Given information:
The weight of the space probe is 3000 lb, the weight of the meteorite is 5 oz, the radius of gyration of space probe along the x-axis is 1.375 ft, the radius of gyration of space probe along the y-axis is 1.425 ft, the radius of gyration along the z-axis is 1.250 ft, and final angular velocity of the space probe is (0.05 rad/s)i−(0.12 rad/s)j+(ωz rad/s)k. The angular momentum of the space probe is reduced by 25%. The resulting change in x component of mass centre of the space probe is −0.675 (in./s).
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 space probe.
mP=WPg ...... (II)
Here, weight of the meteorite is WP.
Write the expression of angular momentum of the meteorite about G.
(HG)M=rA×mMv0 ...... (III)
Here, the distance of point A is rA, the mass of the meteorite is mM, and the initial velocity of the meteorite is v0.
Write the Expression of the moment of inertia along x-axis.
I¯x=mkx2 ...... (IV)
Here, the radius of gyration of the space probe along x-axis is kx.
Write the Expression of the moment of inertia along the y-axis.
I¯y=mky2 ...... (V)
Here, the radius of gyration of the space probe along y-axis is ky.
Write the Expression of the moment of inertia along z-axis.
I¯z=mkz2 ...... (VI)
Here, the radius of gyration of the space probe along z-axis is kz.
Write the expression of angular momentum of space probe.
(HG)P=I¯xωxi+I¯yωyj+I¯zωzk ...... (VII)
Here, moment of inertia along x-axis is Ix, moment of inertia along y-axis is Iy, and moment of inertia along z-axis is Iz.
Substitute mkx2 for I¯x, mky2 for I¯y and mkz2 for I¯z in Equation (VII).
(HG)P=mkx2ωxi+mky2ωyj+mkz2ωzk(HG)P=m(kx2ωxi+ky2ωyj+kz2ωzk) ...... (VIII)
Here, angular velocity along x-axis is ωx, angular velocity along y-axis is ωy, angular velocity along x-axis is ωz, the radius of gyration of space probe along x-axis is kx, the radius of gyration of space probe along y-axis is ky, radius of gyration along z-axis is kz.
Write the expression of kinetic energy of the space probe.
T=12(Ixωx2+Iyωy2+Izωz2) ...... (IX)
Substitute mkx2 for I¯x, mky2 for I¯y and mkz2 for I¯z in Equation (IX).
T=12(mkx2ωx2+mky2ωy2+mkz2ωz2)T=12m(kx2ωx2+ky2ωy2+kz2ωz2) ...... (X)
Calculation:
Substitute 5 oz for WP, and 32.2 ft/s2 for g in Equation (I) for the meteorite.
mM=(5 oz)(32.2 ft/s2)mM=(516 lb)(32.2 ft/s2)
Substitute 3000 lb for WP, and 32.2 ft/s2 for g in Equation (II) for the space probe.
mP=(3000 lb)(32.2 ft/s2)mP=93.167 (lb⋅s2ft)
Write the expression that shows the relation between linear momentum of meteorite and the space probe.
25%(mM)(v0)x=(mP)(Δv)x ...... (XI)
Here, the initial velocity along x-axis is (v0)x, and change in velocity of probe in x-axis is (Δv)x.
Substitute (516 lb)(32.2 ft/s2) for mM, and 93.167 (lb⋅s2ft) for mP in Equation (XI).
25%((516 lb)(32.2 ft/s2))(v0)x=(93.167 (lb⋅s2ft))(−0.675 (in./s))0.25((9.705×10(−3) lb)(ft/s2))(v0)x=(93.167 (lb⋅s2ft))(−0.675 (in./s))2.426×10(−3) (lb)(ft/s2)(v0)x=(−62.887 (lb⋅s2ft))(in./s)(v0)x=−25.92×103 (in./s)
(v0)x=−25.92×103 (112ft)/s=−25.92×10312 (ft/s)=−2160 (ft/s)
Substitute (v0)xi−(v0)yj+(v0)zk for v0 in Equation (III).
