Part One A 3.00m long rod is hinged at one ed. The rod is initially held in the

Question

Part One
A 3.00m long rod is hinged at one ed. The rod is initially held in the horizontal position and then released as the free end is allowed to fall. What is the angular acceleration as it is released? (The moment of inertia of a rod about one end is 1/3ML^2

Part Two
A solid cylinder of mass 10 kg is pivoted about a frctionless axis throughout the center O. A rope wrapped around the outer radius R1=1.0m exerts a force F1=5.0N to the right. A second rope wrapped around another section of radius R2=0.50m exerts a force F2=6.0 N downward.What is the angular acceleration of the disc?

Explanation / Answer

Given Data 1. Length of the rod, L = 3.0 m     Moment of inertia of a rod about one end is , I = 1/3ML2
   Solution: Torque produced is,                          = I                        r F = I               Mg (L/2) = (1/3ML2)                            = (3/2) (9.8 m/s2 / 3.0 m)                               = 4.9 rad /s2   is the angular acceleration ---------------------------------------------------------------------------- 2.Given Data Mass of the solid cylinder, M = 10 kg Outer radius, R1 = 1.0 m Force acting to the right is, F1 = 5.0 N Radius of second section, R2 = 0.5 m Force acting downwards,  F2 = 6.0 N Moment of inertia of the solid cylinder is, I = 1/2 M Solution: Torque produced is,                                                 = I                               R1 F1 - R2F2 = 1/2 M ( R12 +  R22) * (1.0 m) (5.0 N) - (0.5 m) (6.0 N) = 1/2 (10 kg) [(1.0 m)2 + (0.5 m)2] *                                                   = 0.32 rad /s2              
  

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