Problem 16 - Entrance Test

A uniform thin rod of mass M and length L is pivoted at a point L/3 from one end. The rod is initially held horizontal and then released from rest. What is the initial angular acceleration of the rod immediately after it is released?

A. A) α = (3g) / (4L)B. B) α = (g) / LC. C) α = (6g) / (5L)D. D) α = (9g) / (4L)E. E) α = (3g) / (2L)

Correct: E

To find the initial angular acceleration, we need to calculate the net torque acting on the rod about its pivot and its moment of inertia about the same pivot. 1. **Identify the Pivot and Center of Mass (CM):** The rod is uniform, so its center of mass (CM) is at L/2 from either end. The pivot is at L/3 from one end. Let's say it's L/3 from the left end. 2. **Calculate the Distance from Pivot to CM:** Distance d = |CM position - Pivot position| = |L/2 - L/3| d = |3L/6 - 2L/6| = L/6. 3. **Calculate the Moment of Inertia (I_pivot) about the pivot:** First, the moment of inertia of a uniform rod about its CM is I_CM = (1/12)ML². Using the parallel-axis theorem, I_pivot = I_CM + Md². I_pivot = (1/12)ML² + M(L/6)² I_pivot = (1/12)ML² + M(L²/36) I_pivot = (3/36)ML² + (1/36)ML² = (4/36)ML² = (1/9)ML². 4. **Calculate the Net Torque (τ_net) about the pivot:** The only force causing a torque is the weight of the rod (Mg), acting at its center of mass. τ_net = Force × Perpendicular distance from pivot to line of action of force τ_net = Mg * d = Mg * (L/6) 5. **Apply Newton's Second Law for Rotation (τ_net = I_pivot * α):** Mg * (L/6) = (1/9)ML² * α Solve for α: α = (MgL/6) / ((1/9)ML²) α = (MgL/6) * (9 / (ML²)) α = (9g) / (6L) α = (3g) / (2L) The correct choice is E.