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Relative Motion & Constrained Motion Lecture V Constrained MotionRelative Motion.

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Presentation on theme: "Relative Motion & Constrained Motion Lecture V Constrained MotionRelative Motion."— Presentation transcript:

1 Relative Motion & Constrained Motion Lecture V Constrained MotionRelative Motion

2 In previous lectures, the particles motion have been described using coordinates referred to fixed reference axes. This kind of motion analysis is called absolute motion analysis. Not always easy to describe or measure motion by using fixed set of axes. The motion analysis of many engineering problems is sometime simplified by using measurements made with respect to moving reference system. Combining these measurements with the absolute motion of the moving coordinate system, enable us to determine the absolute motion required. This approach is called relative motion analysis. vBvB vAvA v A/B

3 Relative Motion (Cont.) -The motion of the moving coordinate system is specified w.r.t. a fixed coordinate system. -The moving coordinate system should be nonrotating (translating or parallel to the fixed system). -A/B is read as the motion of A relative to B (or w.r.t. B). -The relative motion terms can be expressed in whatever coordinate system (rectangular, polar, n-t). Fixed system Moving system Fixed system Note: r B/A = -r A/B v B/A = -v A/B a B/A = -a A/B Path Note: r A & r B are measured from the origin of the fixed axes X-Y. Note: In relative motion analysis, acceleration of a particle observed in a translating system x-y is the same as observed in a fixed system X-Y, when the moving system has a constant velocity.

4 Relative Motion Exercises

5 Exercise # 1 2/186: The passenger aircraft B is flying east with a velocity v B = 800 km/h. A military jet traveling south with a velocity v A = 1200 km/h passes under B at a slightly lower altitude. What velocity does A appear to have to a passenger in B, and what is the direction of that apparent velocity?.

6 Exercise # 2 At the instant shown, cars A and B are traveling with speeds of 18 m/s and 12 m/s, respectively. Also at this instant, car A has a decrease in speed of 2 m/s 2, and B has an increase in speed of 3 m/s 2. Determine the velocity and acceleration of car B with respect to car A.

7 Exercise # 3 2/191: The car A has a forward speed of 18 km/h and is accelerating at 3 m/s 2. Determine the velocity and acceleration of the car relative to observer B, who rides in a nonrotating chair on the Ferris wheel. The angular rate  = 3 rev/min of the Ferris wheel is constant.

8 Exercise # 4 2/199: Airplane A is flying horizontally with a constant speed of 200 km/h and is towing the glider B, which is gaining altitude. If the tow cable has a length r = 60 m and  is increasing at the constant rate of 5 degrees per second, determine the magnitudes of the velocity v and acceleration a of the glider for the instant when  = 15°.

9 Constrained Motion Here, motions of more than one particle are interrelated because of the constraints imposed by the interconnecting members. In such problems, it is necessary to account for these constraints in order to determine the respective motions of the particles.

10 Constrained Motion (Cont.) Notes: -Horizontal motion of A is twice the vertical motion of B. -The motion of B is the same as that of the center of its pulley, so we establish position coordinates x and y measured from a convenient fixed datum. -The system is one degree of freedom, since only one variable, either x or y, is needed to specify the positions of all parts of the system. One Degree of Freedom System L, r 1, r 2, and b are constants Differentiating once and twice gives: Datum + +

11 Constrained Motion (Cont.) Note: -The positions of the lower pulley C depend on the separate specifications of the two coordinates y A & y B. -It is impossible for the signs of all three terms to be +ve simultaneously. Two Degree of Freedom System Differentiating once gives: Eliminating the terms in gives: Datum + + + +

12 Constrained Motion Exercises

13 Exercise # 5 2/208: Cylinder B has a downward velocity of 0.6 m/s and an upward acceleration of 0.15 m/s 2. Calculate the velocity and acceleration of block A.

14 Exercise # 6 2/210: Cylinder B has a downward velocity in meters per second given by v B = t 2 /2 + t 3 /6, where t is in seconds. Calculate the acceleration of A when t = 2 s.

15 Exercise # 7 2/211: Determine the vertical rise h of the load W during 5 seconds if the hoisting drum wraps cable around it at the constant rate of 320 mm/s.

16 Exercise # 8 2/216: The power winches on the industrial scaffold enable it to be raised or lowered. For rotation in the sense indicated, the scaffold is being raised. If each drum has a diameter of 200 mm and turns at the rate of 40 rev/min, determine the upward velocity v of the scaffold.

17 Exercise # 9 2/218: Collars A and B slide along the fixed right-angle rods and are connected by a cord of length L. Determine the acceleration a x of collar B as a function of y if collar A is given a constant upward velocity v A.


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