An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.52 k

Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.52 kg and length L = 5.08 m to a uniform sphere with mass m_s = 32.6 kg and radius R = 1.27 m. Note m_s = 5m_r and L = 4R. What is the moment of inertia of the object about an axis at the left end of the rod? If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 461 N is exerted perpendicular to the rod at the center of the rod? What is the moment of inertia of the object about an axis at the center of mass of the object? If the object is fixed at the center of mass, what is the angular acceleration if a force F = 461 N is exerted parallel to the rod at the end of rod? What is the moment of inertia of the object about an axis at the right edge of the sphere? Compare the three moments of inertia calculated above: I_CM

Explanation / Answer

1)    moment of inertia of object = 1/3 * 6.52 * 5.082 + 2/5 * 32.6 * 1.272 + 32.6 * 6.352

                                                                          =   1391.63 kg.m2

2)   As,    Force * radius = moment of inertia * angular acceleration

=>        angular acceleration = 461 * 2.54/1391.63

                                             = 0.8414 rad/sec2

3)     moment of inertia = 1/3 * 6.52 * 5.082 + 2/5 * 32.6 * 1.272 + 6.52 * 0.6352 +   32.6 * 0.6352

                                                        =   92.89 kg.m2

4)    angular acceleration = 461 * 5.715/92.89

                                         =   28.36 rad/sec2

5)   moment of inertia   =     1/3 * 6.52 * 5.082 + 2/5 * 32.6 * 1.272 + 6.52 * 2.542 +   32.6 * 1.272

                                                        =   171.76 kg.m2

6)     Icm <   Iright <   ILeft

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