculate \( \sin(74^\circ) \):
\[
\sin(74^\circ) \approx 0.961
\]

Now calculate the torque:
\[
\tau = 4.2 \times 0.98 \times 0.961 = 3.959 \, \text{N} \cdot \text{m}
\]

### Step 3: **Angular Acceleration**
The angular acceleration \( \alpha \) is related to the torque by the equation:

\[
\tau = I \alpha
\]

Solve for \( \alpha \):
\[
\alpha = \frac{\tau}{I}
\]

Substitute the values:
\[
\alpha = \frac{3.959}{0.5447} \approx 7.27 \, \text{rad/s}^2
\]

### Final Answer:
The angular acceleration is approximately **7.27 rad/s²**.

---

Would you like further details or have any questions? Here are five related questions that could expand on this topic:

1. How would the angular acceleration change if the length of the rod increased?
2. What would happen if the force was applied at a different angle?
3. How can we calculate the angular velocity after a certain time, given this acceleration?
4. What would the moment of inertia be if the rod was pivoted at the center instead of one end?
5. How does the direction of the force affect the resulting motion?

**Tip:** Always check whether the force is applied perpendicular to the object or at an angle, as it affects the torque calculation significantly.

A uniform horizontal rod of mass 1.7 kg and length 0.98 m is free to pivot about one end. A 4.2 N force at an angle of 74° acts on the rod. What is the angular acceleration about the pivot point? The acceleration of gravity is 9.8 m/s².

Learn how to calculate the angular acceleration of a rod pivoted at one end, when a force is applied at an angle. This problem involves calculating torque, moment of inertia, and angular acceleration. The rod has a mass of 1.7 kg, a length of 0.98 m, and the applied force is 4.2 N at 74°. Suitable for physics students.

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