A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed just before it hits the ground? (2019)

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Q: A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed just before it hits the ground? (2019)

Solution: Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).

Steps: 8

Step 1: Understand the problem. We have a uniform rod of length L that is pivoted at one end and falls due to gravity.
Step 2: Identify the types of energy involved. Initially, the rod has potential energy when it is held upright. As it falls, this potential energy converts into rotational kinetic energy.
Step 3: Write down the formula for potential energy (PE) at the top position. PE = mgh, where h is the height of the center of mass of the rod. For a rod of length L, h = L/2, so PE = mg(L/2).
Step 4: Write down the formula for rotational kinetic energy (KE) just before it hits the ground. KE = (1/2)Iω², where I is the moment of inertia of the rod about the pivot. For a rod pivoted at one end, I = (1/3)mL².
Step 5: Set the potential energy equal to the rotational kinetic energy because of conservation of energy: mg(L/2) = (1/2)(1/3)mL²ω².
Step 6: Simplify the equation. The mass m cancels out, and we can rearrange the equation to solve for ω²: g(L/2) = (1/6)L²ω².
Step 7: Solve for ω²: ω² = (3g)/L.
Step 8: Take the square root to find the angular speed ω: ω = √(3g/L).