(a) The velocity of the mass centre G .

Question

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Answer 1

Question

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

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Answer 2

Question

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

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Answer 3

Question

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

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Answer 4

Question

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

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Answer 5

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Answer 6

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

Answer 7

Chapter 18.1, Problem 18.28P

To determine

(a)

The velocity of the mass centre G.

Answer 8

Expert Solution

Answer to Problem 18.28P

The velocity of the mass centre G is (0.300 m/s)j.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The mass of the each plate is 4 kg and the corresponding impulse is (2.4 N⋅s)j.

Write the expression for moment of the inertia about the x- axis.

Ix=m(14r2+d2) ...... (I).

Here, the distance of mass centre from the circular plate is d and the radius of the circular plate is r.

Write the expression for moment of the inertia about the y- axis.

Iy=32mr2 ...... (II)

Write the expression for moment of the inertia about the z- axis.

Iz=m(54r2+d2) ...... (III)

Write expression for the product moment of inertia about the plane xy.

Ixy=−mrd ...... (IV)

Write expression for the product moment of inertia about the plane yz.

Iyz=0

Write expression for the product moment of inertia about the plane zx.

Izx=0

The figure below shows the effective kinetic diagram of the system.

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.28P

Figure-(1)

Write the expression for the impulse about point D.

(FΔt)k+(AΔt)j=2m(vxi+vzk) ...... (V)

Write the expression for the velocity of the mass centre of the system.

v=vxi+vyj+vzk ....... (VI)

Here, the velocity of the mass centre in x- direction is vx, the velocity of the mass centre in y- direction is vy and the velocity of the mass centre in z- direction is vz.

Calculation:

For upper plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For lower plate:

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (I).

Ix=(4 kg)(14(180 mm)2+(150 mm)2)=(4 kg)(14(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=(4 kg)(0.25(0.180 m2)+(0.150 m2))=0.1224 kg⋅m2

Substitute 4 kg for m and 180 mm for r in Equation (II).

Iy=32(4 kg)(180 mm)2=32(4 kg)(180 mm(1 m103 mm))2=32(4 kg)(0.180 m2)=0.1944 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (III).

Iz=4 kg(54(180 mm)2+(150 mm)2)=4 kg(54(180 mm(1 m103 mm))2+(150 mm(1 m103 mm))2)=4 kg(54(0.180 m2)+(0.150 m2))=0.252 kg⋅m2

Substitute 4 kg for m, 180 mm for r and 150 mm for d in Equation (IV).

Ixy=−(4 kg)(180 mm)(150 mm)=−(4 kg)(180 mm(1 m103 mm))(150 mm(1 m103 mm))=−(4 kg)(0.180 m)(0.150 m)=−0.108 kg⋅m2

For assembly the inertia of the moment of both the plate is added.

Ix=0.2448 kg⋅m2

Iy=0.3888 kg⋅m2

Iz=0.504 kg⋅m2

Ixy=−0.216 kg⋅m2

Substitute 4 kg for m and (2.4 N⋅s)j for FΔt in Equation (V).

((2.4 N⋅s)j)=2(4 kg)(vxi+vyj+vzk)((2.4 N⋅s)j)=((8 kg)vxi+(8 kg)vyj+(8 kg)vzk) ...... (VII)

Compare coefficient of i in Equation (VII).

(8 kg)vxi=0vx=0

Compare coefficient of k in Equation (VII).

(8 kg)vzi=0vz=0

Compare coefficient of j in Equation (VII).

(8 kg)vyj=((2.4 N⋅s)j)vy=(2.4 N⋅s)(8 kg)vy=0.3 N⋅s/kg(1 N1 kg⋅m/s2)vy=(0.300 m/s)j

Substitute 0 for vz, (0.300 m/s)j for vy and 0 for vx in Equation (VI).

v=(0)i+((0.300 m/s)j)+(0)k=(0.300 m/s)j

Conclusion:

The velocity of the mass centre G is (0.300 m/s)j.

