The largest value of β in the ensuing motion.

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Answer 1

Question

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

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Answer 2

Question

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

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Answer 3

Question

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

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Answer 4

Question

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

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Answer 5

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Answer 6

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)

Consider $β˙=0$ for the maximum value of $β$ .

Substitute $ϕ˙0sec2β$ for $ϕ˙$ , 0 for $(ψ˙−ϕ˙sinβ)$ , and 0 for $β˙$ in Equation (5).

$11(ϕ˙0sec2β)2cos2β+11(0)2−10gasinβ=11ϕ˙0211ϕ˙02sec4β×1sec2β−10gasinβ=11ϕ˙0211ϕ˙02sec2β−11ϕ˙02=10gasinβ11ϕ˙02(1−cos2βcos2β)=10gasinβ$

$ϕ˙02(sin2βcos2β)=1011gasinβϕ˙02=1011gasinβ×cos2βsin2βϕ˙02=1011gacos2βsinβ$ (6)

Substitute $17g/11a$ for $ϕ˙0$ in Equation (6).

$(17g/11a)2=1011gacos2βsinβ17g11a=1011ga(1−sin2β)sinβ17sinβ=10−10sin2β10sin2β+17sinβ−10=0$ (7)

Solve the Equation (7).

The value of $sinβ$ is 0.462 and –2.162.

Determine the largest value of $βmax$ using the relation.

$sinβmax=0.462βmax=sin−1(0.462)βmax=27.5°$

Therefore, the largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Answer 7

Chapter 18.3, Problem 18.141P

To determine

The largest value of $β$ in the ensuing motion.

Answer 8

Expert Solution & Answer

Answer to Problem 18.141P

The largest value of $βmax$ in the ensuing motion is $27.5°_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The position of the sphere $β$ is zero.

The rate of precession $ϕ˙0=17g/11a$ .

Calculation:

Conservation of angular momentum about the Z and z axes:

The only external forces are acting in homogenous sphere is weight of the sphere and reaction at A. Hence, the angular momentum is conserved about the Z and z axes.

Choose the principal axes $Axyz$ with taking y horizontal and pointing into the paper.

Write the expression for the angular velocity $ω$ .

$ω=−ϕ˙cosβi+β˙j+(ψ˙−ϕ˙sinβ)k$

The principal moment of inertia are $Ix=Iy=m(25a2+(2a)2)$ and $Iz=25ma2$ .

Draw the Free body diagram of homogeneous sphere and the forces acting on it as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.3, Problem 18.141P

Write the expression for the angular momentum about point A.

$HA=Ixωxi+Iyωyj+Izωzk$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$HA=m(25a2+(2a)2)(−ϕ˙cosβ)i+m(25a2+(2a)2)β˙j+25ma2(ψ˙−ϕ˙sinβ)k=−225ma2ϕ˙cosβi+225ma2β˙j+25ma2(ψ˙−ϕ˙sinβ)k$

Consider $Hz=constant$ or $HA⋅K=constant$ .

The scalar value of $i⋅K=−cosβ$ , $j⋅K=0$ , and $k⋅K=−sinβ$ .

Determine the conservation of angular momentum about fixed Z axis $HA⋅K$ .

$HA⋅K=constant$

Substitute $−225ma2ϕ˙cosβi+225ma2βj+25ma2(ψ˙−ϕ˙sinβ)k$ for $HA$ , $−cosβ$ for $(i⋅K)$ , 0 for $(j⋅K)$ , and $−cosβ$ for $(k⋅K)$ .

${−225ma2ϕ˙cosβ(i⋅K)+225ma2β˙(j⋅K)+25ma2(ψ˙−ϕ˙sinβ)(k⋅K)}=constant{−225ma2ϕ˙cosβ(−cosβ)+225ma2β˙(0)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant{−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)}=constant$ (1)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (1).

${−225ma2ϕ˙0cos0(−cos0)+25ma2(0−ϕ˙0sin0)(−sin0)}=constantconstant=225ma2ϕ˙0$

Substitute $225ma2ϕ˙0$ for constant in Equation (1).

$−225ma2ϕ˙cosβ(−cosβ)+25ma2(ψ˙−ϕ˙sinβ)(−sinβ)=225ma2ϕ˙025ma2[11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ]=25×11ϕ˙011ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙0$ (2)

Determine the constant value using the angular momentum along z–axis.

$Hz=constantIzωz=constant$

Substitute $25ma2$ for $Iz$ and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$25ma2(ψ˙−ϕ˙sinβ)=constant$ (3).

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $ψ˙$ , and 0 for $β$ in Equation (3).

$25ma2(0−ϕ˙0sin(0))=constantconstant=0$

Substitute 0 for constant in Equation (3).

$25ma2(ψ˙−ϕ˙sinβ)=0ψ˙−ϕ˙sinβ=0$

Substitute 0 for $(ψ˙−ϕ˙sinβ)$ in Equation (2).

$11ϕ˙cos2β−(ψ˙−ϕ˙sinβ)sinβ=11ϕ˙011ϕ˙cos2β−0(sinβ)=11ϕ˙0ϕ˙=11ϕ˙011cos2βϕ˙=ϕ˙0sec2β$

Conservation of energy:

Determine the value of kinetic energy T.

$T=12(Ixωx2+Iyωy2+Izωz2)$

Substitute $m(25a2+(2a)2)$ for $Ix$ , $−ϕ˙cosβ$ for $ωx$ , $m(25a2+(2a)2)$ for $Iy$ , $β˙$ for $ωy$ , , $25ma2$ for $Iz$ , and $(ψ˙−ϕ˙sinβ)$ for $ωz$ .

$T={12(m(25a2+(2a)2)(−ϕ˙cosβ)2+m(25a2+(2a)2)(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)}=12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)$

Select the datum at $β=0$ .

Determine the value of conservation of energy using the relation.

$T+V=constant$

Here, E is the constant and V is the potential energy.

Substitute $12(225ma2ϕ˙2cos2β+225ma2(β)2+25ma2(ψ˙−ϕ˙sinβ)2)$ for T and $−2mgasinβ$ for V.

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=constant$ (4)

Substitute $ϕ˙0$ for $ϕ˙$ , 0 for $β$ , 0 for $β˙$ , and 0 for $ψ˙$ in Equation (4).

${12(225ma2ϕ˙02cos20+225ma2(0)2+25ma2(0−ϕ˙0sin0)2)−2mgasin0}=constantconstant=115ma2ϕ˙02$

Substitute $115ma2ϕ˙02$ for constant in Equation (4).

${12(225ma2ϕ˙2cos2β+225ma2(β˙)2+25ma2(ψ˙−ϕ˙sinβ)2)−2mgasinβ}=115ma2ϕ˙0212×25ma2((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=115ma2ϕ˙02((11ϕ˙2cos2β+11(β˙)2+(ψ˙−ϕ˙sinβ)2)−10gasinβ)=11ϕ˙02$ (5)