(HG)M=rA×mM((v0)xi−(v0)yj+(v0)zk) ...... (XII)
Substitute [(9 ft)i+(0.75 ft)k] for rA, (516 lb)(32.2 ft/s2) for mM, and −2160 (ft/s) for (v0)x in Equation (XII).
(HG)M=[(9 ft)i+(0.75 ft)k]×(516 lb)(32.2 ft/s2)(−2160 (ft/s)i−(v0)yj+(v0)zk)=9.704×10(−3) (lb)(ft/s2)|ijk900.75−2160(v0)y(v0)z|=9.704×10(−3) (lb)(ft/s2)(−0.75 ft(v0)yi−(9 ft(v0)z+1620 (ft2/s))j+9 ft(v0)zk)
Substitute 1.375 ft for kx, 1.425 ft for ky, 1.250 ft for kz, (0.05 rad/s) for ωx, −(0.12 rad/s) for ωy, and 93.167 (lb⋅s2ft) for m in Equation (III).
(HG)P=93.167 (lb⋅s2ft)((1.375 ft)2(0.05 rad/s)i−(1.425 ft)2(0.12 rad/s)j+(1.250 ft)2ωzk)(HG)P=((8.8071 rad/s)i−(22.702 rad/s)j+(145.57)ωzk)(lb⋅s2⋅ft)
The angular momentum of the space probe is 25% of the angular momentum of the meteorite.
Write the expression of the relation between (HG)P and (HG)M.
(HG)P=25%(HG)M(HG)P=(HG)M4 ...... (XIII)
Substitute (9.704×10(−3) (lb)(ft/s2)(−0.75 ft(v0)yi−(9 ft(v0)z+1620 (ft2/s))j+9 ft(v0)zk)) for (HG)M, and ((8.8071 rad/s)i−(22.702 rad/s)j+(145.57)ωzk)(lb⋅s2⋅ft) for (HG)P in Equation (XIII).
Compare x-component of Equation (IX) on both side.
(35.228 rad/s)(lb⋅s2⋅ft)=(7.278×10(−3) ft)(v0)y(lb)(ft/s2)(v0)y=−4840.3 (ft/s)
Compare y-component of Equation (XIV) on both side.
−(90.808) (lb⋅s2⋅ft)s=[−(87.336×10(−3) ft(v0)z+15720.48×10(−3) (ft2/s))](lb)(ft/s2)−(90.808) (lb⋅s2⋅ft)s=[(−87.336×10(−3) (lb⋅s2)(v0)z−15720.48×10(−3) (lb⋅s2⋅ft/s))]−15810.808 (lb⋅s2⋅ft)s=−87.336×10(−3) (lb⋅s2)(v0)z(v0)z=1219.74 (ft/s)
Compare z-component of Equation (XIV) on both side.
((582.28)ωz)(lb⋅s2⋅ft)=[(87.336×10(−3) ft)(v0)z](lb)(ft/s2) ...... (XV)
Substitute 181.039(ft/s) for (v0)z in Equation (XV).
((582.28)ωz)(lb⋅s2⋅ft)=[(87.336 ft)(1219.74 (ft/s))](lb)(ft/s2)ωz=[(87.336)(1219.74)(582.28)]ft(ft/s)(lb⋅s2)ft(lb⋅s2⋅ft)ωz=182.948 (rad/s)
Substitute 1.375 ft for kx, 1.425 ft for ky, 1.250 ft for kz, (0.05 rad/s) for ωx, (−0.12 rad/s) for ωy, 182.948 (rad/s) for ωz, and 93.167 (lb⋅s2ft) for m in Equation (X).
T=12(93.167 (lb⋅s2ft))[(1.375 ft)2(0.05 rad/s)2+(1.425 ft)2(−(0.12 rad/s))2+(1.250 ft)2(182.948 (rad/s))2]=46.583 (lb⋅s2ft)[(4.72×10(−3))+(0.02924)+(52296.82)] (ft/s)2=2.436×106 (ft⋅lb)
Conclusion:
Thus, the kinetic energy is 2.436×106 (ft⋅lb).