Ch 12: Torque & Rotational DynamicsSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (Oscillations)
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Ch 17: Fluid Mechanics
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Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
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Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
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Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
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Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
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Ch 38: Quantum Mechanics

Torque & Acceleration (Rotational Dynamics)

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Solution: 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 Fx 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 mℓ2.  1. Fx = 1/13 mg sin(θ) 2. Fx = 24/13 mg cos(θ) 3. Fx = 24/13 mg sin(θ) 4. Fx = 13/12 mg cos(θ) 5. Fx = 12/13 mg cos(θ) 6. Fx = 12/13 mg sin(θ) 7. Fx = 13/12 mg sin(θ) 8. Fx = mg cos(θ) 9. Fx = 1/13 mg cos(θ) 10. Fx = mg sin(θ)

Problem

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 Fx 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 mℓ2

1. Fx = 1/13 mg sin(θ)

2. Fx = 24/13 mg cos(θ)

3. Fx = 24/13 mg sin(θ)

4. Fx = 13/12 mg cos(θ)

5. Fx = 12/13 mg cos(θ)

6. Fx = 12/13 mg sin(θ)

7. Fx = 13/12 mg sin(θ)

8. Fx = mg cos(θ)

9. Fx = 1/13 mg cos(θ)

10. Fx = mg sin(θ)