🤓 Based on our data, we think this question is relevant for Professor Florin's class at TEXAS.

A uniform rod of mass m and length l is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance l from the center of mass of the rod. The rod is released from rest at an angle of θ with the horizontal, as shown in the figure.

What is the magnitude of the Horizontal force F_{x} exerted on the pivot end of the rod extension at the instant the rod is in a horizontal position? The acceleration due to gravity is g and the moment of inertia of the rod about its center of mass is 1/12 mℓ^{2}.

1. F_{x} = 1/13 mg sin(θ)

2. F_{x} = 24/13 mg cos(θ)

3. F_{x} = 24/13 mg sin(θ)

4. F_{x} = 13/12 mg cos(θ)

5. F_{x} = 12/13 mg cos(θ)

6. F_{x} = 12/13 mg sin(θ)

7. F_{x} = 13/12 mg sin(θ)

8. F_{x} = mg cos(θ)

9. F_{x} = 1/13 mg cos(θ)

10. F_{x} = mg sin(θ)

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Our tutors have indicated that to solve this problem you will need to apply the Conservation of Energy with Rotation concept. You can view video lessons to learn Conservation of Energy with Rotation. Or if you need more Conservation of Energy with Rotation practice, you can also practice Conservation of Energy with Rotation practice problems.

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Based on our data, we think this problem is relevant for Professor Florin's class at TEXAS.