Explanation Given: The width of the thin plate is 400 mm . The total height of the thin plate is 800 mm . The mass per unit area of the thin plate is 10 kg / m 2 . Subdivide the segments of the thin plate into four parts as shown below in Figure 1. Label the holes as segments (3) and (4) and consider them as the negative parts. Show the free body diagram of the thin plate as in Figure 1. From Figure 1, Compute the mass moment of inertia of the thin plate about the z- axis with respect to the centroid G . I z = ∑ ( I z ) G + m d 2 = [ 1 12 m 1 h 2 ] + [ 1 12 m 1 ( h 2 + b 2 ) + m 2 ( b / 2 ) 2 ] − 1 4 m 3 r 2 − [ 1 2 m 3 r 2 + m 4 ( 2 r ) 2 ] (I) Here, the mass of the thin plate is m , the diameter of the hole is d , the moment of inertia about z -axis in reference to the G -axis is ( I z ) G , the mass of the segment 1 is m 1 , the mass of the segment 2 is m 2 , the width of the segment 1 is b , the radius of the segment 2 is r and the height of the segment 1 is h . Compute the mass of the segments (1) and (2). m 1 = m 2 = ( 10 kg / m 2 ) b h (II) Here, the width of the thin plate is b and the height is h . Compute the mass of the segments (3) and (4). m 3 = m 4 = ( 10 kg / m 2 ) π r 2 (III) Here, the radius of the hole is r . Conclusion: Substitute ( 200 mm + 200 mm ) for b and ( 200 mm + 200 mm ) for h in Equation (II). m 1 = m 2 = ( 10 kg / m 2 ) ( 200 mm + 200 mm ) ( 200 mm + 200 mm ) = ( 10 kg / m 2 ) ( 400 mm ) ( 400 mm ) = ( 10 kg / m 2 ) [ ( 400 mm ) 1 m 1000 mm ] [ ( 400 mm ) 1 m 1000 mm ] = ( 10 kg / m 2 ) ( 0.4 m ) ( 0.4 m ) = 1.6 kg Substitute 100 mm for r in Equation (III). m 3 = m 4 = ( 10 kg / m 2 ) π ( 100 mm ) 2 = ( 10 kg / m 2 ) π [ ( 100 mm ) 1 m 1000 mm ] 2 = ( 10 kg / m 2 ) π ( 0.1 m ) 2 = 0

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Chapter 10, Problem 105P

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