these components are and Their sense of direction is established from the

# These components are and their sense of direction is

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these components are and Their sense of direction is established from the diagram such that acts perpendicular to in accordance with the rotational motion of the body, and is directed from B towards A . * Acceleration Equation. Represent the vectors in graphically by showing their magnitudes and directions underneath each term. The scalar equations are determined from the x and y components of these vectors. a B = a A + 1 a B > A 2 t + 1 a B > A 2 n 1 a B > A 2 n A r B > A , 1 a B > A 2 t 1 a B > A 2 n = v 2 r B > A . 1 a B > A 2 t = a r B > A 1 a B > A 2 n 1 a B > A 2 t a B = a A + A * r B > A - v 2 r B > A a B = 1 a B 2 t + 1 a B 2 n . a A = 1 a A 2 t + 1 a A 2 n r B > A . A , V , a B , a A , v B v A V *The notation may be helpful in recalling that A is assumed to be pinned. a B = a A + 1 a B > A 1 pin 2 2 t + 1 a B > A 1 pin 2 2 n The mechanism for a window is shown.Here CA rotates about a fixed axis through C , and AB undergoes general plane motion. Since point A moves along a curved path it has two components of acceleration, whereas point B moves along a straight track and the direction of its acceleration is specified. A C B ( a A ) n ( a A ) t a B V , A
366 C HAPTER 16 P LANAR K INEMATICS OF A R IGID B ODY 16 The rod AB shown in Fig. 16–27 a is confined to move along the inclined planes at A and B . If point A has an acceleration of and a velocity of , both directed down the plane at the instant the rod is horizontal, determine the angular acceleration of the rod at this instant. SOLUTION I (VECTOR ANALYSIS) We will apply the acceleration equation to points A and B on the rod. To do so it is first necessary to determine the angular velocity of the rod. Show that it is d using either the velocity equation or the method of instantaneous centers. Kinematic Diagram. Since points A and B both move along straight-line paths, they have no components of acceleration normal to the paths.There are two unknowns in Fig. 16–27 b , namely, and Acceleration Equation. a B = a A + A * r B > A - v 2 r B > A a . a B v = 0.283 rad > s 2 m > s 3 m > s 2 EXAMPLE 16.14 Carrying out the cross product and equating the i and j components yields (1) (2) Solving, we have d Ans. SOLUTION II (SCALAR ANALYSIS) From the kinematic diagram, showing the relative-acceleration components and Fig. 16–27 c , we have Equating the x and y components yields Eqs. 1 and 2, and the solution proceeds as before. c a B a 45° d = c 3 m > s 2 c 45° d + c a 1 10 m 2 c d + c 1 0.283 rad > s 2 2 1 10 m 2 ; d a B = a A + 1 a B > A 2 t + 1 a B > A 2 n 1 a B > A 2 n , 1 a B > A 2 t a = 0.344 rad > s 2 a B = 1.87 m > s 2 a 45° a B sin 45° = - 3 sin 45° + a 1 10 2 a B cos 45° = 3 cos 45° - 1 0.283 2 2 1 10 2 a B cos 45° i + a B sin 45° j = 3 cos 45° i - 3 sin 45° j + 1 a k 2 * 1 10 i 2 - 1 0.283 2 2 1 10 i 2 10 m B A (a) v A 2 m / s a A 3 m / s 2 45 45 x y A B 45 45 a A 3 m / s 2 r B / A v 0.283 rad / s a B (b) A A B 10 m r B / A ( a B / A ) t a r B / A ( a B / A ) n v 2 r B / A v 0.283 rad / s (c) A Fig. 16–27
16.7 R ELATIVE -M OTION A NALYSIS : A CCELERATION 367 16 EXAMPLE 16.15 At a given instant, the cylinder of radius r , shown in Fig. 16–28 a , has an angular velocity and angular acceleration Determine the velocity and acceleration of its center G and the acceleration of the contact point at A if it rolls without slipping.