moment of inertia of boom about pivot = (1/3) m L^2
= (1/3)(201)(59.05^2)
moment of inertia of man/bucket about pivot = (203 lb *4.45 N/lb)(59.05^2/3.28^2)
add them for total I
Torque of arm about pivot = (201 kg *9.81) (59.05/3.28) (cos 69.1) /2
Torque of man/bucket about pivot = (203*4.45)(59.05/3.28) (cos 69.1)
add them for total torque
a = torque / I
A bucket crane consists of a uniform boom of mass M = 201 kg and length
L = 59.05 ft that pivots at a point on the bed of a fixed truck. The truck supports an elevated bucket with a worker inside at the other end of the boom, as shown in the figure. The bucket and the worker together can be modeled as a point mass of weight 203 lb located at the end point of the boom.
Suppose that when the boom makes an angle of 69.1 degrees with the horizontal truck bed, the bucket crane suddenly loses power, causing the bucket and boom to rotate freely toward the ground. Find the magnitude of the angular acceleration of the system just after the crane loses power. Take the rotation axis to be at the point where the boom pivots on the truck bed. Use g = 9.81 m/s2 for the acceleration due to gravity. For unit conversions, assume that 1 m = 3.28 ft and 1 lb = 4.45 N. Express your answer to at least two decimal places.
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