The mass moment of inertia of the wire with respect to x ,y and z axis.

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Answer 1

Question

Chapter B, Problem B.37P

To determine

The mass moment of inertia of the wire with respect to x ,y and z axis.

Expert Solution & Answer

Answer to Problem B.37P

The mass moment of inertia of the wire with respect to x axis is 39.1721×10−3 lb⋅ft⋅s2.

The mass moment of inertia of the wire with respect to y axis is 36.2542×10−3 lb⋅ft⋅s2.

The mass moment of inertia of the wire with respect to z axis is 30.5983×10−3 lb⋅ft⋅s2

Explanation of Solution

Given information:

The diameter of the formed steel wire is 18 in and the specific weight of the steel is 490 lb/ft3.

The below figure represent the schematic diagram of the wire.

Vector Mechanics For Engineers, Chapter B, Problem B.37P

Figure-(1)

Expression of mass of the steel wire.

m=ρSTV ......(I)

Here, density of the steel wire is ρST and the volume of the steel wire is V.

Expression of density of the steel wire.

ρST=γSTg ......(II)

Here, the specific weight of the steel wire is γST and the acceleration due to gravity is g.

Substitute γSTg for ρST in Equation (I).

m=(γSTg)V ......(III)

For section (1).

Substitute m1 for m and V1 for V in Equation (III).

m1=(γSTg)V1 ......(IV)

Expression of volume of the steel wire after formation for section (1).

V1=(π4d12)(πd) ......(V)

Here, the diameter of the formed wire is d1 and the diameter of the wire is d.

For section (2).

Substitute m2 for m and V2 for V in Equation (III).

m2=(γSTg)V2 ......(VI)

Expression of volume of the steel wire after formation for section (2).

V2=(π4d12)(πd) ......(VII)

For section (3).

Substitute m3 for m and V3 for V in Equation (III).

m3=(γSTg)V3 ......(VIII)

Expression of volume of the steel wire after formation for section (3).

V3=(π4d12)(d) ......(IX)

For section (4).

Substitute m4 for m and V4 for V in Equation (III).

m4=(γSTg)V4 ......(X)

Expression of volume of the steel wire after formation for section (4).

V4=(π4d12)(d) ......(XI)

Expression of Moment of inertia about x axis of the wire.

Ix=(Ix)1+(Ix)2+(Ix)3+(Ix)4 ......(XII)

Here, the moment of inertia of section (1) about the x axis is (Ix)1, the moment of inertia of section (2) about the x axis is (Ix)2 ,the moment of inertia of section (3) about the x axis is (Ix)3 and the moment of inertia of section (4) about the x axis is (Ix)4.

Expression of moment of inertia of section (1) about the x axis.

(Ix)1=12m1d2 ......(XIII)

Expression of moment of inertia of section (2) about the x axis.

(Ix)2=12m2d2+m2d2 ......(XIV)

Expression of moment of inertia of section (3) about the x axis.

(Ix)3=112m3d2+m3{(d2)2+d2} ......(XV)

Expression of moment of inertia of section (4) about the x axis.

(Ix)4=112m4d2+m4{(d2)2+d2} ......(XVI)

Expression of Moment of inertia about y axis of the wire.

Iy=(Iy)1+(Iy)2+(Iy)3+(Iy)4 ......(XVII)

Here, the moment of inertia of section (1) about the y axis is (Iy)1, the moment of inertia of section (2) about the y axis is (Iy)2 ,the moment of inertia of section (3) about the y axis is (Iy)3 and the moment of inertia of section (4) about the y axis is (Iy)4.

Expression of moment of inertia of section (1) about the y axis.

(Iy)1=m1d2 ......(XVIII)

Expression of moment of inertia of section (2) about the y axis.

(Iy)2=m2d2 ......(XIX)

Expression of moment of inertia of section (3) about the y axis.

(Iy)3=m3d2 ......(XX)

Expression of moment of inertia of section (4) about the y axis.

(Iy)4=m4d2 ......(XXI)

Expression of Moment of inertia about z axis of the wire.

Iz=(Iz)1+(Iz)2+(Iz)3+(Iz)4 ......(XXII)

Here, the moment of inertia of section (1) about the z axis is (Iz)1, the moment of inertia of section (2) about the z axis is (Iz)2 ,the moment of inertia of section (3) about the z axis is (Iz)3 and the moment of inertia of section (4) about the z axis is (Iz)4.

Expression of moment of inertia of section (1) about the z axis.

(Iz)1=12m1d2 ......(XXIII)

Expression of moment of inertia of section (2) about the z axis.

(Iz)2=12m2d2+m2d2 ......(XXIV)

Expression of moment of inertia of section (3) about the z axis.

(Iz)3=13m3d2 ......(XXV)

Expression of moment of inertia of section (4) about the z axis.

(Iz)4=13m4d2 ......(XXVI)

Calculation:

Substitute 18 in for d1 and 18 in for d in Equation (V).

