VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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Chapter 18.3, Problem 18.140P
To determine

(a)

The maximum value of the angle θ in the ensuing motion.

Expert Solution
Check Mark

Answer to Problem 18.140P

The maximum value of the angle θ in the ensuing motion is 76.290°.

Explanation of Solution

Given information:

The height of the solid cone is 180mm, the radius of the circular cone is 60mm, the position angle is 30°, the rate of the spin is 300rad/s and the rate of the precession is 4rad/s.

Write the expression for the height of the center of gravity from the apex.

C=34h ...... (I)

Here, the height of the solid cone is h.

Write the expression for the mass moment of the inertia.

I=310mr2 ...... (II)

Here, the mass of the solid cone is m and the radius of the solid cone is r.

Write the expression for the mass moment of the inertia in the transverse axis.

I=320mr2+35mh2 ...... (III)

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

ωx=ϕ˙osinθo ...... (IV)

Here, the rate of the precession is ϕ˙o and the position angle is θo.

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

ωy=θ˙o ...... (V)

Here, the rate of the change of the angle is θ˙o.

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

ωz=ψ˙o+ϕ˙osinθo ...... (VI)

Here, the rate of the spin is ψ˙o.

Write the expression for the kinetic energy.

Tm=12Im(ωx2+ωy2)+12Imωz2 ...... (VII)

Write the expression for the potential energy.

Vm=gCcosθo ...... (VIII)

Write the expression for the total energy.

Em=Tm+Vm ...... (IX)

Write the expression for the constant β.

βm=Iωzm ...... (X)

Write the expression for the constant α.

αm=Imϕ˙osin2θo+βmcosθo ...... (XI)

Write the expression for the force.

F(x)=[2Emβ2m2Im2gC](1x2)mI(αmβxm)20=[2Emβ2m2Im2gC](1x2)mI(αmβxm)2 ...... (XII)

Here, the distance along the x -axis is x.

Write the expression for the maximum angle.

θmax=cos1(xmin) ...... (XIII)

Calculation:

Substitute 180mm for h in Equation (I).

C=34(180mm)=34(180mm(1m103mm))=0.135m

Substitute 60mm for r in Equation (II).

I=310m(60mm)2Im=310(60mm(1m103mm))2Im=1.08×103m2

Substitute 180mm for h and 60mm for r in Equation (III).

I=320m(60mm)2+35m(180mm)2Im=320(60mm(1m103mm))2+35(180mm(1m103mm))2Im=5.4×104m2+0.01944m2Im=19.98×103m2

Substitute 30° for θo and 4rad/s for ϕ˙o in Equation (IV).

ωx=(4rad/s)sin(30°)=(4rad/s)(0.5)=2rad/s

Substitute 0 for θ˙o in Equation (V).

ωy=0

Substitute 300rad/s for ψ˙o, 4rad/s for ϕ˙o and 30° for θo in Equation (VI)

ωz=(300rad/s)+(4rad/s)cos30°=(300rad/s)+(4rad/s)(0.8660)=300rad/s3.464rad/s=296.53rad/s

Substitute 2rad/s for ωx, 0 for ωy, 296.53rad/s for ωz, 19.98×103m2 for Im and 1.08×103m2 for Im in Equation (VII).

Tm=12(19.98×103m2)((2rad/s)2+(0)2)+12(1.08×103m2)(296.53rad/s)2=12(19.98×103m2)(4rad2/s2)+12(1.08×103m2)(87933.53rad2/s2)=3.996×103m2/s2+47484.106×103m2/s2=47.451m2/s2

Substitute 9.81m/s2 for g, 0.135m for C and 30° for θo in Equation (VIII).

Vm=(9.81m/s2)(0.135m)cos30°=(9.81m/s2)(0.135m)(0.8660)=1.146m2/s2

Substitute 1.146m2/s2 for Vm and 47.451m2/s2 for Tm in Equation (IX).

Em=47.451m2/s2+1.146m2/s2=48.597m2/s2

Substitute 296.53rad/s for ωz and 1.08×103m2 for Im in Equation (X).

βm=(296.53rad/s)(1.08×103m2)=0.3202m2/s

Substitute 19.98×103m2 for Im, 0.3202m2/s for βm, 4rad/s for ϕ˙o and 30° for θo in Equation (XI).

αm=(19.98×103m2)(4rad/s)sin2(30°)+(0.3202m2/s)cos(30°)=(19.98×103m2)(4rad/s)(0.25)+(0.3202m2/s)(0.8660)=0.01998m2/s+0.2772m2/s=0.2572m2/s

Substitute 1.08×103m2 for Im, 0.2572m2/s for αm, 0.3202m2/s for βm, 19.98×103m2 for Im, 48.597m2/s2 for Em, 9.81m/s2 for g and 0.135m for C in Equation (XII).

0={[2(48.597m2/s2)(0.3202m2/s)2(1.08×103m2)2(9.81m/s2)(0.135m)](1x2)(119.98×103m2)((0.2572m2/s)(0.3202m2/s)x)2}{[(97.194m2/s2)94.93m2/s22.648m2/s2](1x2)(0.1986m2/s2x2)}=0{(0.384m2/s2)(1x2)((0.1986m2/s2)x2)}=0x=0.237,0.860,1.73

Select the minimum value of x for the maximum angle.

Substitute 0.237 for xmin in Equation (XIII).

θmax=cos1(0.237)=76.290°

Conclusion:

The maximum value of the angle θ in the ensuing motion is 76.290°.

To determine

(b)

The corresponding value of the rate of precession.

The corresponding value of the rate of spin.

Expert Solution
Check Mark

Answer to Problem 18.140P

The corresponding value of the rate of precession is 9.341rad/s.

The corresponding value of the rate of spin is 294.316rad/s.

Explanation of Solution

Write the expression for the rate of the precession.

ϕ˙=αmβmcosθImsin2θ ...... (XIV)

Write the expression for the rate of the spin.

ψ˙=ωzϕ˙cosθ ...... (XV)

Calculation:

Substitute 0.2572m2/s for αm, 0.3202m2/s for βm, 19.98×103m2 for Im and 76.290° for θ in Equation (XIV).

ϕ˙=(0.2572m2/s)(0.3202m2/s)cos(76.290°)(19.98×103m2)sin2(76.290°)=(0.2572m2/s)(0.3202m2/s)(0.2370)(19.98×103m2)(0.9715)=(0.1813m2/s)(19.41×103m2)=9.341rad/s

Substitute 9.341rad/s for ϕ˙, 296.53rad/s for ωz and 76.290° for θ in Equation (XV).

ψ˙=296.53rad/s(9.341rad/s)cos(76.290)=296.53rad/s(2.213rad/s)=294.316rad/s

Conclusion:

The corresponding value of the rate of precession is 9.341rad/s.

The corresponding value of the rate of spin is 294.316rad/s.

To determine

(c)

The value of the angle at which the sense of precession is reversed.

Expert Solution
Check Mark

Answer to Problem 18.140P

The value of the angle at which the sense of precession is reversed is 36.55°.

Explanation of Solution

Write the expression for the rate of the precession.

ϕ˙=αmβmcosθsinθαmβmcosθsinθ=0αmβmcosθ=0cosθ=αmβm ...... (XVI)

Calculation:

Substitute 0.2572m2/s for αm and 0.3202m2/s for βm in Equation (XVI).

cosθ=0.2572m2/s0.3202m2/sθ=cos1(0.8032)θ=36.55°

Conclusion:

The value of the angle at which the sense of precession is reversed is 36.55°.

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Chapter 18 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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