Physics Problem: Angular Speed of a Falling Beam
Question:
A 2.0 m-long horizontal beam is attached to a vertical post 0.5 m from its left end. The right end of the beam is tied to the post with a cable angled 25° above the horizontal. The beam is uniform and has a mass of 60 kg.
Part (c): Suppose the cable breaks and the beam swings freely downward until it becomes vertical. During the swing, the beam’s center of mass moves downward 0.5 m. If mechanical energy is conserved, what is the resulting angular speed of the beam when it reaches vertical? The beam’s moment of inertia about the pivot is 35 kg·m².
Answer:
Step 1: Apply Conservation of Mechanical Energy
Since the beam swings without friction or external forces (other than gravity), mechanical energy is conserved.
Initial Potential Energy = Final Rotational Kinetic Energy
Step 2: Rearranging the Equation for Angular Speed (ω)
Step 3: Substituting Known Values
- m = 60 kg (mass of the beam)
- g = 9.8 m/s² (acceleration due to gravity)
- h = 0.5 m (vertical drop of center of mass)
- I = 35 kg·m² (moment of inertia about pivot)
✅ Final Answer:
The angular speed of the beam when it reaches the vertical position is: ω ≈ 4.10 rad/s