V1=(π4(18 in)2)(π(18 in))=(0.01227 in2)(π(18 in))=(0.694 in3)(1 ft12 in)(1 ft12 in)(1 ft12 in)=4.016×10−4 ft3

Substitute 4.016×10−4 ft3 for V2, 490 lb/ft3 for γST and 32.2 ft/s2 for g in Equation (IV).

m1=(490 lb/ft332.2 ft/s2)(4.016×10−4 ft3)=(15.217 lb⋅s2/ft4)(4.016×10−4 ft3)=6.1112×10−3 lb⋅s2/ft

Substitute 18 in for d1 and 18 in for d in Equation (VII).

V2=(π4(18 in)2)(π(18 in))=(0.01227 in2)(π(18 in))=(0.694 in3)(1 ft12 in)(1 ft12 in)(1 ft12 in)=4.016×10−4 ft3

Substitute 4.016×10−4 ft3 for V2, 490 lb/ft3 for γST and 32.2 ft/s2 for g in Equation (VI).

m2=(490 lb/ft332.2 ft/s2)(4.016×10−4 ft3)=(15.217 lb⋅s2/ft4)(4.016×10−4 ft3)=6.1112×10−3 lb⋅s2/ft

Substitute 18 in for d1 and 18 in for d in Equation (IX).

V3=(π4(18 in)2)(18 in)=(0.01227 in2)(18 in)=(0.22086 in3)(1 ft12 in)(1 ft12 in)(1 ft12 in)=1.2783×10−4 ft3

Substitute 1.2783×10−4 ft3 for V3, 490 lb/ft3 for γST and 32.2 ft/s2 for g in Equation (VIII).

m3=(490 lb/ft332.2 ft/s2)(1.2783×10−4 ft3)=(15.217 lb⋅s2/ft4)(1.2783×10−4 ft3)=1.945×10−3 lb⋅s2/ft

Substitute 18 in for d1 and 18 in for d in Equation (XI).

V4=(π4(18 in)2)(18 in)=(0.01227 in2)(18 in)=(0.22086 in3)(1 ft12 in)(1 ft12 in)(1 ft12 in)=1.2783×10−4 ft3

Substitute 1.2783×10−4 ft3 for V4, 490 lb/ft3 for γST and 32.2 ft/s2 for g in Equation (X).

m4=(490 lb/ft332.2 ft/s2)(1.2783×10−4 ft3)=(15.217 lb⋅s2/ft4)(1.2783×10−4 ft3)=1.945×10−3 lb⋅s2/ft

Substitute 6.1112×10−3 lb⋅s2/ft for m1 and 18 in for d in Equation (XIII).

(Ix)1=12(6.1112×10−3 lb⋅s2/ft)(18 in)2=12(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=12(6.1112×10−3 lb⋅s2/ft)(2.25 ft2)=6.8751×10−3 lb⋅ft⋅s2

Substitute 6.1112×10−3 lb⋅s2/ft for m2 and 18 in for d in Equation (XIV).

(Ix)2=12(6.1112×10−3 lb⋅s2/ft)(18 in)2+(6.1112×10−3 lb⋅s2/ft)(18 in)2=[12(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2+(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2]=(6.8751×10−3 ft⋅s2)+(13.7502×10−3 ft⋅s2)=20.6252×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m3 and 18 in for d in Equation (XV).

(Ix)3=[112(1.945×10−3 lb⋅s2/ft)(18 in)2+(1.945×10−3 lb⋅s2/ft){(18 in2)2+(18 in)2}]=[112(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2+(1.945×10−3 lb⋅s2/ft)[{(18 in2)(1 ft12 in)}2+{(18 in)(1 ft12 in)}2]]=(0.3647×10−3 lb⋅ft⋅s2)+(5.4712×10−3 lb⋅ft⋅s2)=5.8359×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m4 and 18 in for d in Equation (XVI).

(Ix)4=[112(1.945×10−3 lb⋅s2/ft)(18 in)2+(1.945×10−3 lb⋅s2/ft){(18 in2)2+(18 in)2}]=[112(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2+(1.945×10−3 lb⋅s2/ft)[{(18 in2)(1 ft12 in)}2+{(18 in)(1 ft12 in)}2]]=(0.3647×10−3 lb⋅ft⋅s2)+(5.4712×10−3 lb⋅ft⋅s2)=5.8359×10−3 lb⋅ft⋅s2

Substitute 6.8751×10−3 lb⋅ft⋅s2 for (Ix)1, 20.6252×10−3 lb⋅ft⋅s2 for (Ix)2, 5.8359×10−3 lb⋅ft⋅s2 for (Ix)3 and 5.8359×10−3 lb⋅ft⋅s2 for (Ix)4 in Equation (XII).

