(a) Explanation Given information: The mass of the circular plate is m . The falling velocity is ( v ¯ o ) . The line CG makes an angle ( θ ) is 45 ° . Assume the impact to be perfectly plastic that is e = 0 . Calculation: Show the impulse momentum diagram of the circular plate of the system as Figure (1). The expression for the moment of inertia in the x direction ( I ¯ x ) as follows: I ¯ x = 1 4 m R 2 Here, m is the mass and R is the radius of the circular plate. The expression for the moment of inertia in the y direction ( I ¯ y ) as follows: I ¯ y = 1 2 m R 2 The expression for the moment of inertia in the z direction ( I ¯ z ) as follows; I ¯ z = 1 4 m R 2 The expression for the impulse momentum equation by applying conservation of linear momentum as follows; − m v ¯ 0 j + C ( Δ t ) j = m v ¯ x i + m v ¯ y j + m v ¯ z k (1) Here, v ¯ 0 is the velocity with which the plate falls, C ( Δ t ) is the impulse at C , v ¯ x is the velocity in the x direction, v ¯ y is the velocity in the y direction, and v ¯ z is the velocity in the z direction. Take the velocity in the downward direction is negative. Equate the i vector from the Equation (1). m v ¯ x = 0 v ¯ x = 0 Equate the j vector from the Equation (1). − m v 0 + C ( Δ t ) = m v ¯ y C ( Δ t ) = m ( v ¯ 0 + v ¯ y ) (2) Equate the k vector from the Equation (1). m v ¯ z = 0 v ¯ z = 0 The expression for the relative position of C ( r C G ) with respect to the center of mass as follows: r C G = 1 2 R ( i − k ) The impact is assumed to be perfectly elastic; therefore, velocity of disc C after impact in the y direction is 0. The expression for the velocity of disc C after impact according to kinematics as follows: v ¯ C = v ¯ + ω × r C G The scalar quantities are denoted by regular font. Resolve the variables 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 Substitute 0 for ( v C ) y j and 1 2 R ( i − k ) for r C G . ( 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 ) The matrix multiplication for cross product is done. ( 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 (3) Equate j vector from the equation (3). v ¯ y + R 2 ( ω x + ω z ) = 0 v ¯ y = − R 2 ( ω x + ω z ) (4) Take moments about center 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 Substitute 1 2 R ( i − k ) for r C G , 1 4 m R 2 for I ¯ x , 1 2 m R 2 for I ¯ y , and 1 4 m R 2 for I ¯ z . 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 Matrix multiplication for cross product is done (b) To determine Find the impulse ( C Δ t ) exerted on the plate by the obstruction during the impact.

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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

Publisher: McGraw-Hill Education

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Chapter 18.1, Problem 18.30P

(a)

To determine

The velocity (v¯) of its mass center G immediately after the impact.

(b)

To determine

Find the impulse (CΔt) exerted on the plate by the obstruction during the impact.

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Chapter 18.1, Problem 18.29P

Chapter 18.1, Problem 18.31P

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Vector Mechanics for Engineers: Statics and Dynamics

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The velocity ( v ¯ ) of its mass center G immediately after the impact.

Solution Summary: The author shows the impulse momentum diagram of the circular plate as Figure (1).