<LCPO> VECTOR MECH,STAT+DYNAMICS
<LCPO> VECTOR MECH,STAT+DYNAMICS
12th Edition
ISBN: 9781265566296
Author: BEER
Publisher: MCG
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Chapter 18.1, Problem 18.54P

Determine the kinetic energy of the space probe of Prob. 18.34 in its motion about its mass center after its collision with the meteorite.

18.34    The coordinate axes shown represent the principal centroidal axes of inertia of a 3000-lb space probe whose radii of gyration are kx = 1.375 ft, ky = 1.425 ft, and kz = 1.250 ft. The probe has no angular velocity when a 5-oz meteorite strikes one of its solar panels at point A and emerges on the other side of the panel with no change in the direction of its velocity, but with a speed reduced by 25 percent. Knowing that the final angular velocity of the probe is ω = (0.05 rad/s)i − (0.12 rad/s)j + ωzk and that the x component of the resulting change in the velocity of the mass center of the probe is −0.675 in./s, determine (a) the component ωz of the final angular velocity of the probe, (b) the relative velocity v0 with which the meteorite strikes the panel.

Chapter 18.1, Problem 18.54P, Determine the kinetic energy of the space probe of Prob. 18.34 in its motion about its mass center

Fig. P18.33 and P18.34

Expert Solution & Answer
Check Mark
To determine

The kinetic energy (T) of the space probe in its motion about its mass center after its collision with the meteorite.

Answer to Problem 18.54P

The kinetic energy (T) of the space probe is 39.9lbft_.

Explanation of Solution

Given information:

The weight of the space probe (w) is 3,000 lb.

The radius of gyration along x axis (kx) is 1.375 ft.

The radius of gyration along y axis (ky) is 1.425 ft.

The radius of gyration along z axis (kz) is 1.250 ft.

The weight of the meteorite (w) is 5 oz.

The angular velocity (ω) is (0.05rad/s)i(0.12rad/s)j+(ωz)k.

The change in velocity of the mass center of the probe (vx) is –0.675 in./s.

The width of the side panel from center to point A (b) is 9 ft.

The length of the panel from center to point A (l) is 0.75 ft.

The speed is reduced by 25 percent.

Calculation:

Calculate the mass of the space probe (m) using the formula:

w=mgm=wg

Here, g is the acceleration due to gravity.

Substitute 32.2ft/s2 for g and 3,000 lb for w.

m=300032.2=93.17lbs2/ft

Calculate the mass of the meteorite (m) using the formula:

w=mgm=wg

Substitute 32.2ft/s2 for g and 5 oz for w.

m=5oz×116lboz32.2=0.009705lbs2/ft

Write the relative position vector (rA) at the point (A) of impact as follows:

rA=bi+lk

Substitute 9 ft for b and 0.75 ft.

rA=(9ft)i+(0.75ft)k

Write the expression for the velocity (v0) along x, y and z axis as follows:

v0=vxi+vyj+vzk

Calculate the initial liner momentum of the meteorite using the relation:

Initial linear momentum=mv0

Substitute 0.009705lbs2/ft for m and vxi+vyj+vzk for v0.

mv0=(0.009705lbs2/ft)×(vxi+vyj+vzk)

Calculate the moment about origin (HA)O using the relation:

(HA)O=rA×mv0

Substitute (0.009705lbs2/ft)×(vxi+vyj+vzk) for mv0 and (9ft)i+(0.75ft)k for rA.

(HA)O=0.009705|ijk900.75vxvyvz|=0.009705[(0.75vy)i(9vz0.75vx)j+(9vy)k]

The speed is reduced to 25 percent.

Calculate the final liner momentum of the meteorite using the relation:

Final linear momentum=0.75mv0

Substitute 0.009705lbs2/ft for m and vxi+vyj+vzk for v0.

