Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977237
Author: BEER
Publisher: MCG
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Chapter 18.3, Problem 18.129P

An 800-lb geostationary satellite is spinning with an angular velocity ω 0 = ( 1.5 rad/s ) j when it is hit at B by a 6-oz meteorite traveling with a velocity v 0 = ( 1600 ft/s ) i + ( 1300 ft/s ) j + ( 4000 ft/s ) k relative to the satellite. Knowing that b = 20 in. and that the radii of gyration of the satellite are k ¯ x = k ¯ z = 28.8 in. and k ¯ y = 32.4 in., determine the precession axis and the rates of precession and spin of the satellite after the impact.

  Chapter 18.3, Problem 18.129P, An 800-lb geostationary satellite is spinning with an angular velocity 0=(1.5rad/s)j when it is hit

Expert Solution & Answer
Check Mark
To determine

The precession angles.

The rate of precession of the satellite after the impact.

The rate of spin axis of the satellite after the impact.

Answer to Problem 18.129P

Thus, the precession angles about the x-axis, y-axis and z-axis are 1250, 36.700 and 80.70 respectively.

The rate of precession of the satellite after impact is 0.942rad/s.

The rate of spin axis of the satellite after impact is 0.1561rad/s.

Explanation of Solution

Given information:

The weight of the geostationary satellite is 800lb, angular velocity of the satellite is 1.5(rad/s)j, the weight of the meteorite is 6oz, the initial velocity of the meteorite is (1600ft/s)i+(1300ft/s)j+(4000ft/s)k, the radius of gyration of the satellite along x-and z axes are 28.8in., the radius of gyration of the satellite along y-axis is 32.4in., the distance between point B and G is (42in.)i(20in)j.

Write the expression of the mass of the meteorite.

mM=WMg   ....... (I)

Here, weight of the meteorite is WM and gravitational acceleration is g.

Write the expression of the mass of the satellite.

mS=WSg   ....... (II)

Here, weight of the meteorite is WS.

Write the Expression of the moment of inertia along x-axis.

I¯x=mSkx2   ....... (III)

Here, the radius of gyration of the satellite along x-axis is kx.

Write the Expression of the moment of inertia along the y-axis.

I¯y=mSky2   ....... (IV)

Here, the radius of gyration of the satellite along y-axis is ky.

Write the Expression of the moment of inertia along z-axis.

I¯z=mSkz2   ....... (V)

Here, the radius of gyration of the satellite along z-axis is kz.

Write the expression of the initial momentum of the meteorite.

M0=mMv0   ....... (VI)

Here, the initial velocity of meteorite is v0.

Write the expression of initial angular momentum of the satellite before the impact.

(HG)S=I¯x(ωx)0i+I¯y(ωy)0j+I¯z(ωz)0k   ....... (VII)

Here, the initial angular velocity along x-axis is (ωx)0, the initial angular velocity along y-axis is (ωy)0, the initial angular velocity along z-axis is (ωz)0.

Write the expression of the angular momentum about point B before impact.

(HG)B=rBG×M0+(HG)S   ....... (VIII)

Write the expression of angular momentum of the satellite after impact.

(HG)S=I¯xωxi+I¯yωyj+I¯zωzk

Here, the angular velocity after impact along x-axis is ωx, the initial angular velocity after impact along y-axis is ωy, the initial angular velocity after the impact along z-axis is ωz.

According to the momentum equilibrium, the angular momentum before and after the impact of meteorite at point B will be same.

(HG)B=(HG)S   ....... (X)

Write the expression of magnitude net angular velocity after impact of meteorite.

ω=ωxi+ωyj+ωzkω=(ωx)2+(ωy)2+(ωy)2   ....... (XI)

Write the expression of magnitude net angular momentum after impact of meteorite.

HG=(Ixωx)2+(Iyωy)2+(Iyωy)2   ....... (XII)

The satellite is symmetrical about y-axis.

Write the expression of moment of inertia along the symmetry axis.

I=I¯y

Figure-(1) shows the precession axis and the spin axis.

Vector Mechanics For Engineers, Chapter 18.3, Problem 18.129P , additional homework tip  1

Figure-(1)

Here, angle of precession axis with respect to x, y and z axes is defined in cosine, the angular velocity is ω, the angle of angular velocity with spin axis is γ.

