Given information:

The mass of the cylinder is denoted by m.

The length of the circular cylinder is denoted by a.

The diameter OB of the top surface makes 45° with x and z axis.

Calculation:

Show the homogeneous circular cylinder as shown in Figure 1.

Refer Figure 1.

Refer Figure 9.28.

Apply parallel axis theorem

Show the moment of inertia of the circular cylinder about the y axis as follows:

Iy=12ma2+ma2=32ma2

Show the moment of inertia of the circular cylinder about the x and z axis as follows:

Ix=Iz=112m(3a2+L2)+m[(a2)2+(a2)2]} (1)

Here, a is the radius of the cylinder and L is the length of the cylinder.

Substitute a for L in Equation (1).

Ix=Iz=112m(3a2+a2)+m[(a2)2+(a2)2]=13ma2+m[a22+a24]=13ma2+34ma2=1312ma2

The centroidal axis products of inertia are zero due to symmetry.

Ix′y′=Iy′z′=Iz′x′=0

Write the centroidal locations as measured from the origin O along the x, y and z axis as below;

x¯=(a2), y¯=(−a2) and z¯=(a2).

Express the moment of inertia Ixy as follows:

Ixy=Ix′y′+mx¯ y¯=0+m(a2)(−a2)=−122ma2

Express the moment of inertia Iyz as follows:

Iyz=Iy′z′+my¯ z¯=0+m(−a2)(a2)=−122ma2

Express the moment of inertia Izx as follows:

Izx=Iz′x′+m z¯ x¯=0+m(a2)(a2)=12ma2

Show the Equation 9.56 as follows:

{K3−(Ix+Iy+Iz)K2+(IxIy+IyIz+IzIx−Ixy2−Iyz2−Izx2)K−(IxIyIz−IxIyz2−IyIzx2−IzIxy2−2IxyIyzIzx)}=0

Substitute 1312ma2 for Ix, 32ma2 for Iy, 1312ma2 for Iz, −122ma2 for Ixy, −122ma2 for Iyz, and 12ma2 for Izx.

{K3−(1312ma2+32ma2+1312ma2)K2+[(1312ma2)(32ma2)+(32ma2)(1312ma2)+(1312ma2)(1312ma2)−(−122ma2)2−(−122ma2)2−(12ma2)2]K−[(1312ma2)(32ma2)(1312ma2)−(1312ma2)(−122ma2)2−(32ma2)(12ma2)2−(1312ma2)(−122ma2)2−2(−122ma2)(−122ma2)(12ma2)]}=0{K3−(1312+32+1312)K2ma2+[(1312)(32)+(32)(1312)+(1312)(1312)−(−122)2−(−122)2−(12)2]K(ma2)2−[(32)(1312)(1312)−(1312)(−122)2−(32)(12)2−(1312)(−122)2−2(−122)(−122)(12)](ma2)3}=0

Substitute ma2ζ for K.

ζ3−113ζ2+565144ζ−9596=0

Solve the above Equation and get the values of ζ1, ζ2, and ζ3 as follows:

ζ1=0.363ζ2=1.583ζ3=1.72

Show the principal moment of inertia as follows:

K1=0.363ma2K2=1.583ma2K3=1.72ma2

Thus, the principal mass moment of inertia are K1=0.363ma2,K2=1.583ma2, and K3=1.72ma2_

Given information:

Consider the direction cosines of each principal axis are denoted by λx,λy,λz.

Calculation:

Refer Part (a).

Show the Equation 9.54 as follows:

(Ix−K)λx−Ixyλy−Izxλz=0−Izxλx−Iyzλy+(Iz−K)λz=0} (2)

Substitute 1312ma2 for Ix, 1312ma2 for Iz, −122ma2 for Ixy, −122ma2 for Iyz, and 12ma2 for Izx.

(1312ma2−K)λx−(−122)ma2λy−(12ma2)λz=0−(12ma2)λx−(−122)ma2λy+(1312ma2−K)λz=0} (3)

Modify Equation (3).

Consider K=ma2ζ. Then,

(1312−ζ)λx−(−122)λy−(12)λz=0−(12)λx−(−122)λy+(1312−ζ)λz=0} (4)

Solve Equation (4).

1312−ζ=−12ζ=1912

Add both the Equation in Equation (4).

