Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 18.1, Problem 18.54P
To determine

The kinetic energy of the space probe after its collision with the meteorite.

Expert Solution & Answer
Check Mark

Answer to Problem 18.54P

The kinetic energy of the space probe after its collision with the meteorite is 2.436×106(ftlb).

Explanation of Solution

Given information:

The weight of the space probe is 3000lb, the weight of the meteorite is 5oz, the radius of gyration of space probe along the x-axis is 1.375ft, the radius of gyration of space probe along the y-axis is 1.425ft, the radius of gyration along the z-axis is 1.250ft, and final angular velocity of the space probe is (0.05rad/s)i(0.12rad/s)j+(ωzrad/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 5oz for WP, and 32.2ft/s2 for g in Equation (I) for the meteorite.

mM=(5oz)(32.2ft/s2)mM=(516lb)(32.2ft/s2)

Substitute 3000lb for WP, and 32.2ft/s2 for g in Equation (II) for the space probe.

mP=(3000lb)(32.2ft/s2)mP=93.167(lbs2ft)

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 (516lb)(32.2ft/s2) for mM, and 93.167(lbs2ft) for mP in Equation (XI).

25%((516lb)(32.2ft/s2))(v0)x=(93.167(lbs2ft))(0.675(in./s))0.25((9.705×10(3)lb)(ft/s2))(v0)x=(93.167(lbs2ft))(0.675(in./s))2.426×10(3)(lb)(ft/s2)(v0)x=(62.887(lbs2ft))(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 [(9ft)i+(0.75ft)k] for rA, (516lb)(32.2ft/s2) for mM, and 2160(ft/s) for (v0)x in Equation (XII).

(HG)M=[(9ft)i+(0.75ft)k]×(516lb)(32.2ft/s2)(2160(ft/s)i(v0)yj+(v0)zk)=9.704×10(3)(lb)(ft/s2)|ijk900.752160(v0)y(v0)z|=9.704×10(3)(lb)(ft/s2)(0.75ft(v0)yi(9ft(v0)z+1620(ft2/s))j+9ft(v0)zk)

Substitute 1.375ft for kx, 1.425ft for ky, 1.250ft for kz, (0.05rad/s) for ωx, (0.12rad/s) for ωy, and 93.167(lbs2ft) for m in Equation (III).

(HG)P=93.167(lbs2ft)((1.375ft)2(0.05rad/s)i(1.425ft)2(0.12rad/s)j+(1.250ft)2ωzk)(HG)P=((8.8071rad/s)i(22.702rad/s)j+(145.57)ωzk)(lbs2ft)

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)(0.75ft(v0)yi(9ft(v0)z+1620(ft2/s))j+9ft(v0)zk)) for (HG)M, and ((8.8071rad/s)i(22.702rad/s)j+(145.57)ωzk)(lbs2ft) for (HG)P in Equation (XIII).

Vector Mechanics For Engineers, Chapter 18.1, Problem 18.54P

Compare x-component of Equation (IX) on both side.

(35.228rad/s)(lbs2ft)=(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)(lbs2ft)s=[(87.336×10(3)ft(v0)z+15720.48×10(3)(ft2/s))](lb)(ft/s2)(90.808)(lbs2ft)s=[(87.336×10(3)(lbs2)(v0)z15720.48×10(3)(lbs2ft/s))]15810.808(lbs2ft)s=87.336×10(3)(lbs2)(v0)z(v0)z=1219.74(ft/s)

Compare z-component of Equation (XIV) on both side.

((582.28)ωz)(lbs2ft)=[(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)(lbs2ft)=[(87.336ft)(1219.74(ft/s))](lb)(ft/s2)ωz=[(87.336)(1219.74)(582.28)]ft(ft/s)(lbs2)ft(lbs2ft)ωz=182.948(rad/s)

Substitute 1.375ft for kx, 1.425ft for ky, 1.250ft for kz, (0.05rad/s) for ωx, (0.12rad/s) for ωy, 182.948(rad/s) for ωz, and 93.167(lbs2ft) for m in Equation (X).

T=12(93.167(lbs2ft))[(1.375ft)2(0.05rad/s)2+(1.425ft)2((0.12rad/s))2+(1.250ft)2(182.948(rad/s))2]=46.583(lbs2ft)[(4.72×10(3))+(0.02924)+(52296.82)](ft/s)2=2.436×106(ftlb)

Conclusion:

Thus, the kinetic energy is 2.436×106(ftlb).

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Chapter 18 Solutions

Vector Mechanics For Engineers

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