Explanation Given: The distance and position point D is shown in Figure (1). The angular velocity ω = 14 rad/s . The acceleration α = 7 rad/s 2 . Write the expression for unit vector . u O A = r O A | r O A | (I). Refer Figure (1) and consider the shaft position O A and write the relation for position vector along O A . r O A = { − 0.3 i + 0.2 j +0 .6 k } m Refer Figure (1) and write the relation for the position vector for point D . r D = { − 0.3 i + 0.4 j } m Write the expression for angular velocity in Cartesian vector form. ω = ω u O A (II). Here, angular velocity is ω and unit vector is u O A . Write the expression for acceleration in Cartesian vector form. α = α u O A (III). Here, acceleration is α and unit vector is u O A . Write the expression for velocity of point D . v C = ω × r D (IV). Here, angular velocity in Cartesian vector form is ω and position vector of point D is r D . Write the expression for acceleration of point D . a D = α × r D + ω × ( ω × r D ) (V). Here the acceleration in Cartesian vector form is α , position vector of point D is r D and angular velocity in Cartesian vector form is ω . Conclusion: Substitute − 0.3 i + 0.2 j +0 .6 k for r O A in Equation (I). u O A = r O A | r O A | = − 0.3 i + 0.2 j +0 .6 k | ( − 0.3 ) 2 + ( 0.2 ) 2 + ( 0 .6 ) 2 | = − 3 7 i + 2 7 j + 6 7 k Substitute − 3 7 i + 2 7 j + 6 7 k for u O A and 14 rad/s for ω in Equation (II). ω = ω u O A = 14 ( − 3 7 i + 2 7 j + 6 7 k ) = − 6 i + 4 j + 12 k rad/s Substitute − 3 7 i + 2 7 j + 6 7 k for u O A and 7 rad/s 2 for α in Equation (III)

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 16, Problem 36P

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The velocity and acceleration of point D .

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The velocity and acceleration of point D .

Solution Summary: The angular velocity and acceleration of point D are shown in Figure (1).

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