This problem describes rigid body rotation. A uniform rod of mass m and length 2a is...

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This problem describes rigid body rotation. A uniform rod of mass m and length 2a is held so that one-third of its length rests on a horizontal table at right angles to an edge, the other two- thirds projecting beyond the edge. It is then released from rest. The rod will rotate about an edge at O through an angle before it begins to slip as shown. The coefficient of static friction between the rod and the edge is u. (a) (b) (c) (d) mg Implement the parallel axis theorem, find the moment of inertia of the rod of length 2a about an axis perpendicular to the rod and through a point O, one-third from one end. [Given IG = 1/2 m/², where / is the length of the rod] Implement the conservation of mechanical energy to show that while the rod is rotating without slipping about the edge of the table, aġ² =g sin where g is the magnitude of the acceleration due to gravity. (6 marks) The reaction of the edge on the rod has two components, namely the frictional component of magnitude S preventing slipping, and the normal component of magnitude N acting perpendicular to the rod. By implementing the general equation of circular motion of the rod's centre of mass G, show that S = mg sino + N = mg cose while the rod is rotating without slipping. and (4 marks) maġ² ma (5 marks) By using the equation derived in part (b) express S and N in terms of m, a, g and 0. (5 marks) This problem describes rigid body rotation. A uniform rod of mass m and length 2a is held so that one-third of its length rests on a horizontal table at right angles to an edge, the other two- thirds projecting beyond the edge. It is then released from rest. The rod will rotate about an edge at O through an angle before it begins to slip as shown. The coefficient of static friction between the rod and the edge is u. (a) (b) (c) (d) mg Implement the parallel axis theorem, find the moment of inertia of the rod of length 2a about an axis perpendicular to the rod and through a point O, one-third from one end. [Given IG = 1/2 m/², where / is the length of the rod] Implement the conservation of mechanical energy to show that while the rod is rotating without slipping about the edge of the table, aġ² =g sin where g is the magnitude of the acceleration due to gravity. (6 marks) The reaction of the edge on the rod has two components, namely the frictional component of magnitude S preventing slipping, and the normal component of magnitude N acting perpendicular to the rod. By implementing the general equation of circular motion of the rod's centre of mass G, show that S = mg sino + N = mg cose while the rod is rotating without slipping. and (4 marks) maġ² ma (5 marks) By using the equation derived in part (b) express S and N in terms of m, a, g and 0. (5 marks)