A uniform rod of length L is pivoted at one end. If it is allowed to fall freely, what is its angular acceleration just after it is released?

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Q: A uniform rod of length L is pivoted at one end. If it is allowed to fall freely, what is its angular acceleration just after it is released?

Solution: The angular acceleration α = τ/I = (MgL/2)/(1/3 ML^2) = 3g/2L.

Steps: 7

Step 1: Understand that the rod is pivoted at one end and will rotate around that pivot when it falls.
Step 2: Identify the forces acting on the rod. The weight of the rod (Mg) acts at its center of mass, which is located at L/2 from the pivot.
Step 3: Calculate the torque (τ) caused by the weight of the rod. Torque is given by τ = force × distance from pivot. Here, τ = Mg × (L/2).
Step 4: Calculate the moment of inertia (I) of the rod about the pivot. For a uniform rod pivoted at one end, I = (1/3)ML².
Step 5: Use the formula for angular acceleration (α), which is α = τ/I. Substitute the values from Steps 3 and 4 into this formula.
Step 6: Simplify the expression: α = (Mg(L/2)) / ((1/3)ML²). The M cancels out, and you are left with α = (3g)/(2L).
Step 7: Conclude that the angular acceleration just after the rod is released is α = (3g)/(2L).