Answer 13

To determine

(b)

The angular velocity of the assembly.

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s

Substitute −0.860 rad/s for ωz, −3.46 rad/s for ωx and 1.923 rad/s for ωy in Equation (X).

ω=(−3.46 rad/s)i+(1.923 rad/s)j+(−0.860 rad/s)k=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k

Conclusion:

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Answer 14

To determine

(b)

The angular velocity of the assembly.

Answer 15

Expert Solution

Answer to Problem 18.28P

The angular velocity of the assembly is ω=(−0.9154 rad/s)i+(0.5769 rad/s)j−(0.860 rad/s)k.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the moments about centre point G.

[−ri+dj]×FΔt=Hxi+Hyj+Hzk ...... (VIII)

Here, the angular momentum about x- direction is Hx, the angular momentum about y- direction is Hy and the angular momentum about z- direction is Hz.

Write the expression for the angular momentum in x- direction.

Hx=Ixωx−Ixyωy−Ixzωz ...... (IX)

Write the expression for the angular momentum in y- direction.

Hy=−Ixyωx+Iyωy−Ixzωz ...... (X)

Write the expression for the angular momentum in z- direction.

Hz=−Ixyωx+Iyzωy−Izωz ...... (XI)

Write the expression for the angular velocity of the assembly.

ω¯=ωxi+ωyj+ωzk ...... (XII)

Calculation:

Substitute (2.4 N⋅s)j for FΔt, 180 mm for r and 150 mm for d in Equation (VIII).

[−(180 mm)i+(180 mm)k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk{[−(180 mm(1 m103 mm))i+(180 mm(1 m103 mm))k]×((2.4 N⋅s)j)=Hxi+Hyj+Hzk}[0.432 N⋅s⋅m(1 kg⋅m/s1 N)]×(−k−i)=Hxi+Hyj+Hzk[0.432 kg⋅m2]×(−k−i)=Hxi+Hyj+Hzk .... (XIII)

Compare coefficient of i in Equation (XIII).

Hx=−0.432 kg⋅m2

Compare coefficient of j in Equation (XIII).

Hy=0 kg⋅m2

Compare coefficient of k in Equation (XIII).

Hz=−0.432 kg⋅m2

Substitute −0.432 kg⋅m2 for Hx, 0.2448 kg⋅m2 for Ix, 0 for Izx and 0.216 kg⋅m2 for Ixy in Equation (IX).

−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy−(0)ωz−0.432 kg⋅m2=(0.2448 kg⋅m2)ωx−(0.2448 kg⋅m2)ωy ...... (XIV)

Substitute 0 for Hy, 0.3888 kg⋅m2 for Iy, 0.216 kg⋅m2 for Ixy and 0 for Ixz in Equation (X).

0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωy−(0)ωz0=(0.216 kg⋅m2)ωx+(0.3888 kg⋅m2)ωyωx=−(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) ..... (XV)

Substitute −0.432 kg⋅m2 for Hz, 0.504 kg⋅m2 for Iz, 0.216 kg⋅m2 for Ixy and 0 for Iyz in Equation (X).

−0.432 kg⋅m2=−0−0−(0.502 kg⋅m2)ωzωz=−0.432 kg⋅m20.502 kg⋅m2ωz=−0.860 rad/s

Substitute −(0.3888 kg⋅m2)ωy(0.216 kg⋅m2) for ωx in Equation (XIV).

Substitute 1.9231 rad/s for ωy in Equation (XV).

ωx=−(0.3888 kg⋅m2)(1.9231 rad/s)(0.216 kg⋅m2)=−(0.3888 kg⋅m2(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m))(1.9231 rad/s)(0.216 kg⋅m2)(1 N1 kg⋅m/s2)(1 (m/s)(rad/s)1 N⋅m)=−0.7447 rad/s0.216 rad/s=−3.46 rad/s