HW13Sol(2)

# HW13Sol(2) - 2 . The total moment of inertia is then I = I...

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Jim Guinn’s PHYS2211 Assignment #13 Solutions Due, Wednesday, Mar. 24, 2010 1. Two spheres, each with a mass of 1kg and radius of 3cm are attached to the end of a 1m long thin rod so that the end of the rod is attached to the surface of the spheres. The rod has a mass of 2kg. What is the moment of inertia of this arrangement about an axis through the center of, and perpendicular to, the rod? The moment of inertia of the system is the sum of the moments of inertia of each part. The moment of inertia of rod is I rod = (1/12) M L 2 = (1/12) (2kg) (1m) 2 = 0.167kg m 2 . The moment of inertia of one sphere is given by the parallel-axis theorem, so I sphere = M d 2 + I cm = (1kg) (0.53m) 2 + (2/5) (1kg) (0.03m) 2 = 0.281kg m
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Unformatted text preview: 2 . The total moment of inertia is then I = I rod + I sphere + I sphere = 0.167kg m 2 + 0.281kg m 2 + 0.281kg m 2 = I = 0.729kg m 2 . 2. Bijin holds the end of a 1.5m long stick with a 2kg mass hanging off the end of it as in the diagram. The stick makes an angle of 20 o with the horizontal. What is the torque exerted on the stick, about Bijins hands, by the hanging mass? With = r F sin we have F = mg = (2kg) (9.80m/s 2 ) = 19.6N , so = (1.5m) (19.6N) sin(110 o ) = = 27.6N m . 3. What would the torque be if Bijin lowered the mass so that the stick was horizontal? Now = (1.5m) (19.6N) sin(90 o ) = = 29.4N m . 0.500m 0.530m 2kg 20 o 70 o 110 o...
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