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Relative Motion & Constrained Motion. Lecture V. Relative Motion. Constrained Motion. Relative Motion. v B. v A. v A/B.
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Relative Motion & Constrained Motion Lecture V Relative Motion Constrained Motion
Relative Motion vB vA vA/B • 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.
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). Path Path Moving system Moving system Path Path Fixed system Fixed system 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. Note:rA & rBare measured from the origin of the fixed axes X-Y. Note:rB/A = -rA/B vB/A = -vA/B aB/A = -aA/B
Exercise # 1 2/186: The passenger aircraft B is flying east with a velocity vB = 800 km/h. A military jet traveling south with a velocity vA = 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?.
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/s2, and B has an increase in speed of 3 m/s2. Determine the velocity and acceleration of car B with respect to car A.
Exercise # 3 2/191: The car A has a forward speed of 18 km/h and is accelerating at 3 m/s2. 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 W = 3 rev/min of the Ferris wheel is constant.
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 q 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 q = 15° .
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.
Constrained Motion (Cont.) + Datum + One Degree of Freedom System • 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. L, r1, r2, and b are constants Differentiating once and twice gives:
Constrained Motion (Cont.) Datum Datum + + + + Two Degree of Freedom System • Note: • The positions of the lower pulley C depend on the separate specifications of the two coordinates yA & yB. • It is impossible for the signs of all three terms to be +ve simultaneously. Differentiating once gives: Differentiating once gives: Eliminating the terms in gives:
Exercise # 5 2/208: Cylinder B has a downward velocity of 0.6 m/s and an upward acceleration of 0.15 m/s2. Calculate the velocity and acceleration of block A.
Exercise # 6 2/210: Cylinder B has a downward velocity in meters per second given by vB = t2/2 + t3/6, where t is in seconds. Calculate the acceleration of A when t = 2 s.
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.
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.
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 axof collar B as a function of y if collar A is given a constant upward velocity vA.