Explanation Given information: The mass of the circular plate is m , the velocity of the circular plate is , the angle on the x -axis is 45 ° and the eccentricity for the perfectly plastic is 0 . Expression for moment of the inertia about the x- axis. I x = 1 4 m r 2 ...... (I). Here, the radius of the circular plate is r . Expression for moment of the inertia about the y- axis. I y = 1 2 m r 2 ...... (II) Expression for moment of the inertia about the z- axis. I z = 1 4 m r 2 ...... (III) Expression for the conservation of the linear momentum. − m v ¯ 0 j + C ( Δ t ) j = m v ¯ x i + m v ¯ y j + m v ¯ z k ...... (IV) Expression for the relative position of C with respect to the centre of mass. r C G = 1 2 r ( i − k ) ...... (V) Expression for the disc C after the impact. v ¯ C = v ¯ + ω × r C G ...... (VI) Here, the angular velocity is ω . Expression to show the resolution of the variable along x , y and z direction. ( v C ) x i + ( v C ) y j + ( v C ) z k = v ¯ y j + ( ω x i + ω y j + ω z k ) × r C G ...... (VII) Here, the velocity with which the plate fall is v ¯ 0 , the impulse of mass centre C is C ( Δ t ) , the velocity in the x direction is v ¯ x , the velocity in the y direction is v ¯ y and the velocity in the z direction is v ¯ z . Expression for the moment about centre of mass G . r C G × C ( Δ t ) j = I x ω x i + I y ω y j + I z ω z k ...... (VIII) Expression for the velocity of the mass centre G . v ¯ = v ¯ x i + v ¯ y j + v ¯ z k ...... (IX) Calculation: Equate the vector i in Equation (IV). m v ¯ x = 0 v ¯ x = 0 Equate the vector j in Equation (IV). − m v ¯ 0 + C ( Δ t ) = m v ¯ y C ( Δ t ) = m ( v ¯ 0 + v ¯ y ) ......(X) Equate the vector k in Equation (IV). m v ¯ z = 0 v ¯ z = 0 Substitute 0 for ( v C ) y j and 1 2 r ( i − k ) for r C G in Equation (VII). ( v C ) x i + 0 + ( v C ) z k = v ¯ y j + ( ω x i + ω y j + ω z k ) × ( 1 2 r ( i − k ) ) ( v C ) x i + ( v C ) z k = v ¯ y j + ( ω x i + ω y j + ω z k ) × ( 1 2 r ( i − k ) ) ( v C ) x i + ( v C ) z k = v ¯ y j + ( r 2 ) ω x j − ( r 2 ) ω y ( k + i ) + ( r 2 ) ω z j ...... (XI) Equate the vector j in Equation (X). v ¯ y + r 2 ( ω x + ω z ) = 0 v ¯ y = − r 2 ( ω x + ω z ) ...... (XII) Substitute 1 4 m r 2 for I x , 1 2 m r 2 for I y , 1 4 m r 2 for I z and 1 2 r ( i − k ) for r C G in Equation (VIII). ( 1 2 r ( i − k ) ) × C ( Δ t ) j = ( 1 4 m r 2 ) ω x i + ( 1 2 m r 2 ) ω y j + ( 1 4 m r 2 ) ω z k r 2 C ( Δ t ) ( i + k ) = ( 1 4 m r 2 ) ω x i + ( 1 2 m r 2 ) ω y j + ( 1 4 m r 2 ) ω z k ...... (XIII) Equate the vector i in Equation (XIII). r 2 C ( Δ t ) = ( 1 4 m r 2 ) ω x C ( Δ t ) = 2 r ( 1 2 × 2 × 2 m r 2 ) ω x C ( Δ t ) = ( 1 2 2 m r ) ω x ...... (XIV) Equate the vector j in Equation (XIII). ( 1 2 m r 2 ) ω y = 0 ω y = 0 Equate the vector k in Equation (XIII) To determine (b) The impulse generated on the plate.

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 18.30P

To determine

(a)

The velocity of the mass centre G immediately after the impact.

To determine

(b)

The impulse generated on the plate.

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Chapter 18.1, Problem 18.29P

Chapter 18.1, Problem 18.31P

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Chapter 18 Solutions

Vector Mechanics for Engineers: Dynamics

Ch. 18.1 - Prob. 18.1P Ch. 18.1 - A thin rectangular plate of weight 15 Ib rotates...Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W=6 lb rotates at the...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Solve Prob. 18.6, assuming that the solid...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P

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(a) The velocity of the mass centre G immediately after the impact.

Solution Summary: The author explains the velocity of the mass centre G immediately after the impact, the angle on the x -axis, and the eccentricity for the perfectly plastic.