Write the expression of the angle of precession axis with respect to x-axis.

cosθx=I¯xωxHG   ....... (XIII)

Here, the angle of precession axis with respect to x-axis is θx.

Write the expression of the angle of precession axis with respect to y-axis.

cosθy=I¯yωyHG   ....... (XIV)

Here, the angle of precession axis with respect to x-axis is θy.

Write the expression of the angle of precession axis with respect to y-axis.

cosθz=I¯zωzHG   ....... (XV)

Here, the angle of precession axis with respect to x-axis is θz.

Figure-(2) shows the angle of spin axis with respect to precession axis is constant.

Vector Mechanics For Engineers, Chapter 18.3, Problem 18.129P , additional homework tip  2

Figure-(2)

Here, the rate of precession is φ˙, and rate of spin is ψ˙.

Write the expression of the angle of precession axis with respect to spin axis is constant from figure (2).

θ=θy   ....... (XVI)

Write the expression of the angle of angular velocity with respect to spin axis.

cosγ=ωyω   ....... (XVII)

Here, the angle of angular velocity with respect to spin axis is γ.

Write the expression for relation between φ˙, ψ˙ and ω according to sine rule.

ψ˙sin(γθ)=ωsinθ=φ˙sinγ   ....... (XVIII)

Calculation:

Substitute 800lb for WS and 32.2(ft/s2) for g in Equation (II).

mS=800lb32.2(ft/s2)=24.844(lbs2/ft)

Substitute 6oz for WM, and 32.2(ft/s2) in Equation (I).

mM=6oz(1lb16oz)32.2(ft/s2)=0.375lb32.2(ft/s2)=0.0116(lb×s2/ft)

Substitute 28.8in. for kx, and 24.844(lbs2/ft) for mS in Equation (III).

I¯x=24.844(lbs2/ft)(28.8in.)2=24.844(lbs2/ft)(28.8in.(1ft12in.))2=24.844(lbs2/ft)(2.4ft)2=143.101(lbs2ft)

Substitute 32.4in. for ky, and 24.844(lbs2/ft) for mS. in Equation (IV).

I¯y=24.844(lbs2/ft)(32.4in.)2=24.844(lbs2/ft)(32.4in.(1ft12in.))2=24.844(lbs2/ft)(2.7ft)2=181.112(lbs2ft)

Substitute 28.8in. for kz, and 24.844(lbs2/ft) for mS. in Equation (V).

I¯z=24.844(lbs2/ft)(28.8in.)2=24.844(lbs2/ft)(28.8in.(1ft12in.))2=24.844(lbs2/ft)(2.4ft)2=143.101(lbs2ft)

Substitute in (1600ft/s)i+(1300ft/s)j+(4000ft/s)k for v0, and 0.0116(lb×s2/ft) for mM Equation (VI).

M0=0.0116(lbs2/ft)[(1600ft/s)i+(1300ft/s)j+(4000ft/s)k]=[(18.56)i+(15.08)j+(46.4)k](lbs2ftsft)=[(18.56)i+(15.08)j+(46.4)k](lbs)

Substitute 143.101(lbs2ft) for I¯x, 181.112(lbs2ft) for I¯y, 143.101(lbs2ft) for I¯z, 0 for ωx, 0 for ωy, 1.5(rad/s)j for ωz in Equation (VII).

(HG)S=[(143.101(lbs2ft))(0)i+(181.112(lbs2ft))(1.5(rad/s))j+(143.101(lbs2ft))(0)k]=271.668(lbsft)j

Substitute (42in.)i(20in)j for rBG, [(18.56)i+(15.08)j+(46.4)k](lbs) for M0, and 271.668(lbsft)j for (HG)S in Equation (VIII).