[1312−ζ−(−12)]λx+[−12−(1312−ζ)]λz=0λx=λz

Substitute λx for λz in Equation (4).

(1312−ζ)λx−(−122)λy−(12)λx=0−(−122)λy=(12−1312+ζ)λx(122)λy=(ζ−712)λxλy=22(ζ−712)λx

Show the Equation 9.57 as follows:

λx2+λy2+λz2=1

Substitute 22(ζ−712)λx for λy, and λx for λz.

λx2+[22(ζ−712)λx]2+λx2=1λx2+8(ζ−712)2λx2+λx2=1[2+8(ζ−712)2]λx2=1 (5)

Consider K1.

Substitute 0.363 for ζ1 in Equation (5).

[2+8(0.363−712)2](λx)12=1(λx)1=(λz)1=0.647249

Calculate the value of (λy)1 as follows:

(122)(λy)1=(ζ1−712)(λx)1

Substitute 0.647249 for (λx)1 and 0.363 for ζ1.

(122)(λy)1=(0.363−712)×0.647249(λy)1=−0.402662

Show the direction cosines (θx)1,(θy)1,(θz)1 using the relation:

cos(θx)1=(λx)1=0.647249(θx)1=cos−1(0.647249)=49.7°

cos(θz)1=(λz)1=0.647249(θz)1=cos−1(0.647249)=49.7°

cos(θy)1=(λy)1=−0.402662(θy)1=cos−1(−0.402662)=113.7°

Conisder K3.

Substitute 1.720 for ζ3 in Equation (5).

[2+8(1.720−712)2](λx)32=1(λx)3=(λz)3=0.284726

Calculate the value of (λy)3 as follows:

(122)(λy)3=(ζ3−712)(λx)3

Substitute 0.284726 for (λx)3 and 1.720 for ζ3.

(122)(λy)3=(1.720−712)×0.647249(λy)3=0.915348

Show the direction cosines (θx)3,(θy)3,(θz)3 using the relation:

cos(θx)3=(λx)3=0.284726(θx)3=cos−1(0.284726)=73.5°

cos(θz)3=(λz)3=0.284726(θz)3=cos−1(0.284726)=73.5°

cos(θy)3=(λy)3=0.915348(θy)3=cos−1(0.915348)=23.7°

Consider K2.

Show the Equation 9.54b as follows:

−Ixyλx+(Iy−K)λy−Iyzλz=0

Substitute 32ma2 for Iy, −122ma2 for Ixy, and −122ma2 for Iyz.

−(−122ma2)λx+(32ma2−K)λy+(122ma2)λz=0122λx+(32−ζ)λy+(122)λz=0 (6)

Refer Equation (3) and (6).

(1312−ζ)λx−(−122)λy−(12)λz=0122λx+(32−ζ)λy+(122)λz=0} (7)

Substitute 1912 for ζ in Equation (4).

(1312−1912)(λx)2−(−122)(λy)2−(12)(λz)2=0122(λx)2+(32−1912)(λy)2+(122)(λz)2=0

Modify above Equations as follows:

112(λx)2+(122)(λy)2−(12)(λz)2=0122(λx)2−112(λy)2+(122)(λz)2=0} (8)

Solve Equation (8) and get the value of (λy)2=0.

Show the Equation 9.57 as follows:

(λx)22+(λy)22+(λz)22=1(λx)22+0+(λz)22=1(λx)22+(λz)22=1

Substitute λx for λz.

(λx)22+(λx)22=1(λx)2=12

(λz)2=−12

Show the direction cosines (θx)2,(θy)2,(θz)2 using the relation:

cos(θx)2=(λx)2=12(θx)2=cos−1(12)=45°

cos(θz)2=(λz)2=−(12)(θz)2=cos−1(−12)=135°

cos(θy)2=(λy)2=0(θy)2=cos−1(0)=90°

Thus, the velocity of the point B is 1.537 ft/s(Acting 77.48° Clockwise above the horizontal)_.

Given information:

Calculation:

Refer Part (a) and (b).

Sketch the body and show the orientation of the principal axis of inertia relative to x, y, and z axis as shown in Figure 2.

Refer Figure 2.

The principal axis 1 and 3 lies on the vertical plane of symmetry passing through OB.

The principal axis 2 lies in xz plane.