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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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- Spring '11