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Course: AME 301, Fall 2010
School: USC
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Word Count: 585

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8. 336 CHAPTER RIGID-BODY KINETICS 8.7 Chapter 8, Problem 7 Problem: Gear A of mass m and radius r is constrained to roll on fixed Gear B. It rotates with counterclockwise angular velocity about Axle AD, which has negligible mass and length L. Axle AD is connected by a clevis to vertical Shaft DE, which rotates counterclockwise with constant angular velocity . You can ignore effects of friction. HINT: This...

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8. 336 CHAPTER RIGID-BODY KINETICS 8.7 Chapter 8, Problem 7 Problem: Gear A of mass m and radius r is constrained to roll on fixed Gear B. It rotates with counterclockwise angular velocity about Axle AD, which has negligible mass and length L. Axle AD is connected by a clevis to vertical Shaft DE, which rotates counterclockwise with constant angular velocity . You can ignore effects of friction. HINT: This problem is most conveniently solved in a coordinate system (x y z ) in which z is aligned with Axle AD. Gear A can be represented as a thin disk. In its principal axis system, the inertia tensor is 1 2 0 0 4 mr 1 2 0 [I ] = 0 4 mr 1 2 0 0 2 mr (a) The two gears make contact at Point C. Why is the absolute velocity of Point C zero? (b) Determine = | | as a function of L, r, = ||, and angle . (c) Compute the absolute angular-momentum vector relative to Gear As center of mass, HG , as a function of m, L, r, and . (d) Compute HG , the absolute rate of change of HG , as a function of m, L, r, and . Solution: The first thing we must do is take care of kinematical considerations. For the problem at hand, we will have to transform from the given xyz coordinates to principal axis coordinates x y z . The figure below illustrates the relation between the xyz and x y z coordinate systems (the y and y axes are aligned). z ... . .. . . .. .. .. .. . . .. . .. . . .. . .. . . .. . .... .. .. . . .... .. . ..... . .. . .... .... ... ....... . ... ...... . ..... . ... .. ..................... .... . ...... ...... ............... .. . .. ... ...... ..... ....... . . ....... . ....... ... .. . ....... . ....... . ......... . ...... . ........ . . ....... . ...... . ...... . . ....... ... . ..... . . . ........ ... ....... . . .. . ........ .... . ...... .. .... . . ...... . .. ....... .. .... ............ . ......... .... ....... ....... .. ...... ..... ...... ....... ...... .... ..... . ... z x Obviously, x j = j . The other unit vectors in the two coordinate systems are related as follows. i = cos i + sin k i = cos i sin k and and k = sin i + cos k k = sin i + cos k 8.7. CHAPTER 8, PROBLEM 7 Also, we will need the vectors rA/D and rC/A . Inspection of the geometry shows that rA/D = L k and rC/A = r i 337 (a) Gear B is stationary, so that Point As absolute velocity must be zero. (b) We can determine by computing the velocities of Points A and C. First, note the velocity of Point A is where tot = + vA = vD + tot rA/D Now, Point D is stationary so that vD = 0, and we are given = k Thus, the velocity of Point A is vA = sin i + ( + cos ) k Now, i k = j and k k = 0, wherefore vA = L sin j The velocity of Point C is vC = vA + tot rC/A = L sin j + sin i + ( + cos ) k Evaluating the cross product, we arrive at the following. vC = [L sin r cos r] j But, we know from Part (a) that vC = 0. Therefore, = L sin r cos r r i [L k ] and = k = sin i + cos k (c) As noted in the problem statement, the moment of inertia tensor for the center-of-mass based principal axis system is 1 2 0 0 4 mr 1 2 0 [I ] = 0 4 mr 1 2 0 0 2 mr But, we know from Part (b) that The angular-momentum vector is the product of the moment of inertia tensor and the angular-velocity vector, tot , i.e., 1 2 1 2 0 0 sin 4 mr 4 mr sin 1 0 0 mr2 0 = HG = [I ] tot = 0 4 1 2 1 2 + cos 0 0 2 mr 2 mr ( + cos ) + cos = L L sin r cos + cos = sin r r 338 Hence, in standard vector form, the angular-momentum vector is HG = CHAPTER 8. RIGID-BODY KINETICS 1 mr sin r i + 2L k 4 (d) Since the principle-axis system rotates about the z axis, the absolute rate of change of HG is given by the Coriolis Theorem, i.e., dHG d HG = + HG = 0 + sin i + cos k dt dt Evaluating the cross product yields 1 HG = mr2 sin (r cos 2L sin ) j 4 1 mr sin r i + 2L k 4
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