(HG)B=[(42in.)i(20in)j]×[[(18.56)i+(15.08)j+(46.4)k](lbs)]+[271.668(lbsft)j]

(HG)B=[(42in.(1ft12in.))i(20in.(1ft12in.))j]×[[(18.56)i+(15.08)j+(46.4)k](lbs)]+[271.668(lbsft)j]

(HG)B=[(3.5ft)i(1.67ft)j]×[[(18.56)i+(15.08)j+(46.4)k](lbs)]+[271.668(lbsft)j]

(HG)B=|ijk3.51.67018.5615.0846.4|(lbsft)+[271.668(lbsft)j]

(HG)B=[(77.64)i+(163.044)j+(21.935)k](lbsft)+[271.668(lbsft)j](HG)B=[(77.64)i+(108.624)j+(21.935)k](lbsft)

Substitute [(77.64)i+(108.624)j+(21.935)k](lbsft) for (HG)B, and I¯xωxi+I¯yωyj+I¯zωzk for (HG)S in Equation (X).

[(77.64)i+(108.624)j+(21.935)k](lbsft)=I¯xωxi+I¯yωyj+I¯zωzk   ....... (XIX)

Equate x-components in Equation (XIX) on both side and substitute 143.101(lbs2ft) for I¯x.

(77.64)(lbsft)=(143.101(lbs2ft))ωxωx=(77.64)(lbsft)143.101(lbs2ft)ωx=0.542rad/s

Equate y-components in Equation (XIX) on both side and substitute 181.112(lbs2ft) for I¯y.

(108.624)j(lbsft)=(181.112(lbs2ft))ωyjωy=(108.624)(lbsft)181.112(lbs2ft)ωy=0.599rad/s

Equate z-components in Equation (XIX) on both side and substitute 43.101(lbs2ft) for I¯z.

(21.935)(lbsft)=(143.101(lbs2ft))ωzωz=(21.935)(lbsft)143.101(lbs2ft)ωz=0.153rad/s

Substitute 0.542rad/s for ωx, 0.599rad/s for ωy, and 0.153rad/s for ωz in Eqaution (XI).

ω=(0.542rad/s)2+(0.599rad/s)2+(0.153rad/s)2ω=0.822rad/s

Substitute (77.64)(lbsft) for (Ixωx), (108.624)(lbsft) for (Iyωy), and (21.935)(lbsft) for (Izωz) in Eqaution (XII).

HG=(77.64(lbsft))2+(108.624(lbsft))2+(21.935(lbsft))2=6027.969+11799.173+481.144(lbsft)=135.308(lbsft)

Substitute 143.101(lbs2ft) for I¯x, 0.542rad/s for ωx, 135.308(lbsft) for HG in Equation (XIII).

cosθx=143.101(lbs2ft)×0.542rad/s135.308(lbsft)cosθx=0.5732θx=124.9701250

Substitute 181.112(lbs2ft) for I¯y, 0.599rad/s for ωy, 135.308(lbsft) for HG in Equation (XIV).

cosθy=181.112(lbs2ft)×0.599rad/s135.308(lbsft)cosθy=0.80177θy=36.700

Substitute 143.101(lbs2ft) for I¯z, 0.153rad/s for ωz, 135.308(lbsft) for HG in Equation (XIV).

cosθz=143.101(lbs2ft)×0.153rad/s135.308(lbsft)cosθz=0.1618θz=80.687080.70

Substitute 36.700 for θy in Equation (XVI).

θ=36.700

Substitute 0.599rad/s for ωy, and 0.822rad/s for ω in Equation (XVII).

cosγ=0.599rad/s0.822rad/scosγ=0.7287γ=43.220

Compare first and second term of Equation (XVIII).

ψ˙sin(γθ)=ωsinθ   ....... (XX)

Substitute 43.220 for γ, 36.700 for θ, 0.822rad/s for ω in Equation (XX).

ψ˙sin(43.22036.700)=0.822rad/ssin36.700ψ˙=(0.822rad/s)(0.1135)0.5976ψ˙=0.1561rad/s

Compare second and third term of Equation (XVIII).

ωsinθ=φ˙sinγ   ....... (XXI)

Substitute 43.220 for γ, 36.700 for θ, 0.822rad/s for ω in Equation (XXI).

0.822rad/ssin36.700=φ˙sin43.220φ˙=(0.822rad/s)(0.6848)0.5976φ˙=0.942rad/s

Conclusion:

Thus, the precession angle about the x-axis, y-axis and z-axis are 1250, 36.700 and 80.70.

The rate of precession of the satellite after impact is 0.942rad/s.

The rate of spin axis of the satellite after impact is 0.1561rad/s.

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Chapter 18 Solutions

Vector Mechanics For Engineers

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