Explanation The free body diagram of system is shown below. Figure-(1) Write the expression for the angular velocity of crank C A . θ ˙ = 2 π N 60 (I) Here, the angular velocity of crank C A is θ ˙ and the speed of crack C A is N . Write the expression for tan β in Δ O D A tan β = A D O D (II) Here, the angle between link O C and O A is β . Conclusion: Substitute 120 rev / min for N in Equation (I). θ ˙ = 2 π 60 ( 120 rev / min ) = 4 π rad / s Substitute 5 sin θ for A D and ( 25 − 5 cos θ ) for O D in Equation (II). tan β = 5 sin θ 25 − 5 cos θ = sin θ 5 − cos θ (III) Differentiate Equation (III). β ˙ sec 2 β = ( 5 − cos θ ) θ ˙ cos θ − sin θ ( θ ˙ sin θ ) ( 5 − cos θ ) 2 β ˙ sec 2 β = ( 5 − cos θ ) θ ˙ cos θ − θ ˙ sin 2 θ ( 5 − cos θ ) 2 β ˙ ( 1 + tan 2 β ) = 5 θ ˙ cos θ − θ ˙ ( 5 − cos θ ) 2 β ˙ = θ ˙ ( 5 cos θ − 1 ) ( 1 + tan 2 β ) ( 5 − cos θ ) 2 (IV) Here, the angular velocity of the link O B is β ˙ . Substitute 4 π rad / s for θ ˙ and sin θ 5 − cos θ for tan β in Equation (IV)

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Chapter 5.8, Problem 204RP

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The angular velocity β˙ of arm OB, the graph of the angular velocity β˙ of arm OB as a function of θ through 360° and value of θ for zero angular velocity of OB.

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Chapter 5.8, Problem 203RP

Chapter 5.8, Problem 206RP

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The angular velocity β ˙ of arm O B , the graph of the angular velocity β ˙ of arm O B as a function of θ through 360 ° and value of θ for zero angular velocity of O B .

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The angular velocity β˙ of arm OB, the graph of the angular velocity β˙ of arm OB as a function of θ through 360° and value of θ for zero angular velocity of OB.

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Chapter 5 Solutions