Ix=[(6.8751×10−3 lb⋅ft⋅s2)+(20.6252×10−3 lb⋅ft⋅s2)+(5.8359×10−3 lb⋅ft⋅s2)+(5.8359×10−3 lb⋅ft⋅s2)]=(27.5003×10−3 lb⋅ft⋅s2)+(11.6718×10−3 lb⋅ft⋅s2)=39.1721×10−3 lb⋅ft⋅s2

Thus, the mass moment of inertia of the wire with respect to x axis is 39.1721×10−3 lb⋅ft⋅s2

Substitute 6.1112×10−3 lb⋅s2/ft for m1 and 18 in for d in Equation (XVIII).

(Iy)1=(6.1112×10−3 lb⋅s2/ft)(18 in)2=(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=(6.1112×10−3 lb⋅s2/ft)(2.25 ft2)=13.7502×10−3 lb⋅ft⋅s2

Substitute 6.1112×10−3 lb⋅s2/ft for m2 and 18 in for d in Equation (XIX).

(Iy)2=(6.1112×10−3 lb⋅s2/ft)(18 in)2=(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=(6.1112×10−3 lb⋅s2/ft)(2.25 ft2)=13.7502×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m3 and 18 in for d in Equation (XX).

(Iy)3=(1.945×10−3 lb⋅s2/ft)(18 in)2=(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=4.3769×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m4 and 18 in for d in Equation (XXI).

(Iy)4=(1.945×10−3 lb⋅s2/ft)(18 in)2=(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=4.3769×10−3 lb⋅ft⋅s2

Substitute 13.7502×10−3 lb⋅ft⋅s2 for (Iy)1, 13.7502×10−3 lb⋅ft⋅s2 for (Iy)2, 4.3769×10−3 lb⋅ft⋅s2 for (Ix)3 and 4.3769×10−3 lb⋅ft⋅s2 for (Ix)4 in Equation (XVII).

Iy=[(13.7502×10−3 lb⋅ft⋅s2)+(13.7502×10−3 lb⋅ft⋅s2)+(4.3769×10−3 lb⋅ft⋅s2)+(4.3769×10−3 lb⋅ft⋅s2)]=(27.5004×10−3 lb⋅ft⋅s2)+(8.7538×10−3 lb⋅ft⋅s2)=36.2542×10−3 lb⋅ft⋅s2

Thus, the mass moment of inertia of the wire with respect to y axis is 36.2542×10−3 lb⋅ft⋅s2.

Substitute 6.1112×10−3 lb⋅s2/ft for m1 and 18 in for d in Equation (XXIII).

(Iz)1=12(6.1112×10−3 lb⋅s2/ft)(18 in)2=12(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=12(6.1112×10−3 lb⋅s2/ft)(2.25 ft2)=6.8751×10−3 lb⋅ft⋅s2

Substitute 6.1112×10−3 lb⋅s2/ft for m2 and 18 in for d in Equation (XXIV).

(Iz)2=12(6.1112×10−3 lb⋅s2/ft)(18 in)2+(6.1112×10−3 lb⋅s2/ft)(18 in)2=[12(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2+(6.1112×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2]=(6.8751×10−3 ft⋅s2)+(13.7502×10−3 ft⋅s2)=20.6252×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m3 and 18 in for d in Equation (XXV).

(Iz)3=13(1.945×10−3 lb⋅s2/ft)(18 in)2=13(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=1.4590×10−3 lb⋅ft⋅s2

Substitute 1.945×10−3 lb⋅s2/ft for m4 and 18 in for d in Equation (XXVI).

(Iz)4=13(1.945×10−3 lb⋅s2/ft)(18 in)2=13(1.945×10−3 lb⋅s2/ft){(18 in)(1 ft12 in)}2=1.4590×10−3 lb⋅ft⋅s2

Substitute 6.8751×10−3 lb⋅ft⋅s2 for (Iz)1, 20.6252×10−3 lb⋅ft⋅s2 for (Iz)2, 1.4590×10−3 lb⋅ft⋅s2 for (Iz)3 and 1.4590×10−3 lb⋅ft⋅s2 for (Iz)4 in Equation (XXII).

Iz=[(6.8751×10−3 lb⋅ft⋅s2)+(20.6252×10−3 lb⋅ft⋅s2)+(1.4590×10−3 lb⋅ft⋅s2)+(1.4590×10−3 lb⋅ft⋅s2)]=(27.5003×10−3 lb⋅ft⋅s2)+(3.098×10−3 lb⋅ft⋅s2)=30.5983×10−3 lb⋅ft⋅s2

Thus, the mass moment of inertia of the wire with respect to z axis is 30.5983×10−3 lb⋅ft⋅s2

Conclusion:

The mass moment of inertia of the wire with respect to x axis is 39.1721×10−3 lb⋅ft⋅s2.

The mass moment of inertia of the wire with respect to y axis is 36.2542×10−3 lb⋅ft⋅s2.

The mass moment of inertia of the wire with respect to z axis is 30.5983×10−3 lb⋅ft⋅s2

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