A telephone pole has been knocked over by the wind so that it makes an angle of  theta= 15 degrees with the vertical. The wind has stopped blowing and the pole is to be cut down. Once the cut is across most of the thickness of the pole, the pole begins to tip over. As the pole tips, the bottom of the pole stays attached to the base by the remaining part of the pole that was not cut (but the torque from the base on the tipping pole is negligible). There is a nail (of negligible mass) on the pole located a distance x = 47 m from the cut. The length of the pole from the cut to the top is L=69m. Remember that the moment of inertia of a stick of mass m and length L about its end is (13)mL2. If  theta= 67.5 degrees, what is the magnitude of the angular acceleration of the nail?

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A telephone pole has been knocked over by the wind so that it makes an angle of  theta= 15 degrees with the vertical. The wind has stopped blowing and the pole is to be cut down. Once the cut is across most of the thickness of the pole, the pole begins to tip over. As the pole tips, the bottom of the pole stays attached to the base by the remaining part of the pole that was not cut (but the torque from the base on the tipping pole is negligible). There is a nail (of negligible mass) on the pole located a distance x = 47 m from the cut. The length of the pole from the cut to the top is L=69m. Remember that the moment of inertia of a stick of mass m and length L about its end is (13)mL2. If  theta= 67.5 degrees, what is the magnitude of the angular acceleration of the nail?

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