Unformatted text preview: Problem Set 8: Problem 4.
Problem: In the position shown in the figure, bar AB has a clockwise angular velocity . For the indicated lengths, determine the angular velocities of bars BD and DE. Solution: Focusing first on bar BD, we know that its velocity is vD = vB + BD rD/B where vB is the velocity of Point B and BD is the angular velocity of bar BD about Point B. From the geometry, clearly rD/B = 3L i. Also, velocity vB is the sum of the translation velocity of Point A and a rotation about A. However, Point A is fixed so that its translation velocity is zero. Thus, since rod AB rotates clockwise with angular velocity and rB/A = -4L j, vB = (-k) rB/A = (-k) (-4L j) = -4L i Therefore, letting BD = BD k, the velocity of Point D is vD = -4L i + (BD k) (3L i) = -4L i + 3BD L j We can develop a second equation for vD relative to Point E, viz., vD = vE + DE rD/E Now, Point E is not translating so that vE = 0. Letting DE = DE k and noting that rD/E = -L i - 2L j, there follows vD = (DE k) (-L i - 2L j) = 2DE L i - DE L j -4L = 2DE L The solution for BD and DE is BD = 2 3 and DE = -2 and 3BD L = -DE L Hence, we have the following two equations (one for each velocity component). Thus, bar BD rotates counterclockwise with angular velocity 2 and bar DE rotates clockwise with angular 3 velocity 2. ...
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- Spring '06
- Angular Momentum, Relative velocity, 3L, 4L, 2L