VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 18.3, Problem 18.129P
To determine

The precession axis, rates of precession, and the spin rate of the satellite after the impact.

Expert Solution & Answer
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Answer to Problem 18.129P

The precession along x, y, and z axis (θx,θy,andθz) is 125°_, 36.4°_, and 80.7°_ respectively.

The precession rate (ϕ˙) is 0.949rad/s_.

The spin rate (ψ˙) is 0.16rad/s_.

Explanation of Solution

Given information:

The weight of geostationary satellite (W) is 800 lb.

The angular velocity of the satellite (ω0) is (1.5rad/s)j.

The weight of a meteorite W is 6 oz.

The travelling velocity of the meteorite v0 is (1,600ft/s)i+(1,300ft/s)j+(4,000ft/s)k.

The distance value b is 20 in..

The radii of gyration of the satellite along x, y, and z direction (k¯x, k¯y, and k¯z) is 28.8 in., 32.4 in., and 28.8 in. respectively.

Calculation:

Determine the mass of the satellite (m).

m=Wg

Here, g is the acceleration due to gravity.

Substitute 800 lb for W and 32.2ft/s2 for g.

m=80032.2=24.845lbs2/ft

Determine the principal moment of inertia along x axis.

I¯x=mk¯x2

Substitute 24.845lbs2/ft for m and 28.8 in. for k¯x.

I¯x=24.845×(28.8in.×1ft12in.)2=143.11lbs2ft

Determine the principal moment of inertia along y axis.

I¯y=mk¯y2

Substitute 24.845lbs2/ft for m and 32.4 in. for k¯y.

I¯y=24.845×(32.4in.×1ft12in.)2=181.12lbs2ft

Determine the principal moment of inertia along z axis.

I¯z=mk¯z2

Substitute 24.845lbs2/ft for m and 28.8 in. for k¯z.

I¯z=24.845×(28.8in.×1ft12in.)2=143.11lbs2ft

Determine the mass of the meteorite m.

m=Wg

Substitute 6 oz for W and 32.2ft/s2 for g.

m=6oz×0.0625lb1oz32.2=0.0116lbs2/ft

Determine the initial moment of meteorite (p0).

p0=mv0

Substitute 0.0116lbs2/ft for m and (1,600ft/s)i+(1,300ft/s)j+(4,000ft/s)k for v0.

p0=0.01164×((1,600ft/s)i+(1,300ft/s)j+(4,000ft/s)k)=(18.62lbs)i+(15.13lbs)j+(46.56lbs)k

Consider that the position of the satellite mass center plus the meteorite is essentially that of the satellite alone.

Determine the position of the point B relative to the mass center.

rB/G=xiyj

Here, x is the horizontal distance and y is the vertical distance.

Substitute 42 in. for x and 20 in. for y.

rB/G=(42in.×1ft12in.)i(20in.×1ft12in.)j=(3.5ft)i(1.667ft)j

The angular velocity of satellite before impact (ω0) is (1.5rad/s)j.

The angular velocity of satellite before impact along x, y, and z axis is (ω0)x=0, (ω0)y=1.5rad/s, and (ω0)z=0.

Determine the angular momentum of satellite–meteorite system before impact (HG)0.

(HG)0=I¯yω0j+rB/G×p0

Substitute 181.12lbs2ft for I¯y, 1.5rad/s for ω0, (3.5ft)i+(1.667ft)j for rB/G, and (18.62lbs)i+(15.13lbs)j+(46.56lbs)k for p0.

(HG)0=(181.12×1.5)j+|ijk3.51.667018.6215.1346.56|={271.68j+i(1.667×46.560)j(3.5×46.560)+k(3.5×15.13(1.667×18.62))}=271.68j77.61i162.96j+21.92k=(77.61i+108.72j+21.92k)lbsft

Principle of impulse and momentum for satellite–meteorite system:

The value of moments about G is (HG)1=(HG)0=HG.

The expression for HG is HG=I¯xωxi+I¯yωyj+I¯zωzk.

Determine the angular velocity about x axis.

ωx=(HG)xIx

Substitute 77.61lbsft for (HG)x and 143.11lbs2ft for Ix.

