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# Test 3 Spring 2005-with answers - α AB = 2 rad/sec(CCW as...

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AEM 264-002 Test#3 4/14/05 Instructions: One 8½ inch by 11 inch sheet of handwritten formulas allowed. Otherwise, this is a closed book, closed notes test. 75 minute time limit – strictly enforced. Show all of your work . 1. [30pts] A machine component with a uniform thickness of 1.0 inch is shown below. The component is made of steel with a density of 0.283 lbm/in 3 . Determine the mass moment of inertia I G about the Z axis at the center of the hole . Express I G in two different sets of correct units. 5.0 in 12.0 in 2.5 in 9.0 in 3.5 in dia X Y 2. [35pts] A pulley with a mass of 12 kg is connected to two blocks as shown below. The pulley has a mass moment of inertia about the center of mass of I G =0.09 kg-m 2 . Assume no friction at the axle. Determine the angular acceleration of the pulley and the linear acceleration of each block. 10 cm 6 cm 5 kg 10 kg

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3. [35pts] At this instant for link AB the angular velocity is ω AB = -2 rad/sec (CW) and the angular acceleration is
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Unformatted text preview: α AB = +2 rad/sec (CCW) as shown in the figure above. • Determine numerical values for the angular velocities ω BC and ω CD • Determine numerical values for the linear velocities v B and v C • Write the equations necessary to find angular acceleration α BC and α CD , substitute in all known values, but do not solve the equations . AEM 264-002 Test#3 - Answers Spring 2005 1. 2 2 2 , 00315 . 0427 . 7 . 42 s ft lb m kg cm kg I hole ZZ ⋅ ⋅ = ⋅ = ⋅ = 2. ( ) 2 sec 557 . m j a A r r + = , ( ) 2 sec 334 . m j a B r r − = , ( ) ) ( sec 52 . 5 2 CCW rad k r + = α 3. ( ) ) ( sec 3 CCW rad k BC r + = ω , ( ) ) ( sec 5 . 1 CCW rad k CD r + = , ( ) sec 12 in j i v B r r r − = , ( ) sec 9 in j i v C r r r + − = j s in i i j j s in i s in a CD BC BC C r r r r r r r 2 2 2 5 . 13 6 3 4 15 60 − − = − + − − = BC CD BC j s in j s in i 4 5 . 1 : 6 3 60 : 2 2 = − = − r r r...
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Test 3 Spring 2005-with answers - α AB = 2 rad/sec(CCW as...

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