Explanation Given information: The velocity and acceleration of the ankle is zero. During a jump, The velocity of the ankle is zero. The angular velocity of the tibia AK is ω = 1.5 rad/s ( Counterclockwise ) . The angular acceleration of the tibia AK is α = 1 rad/s 2 ( Counterclockwise ) . Calculation: Show the unit vectors as follows: i = 1 → j = 1 ↑ k=1 ( Counterclockwise ) Consider the position of the point K with respect to A is denoted by r K / A . Consider the position of the point H with respect to K is denoted by r H / K . Consider the position of the point H with respect to A is denoted by r H / A . Calculate the distances r K / A , r H / K , and r H / A as follows: r K / A = 12 i + 12 j r H / K = − 14 i + 14 j r H / A = r K / A + r K / H = ( 12 i + 12 j ) + ( − 14 i + 14 j ) = − 2 i + 26 j Take the frame of reference moving with the lower leg AK with angular velocity Ω and angular acceleration Ω ′ . Show the angular velocity Ω and Ω ′ as follows: Ω = ( 1.5 rad/s ) k ( Counterclockwise ) Ω ˙ = ( 1 rad/s 2 ) k ( Counterclockwise ) Show the motion of the hip H relative to the frame. The angular velocity and angular acceleration of H is denoted by ω = ω k and α = α k . The angular velocity and angular acceleration measured relative to lower leg AK . Consider the motion of the point H ′ coincides with H . Consider the velocity and acceleration on the point v H ′ are v H ′ and a W ′ . Calculate the value of v H ′ and v H ′ as follows: Substitute ( 1 rad/s 2 ) k for Ω ˙ , ( 1.5 rad/s ) k for Ω and ( − 2 i + 26 j ) for r H / A . v H ′ = Ω × r H / A = ( 1.5 rad/s ) k × ( − 2 i + 26 j ) = − 3 j − 39 i = − 39 i − 3 j a H ′ = Ω ˙ × r H / A − Ω 2 × r H / A = 1 k × ( − 2 i + 26 j ) − ( 1.5 ) 2 × ( − 2 i + 26 j ) = − 2 j − 26 i+4 .5i − 58 .5j = − 21.5 i − 60.5 j Consider the motion of H relative to the frame. Consider the velocity and acceleration on the point H relative to the frame are v H / A K and a H / A K . Calculate the value of v H / A K and a H / A K as follows: Substitute α k for α , ω k for ω and ( − 14 i + 14 j ) for r H / K . v H / A K = ω × r H / K = ω k × ( − 14 i + 14 j ) = − 14 ω j − 14 ω i = − 14 ω i − 14 ω j a H / A K = α × r H / K − ω 2 × r H / K = α k × ( − 14 i + 14 j ) − ω 2 ( − 14 i + 14 j ) = − 14 α j − 14 α i+ 14 ω 2 i − 14 ω 2 j Calculate the velocity of H using the relation: v H = v H ′ + v H / A K Substitute ( − 39 i − 3 j ) for v H ′ and ( − 14 ω i − 14 ω j ) for v W / S E

Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977206

Author: BEER, Ferdinand P., Johnston Jr., E. Russell, Mazurek, David, Cornwell, Phillip J., SELF, Brian

Publisher: McGraw-Hill Education

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Chapter 15.5, Problem 15.171P

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Find the relative angular velocity and relative angular acceleration of KH such that the velocity and acceleration of H both are straight up.

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Chapter 15.5, Problem 15.170P

Chapter 15.5, Problem 15.172P

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Find the relative angular velocity and relative angular acceleration of KH such that the velocity and acceleration of H both are straight up.

Solution Summary: The author explains the relative angular velocity and relative acceleration of KH, such that they are straight up.

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Find the relative angular velocity and relative angular acceleration of KH such that the velocity and acceleration of H both are straight up.

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