Explanation Given information: The mass of the circular plate is m , the side of the square plate is a , the angular velocity is ω 0 j and the eccentricity for the perfectly plastic is 0 . Draw the free body diagram for the plate. Figure-(1) Write the expression for the angle θ . θ = tan − 1 ( a / 2 a / 2 ) = tan − 1 ( 1 ) = 45 ° Write the expression for the angular velocity in the x- direction. ω x = ω 0 cos θ ...... (I) Write the expression for the angular velocity in the y- direction. ω y = ω 0 sin θ ...... (II) Since, the initial velocity of the mass centre is zero hence the initial angular acceleration is also zero that is r G / A × m V ¯ 0 = 0 . Write the expression for the initial angular momentum about the mass centre. ( H G ) 0 = I x ω x i + I y ω y j ...... (III) The angular momentum is zero in the z- direction. Write the expression for the moment of the inertia of a square plate about its diagonal. I x = m a 2 12 ...... (IV) Write the expression for the moment of the inertia of a square plate about its centroidal axis. I y = m a 2 12 ...... (V) Write the expression for the velocity of the mass centre at point B after the impact. V B = ω × r B / A ....... (VI) Here, the position vector of point B with respect A is r B / A and the angular velocity of the mass centre after the impact is ω . Write the expression for the velocity of the mass centre after the impact. V ¯ = ω × r G / A ....... (VII) Here, the position vector of point G with respect A is r G / A and the angular velocity of the mass centre after the impact is ω . Write the expression for the angular momentum of the linear moment with respect to A after the impact. H A = r G / A × m V ¯ ...... (VIII) The angular velocity in the x- direction is zero. Write the expression for the angular momentum of the mass centre after the impact. H G = I x ω x i + I y ω y j + I z ω z k = I y ω y j + I z ω z k ...... (IX) Write the expression for moment of the inertia about the z- axis. I z = 1 6 m a 2 ...... (X) Write the expression for the impulse momentum of the system. H G + H A = ( H G ) 0 + r G / A × m V ¯ 0 − ( a F Δ t ) i ...... (XI) Write the expression for the angular velocity of the assembly. ω = ω x i + ω y k + ω z k ...... (XII) Calculation: Substitute 45 ° for θ in Equation (I). ω x = ω 0 cos 45 ° = ω 0 ( 0.7071 ) = 0.7071 ω 0 Substitute 45 ° for θ in Equation (II). ω y = ω 0 cos 45 ° = ω 0 ( 0.7071 ) = 0.7071 ω 0 Substitute 0.7071 ω 0 for ω x , 0.7071 ω 0 for ω y , 1 12 m a 2 for I y and 1 12 m a 2 for I x in Equation (III). ( H G ) 0 = ( 1 12 m a 2 ) ( 0.7071 ω 0 ) i + ( 1 12 m a 2 ) ( 0.7071 ω 0 ) j = ( 1 12 m a 2 ) ( 0.7071 ω 0 ) ( i + j ) Since, the initial velocity is zero hence the angular momentum is also zero. Substitute ω x i + ω y j + ω z k for ω and − a j for r B / A in Equation (VI). V B = ( ω x i + ω y j + ω z k ) ( − a j ) = a ( ω z i + ω x k ) ....... (XIII) Substitute ω y j + ω z k for ω and ( 1 2 a ) ( − i − j ) for r G / A in Equation (VII). V ¯ = ( ω y j + ω z k ) ( ( 1 2 a ) ( − i − j ) ) = a 2 ( ω z i − ω z j + ω y k ) Substitute a 2 ( ω z i − ω z j + ω y k ) for V ¯ and ( 1 2 a ) ( − i − j ) for r G / A in Equation (VIII) To determine (b) The velocity of the mass centre after the impact.

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 18.1, Problem 18.31P

To determine

(a)

The angular velocity of the plate.

To determine

(b)

The velocity of the mass centre after the impact.

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Chapter 18.1, Problem 18.30P

Chapter 18.1, Problem 18.32P

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(a) The angular velocity of the plate.

Solution Summary: The author explains the angular velocity of the circular plate and the eccentricity for the perfectly plastic plate.