A uniform rod of length L and mass M is pivoted at one end and allowed to fall under gravity. What is the angular acceleration of the rod just after it is released? (2019)
Practice Questions
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Q1
A uniform rod of length L and mass M is pivoted at one end and allowed to fall under gravity. What is the angular acceleration of the rod just after it is released? (2019)
g/L
2g/L
3g/L
g/2L
The torque τ = Mg(L/2) and moment of inertia I = (1/3)ML². Using τ = Iα, we find α = 3g/2L.
Questions & Step-by-step Solutions
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Q
Q: A uniform rod of length L and mass M is pivoted at one end and allowed to fall under gravity. What is the angular acceleration of the rod just after it is released? (2019)
Solution: The torque τ = Mg(L/2) and moment of inertia I = (1/3)ML². Using τ = Iα, we find α = 3g/2L.
Steps: 10
Step 1: Identify the pivot point of the rod. The rod is pivoted at one end.
Step 2: Understand that the rod is uniform, meaning its mass is evenly distributed along its length.
Step 3: Determine the force acting on the rod due to gravity. The weight of the rod is Mg, where M is the mass and g is the acceleration due to gravity.
Step 4: Find the location of the center of mass of the rod. For a uniform rod, the center of mass is located at L/2 from the pivot.
Step 5: 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 6: Calculate the moment of inertia (I) of the rod about the pivot point. For a uniform rod pivoted at one end, I = (1/3)ML².
Step 7: Use the relationship between torque, moment of inertia, and angular acceleration: τ = Iα, where α is the angular acceleration.
Step 8: Substitute the values of τ and I into the equation: Mg(L/2) = (1/3)ML² × α.
Step 9: Solve for α by rearranging the equation: α = (Mg(L/2)) / ((1/3)ML²).
Step 10: Simplify the equation to find α = (3g)/(2L).