ωx=77.61143.11=0.5423rad/s

Determine the angular velocity about y axis.

ωy=(HG)yIy

Substitute 108.72lbsft for (HG)y and 181.12lbs2ft for I¯y.

ωy=108.72181.12=0.6rad/s

Determine the angular velocity about z axis.

ωz=(HG)zIz

Substitute 21.92lbsft for (HG)z and 143.11lbs2ft for I¯z.

ωz=21.92143.11=0.153rad/s

Determine the vector format of angular velocity using the relation;

ω=ωxi+ωyj+ωzk

Substitute 0.5423rad/s for ωx, 0.6rad/s for ωy, and 0.153rad/s for ωz.

ω=(0.5423rad/s)i+(0.6rad/s)j+(0.153rad/s)k

Determine the magnitude of the angular velocity ω.

ω=ωx2+ωy2+ωz2

Substitute 0.5423rad/s for ωx, 0.6rad/s for ωy, and 0.153rad/s for ωz.

ω=(0.5423)2+(0.6)2+(0.153)2=0.823rad/s

Determine the magnitude of angular momentum HG using the relation.

HG=(HG)x2+(HG)y2+(HG)z2

Substitute 77.61lbsft for (HG)x, 108.72lbsft for (HG)y, and 21.92lbsft for (HG)z.

HG=(77.61)2+(108.72)2+(21.92)2=135lbsft

Motion after impact:

The moment of inertia about x and y axis is equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis.

The moment of inertia about the symmetry axis is I=I¯y=181.12lbs2ft.

The moment of inertia about a transverse axis through G is I=I¯x=I¯z=143.11lbs2ft.

The precession is retrograde when the value of I>I.

Determine the angle θ about x axis.

cosθx=(HG)xHG

Substitute 77.61lbsft for (HG)x and 135lbsft for HG.

cosθx=77.61135θx=cos1(0.575)θx=125°

Determine the angle θ about y axis.

cosθy=(HG)yHG

Substitute 108.72lbsft for (HG)y and 135lbsft for HG.

cosθy=108.72135θy=cos1(0.805)θy=36.4°

Determine the angle θ about z axis.

cosθz=(HG)zHG

Substitute 21.92lbsft for (HG)z and 135lbsft for HG.

cosθz=21.92135θz=cos1(0.162)θz=80.7°

Thus, the precession along x, y, and z axis (θx,θy,andθz) is 125°_, 36.4°_, and 80.7°_ respectively.

The angle θ between the spin axis (y–axis) and the precession axis remains constant θ=θy=36.4°.

Determine the angle γ between the angular velocity and the spin axis.

cosγ=ωyω

Substitute 0.6rad/s for ωy and 0.823rad/s for ω.

cosγ=0.60.823γ=cos1(0.729)γ=43.2°

Determine the value of (γθ).

Substitute 43.2° for γ and 36.4° for θ.

(γθ)=43.2°36.4°=6.8°

Draw the free body diagram of precession and spin axis as in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 18.3, Problem 18.129P , additional homework tip  1

Draw the free body diagram of triangle of vector addition as in Figure (2).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 18.3, Problem 18.129P , additional homework tip  2

Write the relation between the angles using the sine law.

ϕ˙sinγ=ψ˙sin(γθ)=ωsinθ

Determine the precession rate (ϕ˙) using law of sines.

ϕ˙=ωsinγsinθ

Substitute 0.823rad/s for ω, 43.2° for γ, and 36.4° for θ.

ϕ˙=0.823×sin43.2°sin36.4°=0.949rad/s

Therefore, the precession rate (ϕ˙) is 0.949rad/s_.

Determine the rate of spin ψ˙ using the relation.

ψ˙=ωsin(γθ)sinθ

Substitute 0.823rad/s for ω, 6.8° for (γθ), and 36.4° for θ.

ψ˙=0.823sin6.8°sin36.4°=0.16rad/s

The precession is retrograde due to value of γ is greater than the angle θ.

Therefore, the spin rate (ψ˙) is 0.16rad/s_.

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Chapter 18 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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