0.75mv0=0.75(0.009705lbs2/ft)×vxi+vyj+vzk=(0.007279)×(vxi+vyj+vzk)

Calculate the final linear momentum of meteorite and its moment about the origin using the relation:

M=rA×(0.75mv0)

Substitute (0.007279)×(vxi+vyj+vzk) for 0.75mv0 and (9ft)i+(0.75ft)k for rA.

M=0.007278|ijk900.75vxvyvz|=0.007278[(0.75vy)i(9vz0.75vx)j+(9vy)k]

The initial linear momentum of the space probe (mv0) is zero.

Calculate the final linear momentum of the space probe using the relation:

Final linear momentum=mv0

Substitute vxi+vyj+vzk for v0 and 93.17lbs2/ft for m.

mv0=93.17(vxi+vyj+vzk)

Substitute -0.675 in./s for vx.

mv0=93.17((0.675in.12ftin.)i+vyj+vzk)=93.17(0.05625i+vyj+vzk)

Calculate the final angular momentum of the space probe (HA) using the formula:

HA=m(kx2ωxi+ky2ωyj+kz2ωzk)

Substitute 93.17lbs2/ft for m, 1.375 ft for kx, 1.425 ft for ky, 1.25 ft for kz, 0.05 rad/s for ωx, and –0.12 rad/s for ωy.

HA=93.17[(1.3752×0.05)i+(1.4252×0.12)j+(1.252×ωz)k]=93.17[0.09453i0.243675j+1.5625ωzk]=8.8074i22.703j+145.58ωzk

Write the expression for the conservation of linear momentum of the probe plus the meteorite as follows:

mv0=0.75mv0+mv0

Substitute (0.009705lbs2/ft)×(vxi+vyj+vzk) for mv0, (0.007279)×(vxi+vyj+vzk) for 0.75mv0 and 93.17(0.05625i+vyj+vzk) for mv0.

{(0.009705lbs2/ft)×(vxi+vyj+vzk)[(0.007279)×(vxi+vyj+vzk)]}=[93.17(0.05625i+vyj+vzk)]{0.009705vxi+0.009705vyj+0.009705vzk(0.007279vxi+0.007279vyj+0.007279vzk)}=(5.2408i+93.17vyj+93.17vzk)5.2408i+93.17vyj+93.17vzk=(0.009705vxi+0.009705vyj+0.009705vzk0.007279vxi0.007279vyj0.007279vzk)5.2408i+93.17vyj+93.17vzk=(0.002426vxi+0.002426vyj+0.0024265vzk) (1)

Equate the i component from the Equation (1).

5.2408=0.002426vxvx=5.24080.002426vx=2160ft/s

Equate j component from the Equation (1).

93.17vy=0.002426vy

Equate k component from the Equation (1).

93.17vz=0.002426vz

Write the expression for the conservation of angular momentum about the origin as follows:

(HA)O=M+HA

Substitute 0.009705[(0.75vy)i(9vz0.75vx)j+(9vy)k] for (HA)O, 0.007278[(0.75vy)i(9vz0.75vx)j+(9vy)k] for M, and 8.8074i22.703j+145.58ωzk for HA.

{0.009705[(0.75vy)i(9vz0.75vx)j+(9vy)k]}={0.007278[(0.75vy)i(9vz0.75vx)j+(9vy)k]}+(8.8074i22.703j+145.58ωzk)(0.00727875vyi0.087345vzj+0.00727875vxj+0.087345vyk)=(0.0054585vyi0.065502vzj+0.0054585vxj+0.065502vyk)+(8.8074i22.703j+145.58ωzk)(0.00727875vyi0.087345vzj+0.00727875vxj+0.087345vyk+0.0054585vyi+0.065502vzj0.0054585vxj0.065502vyk)=(8.8074i22.703j+145.58ωzk)(0.00182vyi0.02184vzj+0.00182vxj+0.02184vyk)=(8.8074i22.703j+145.58ωzk) (2)

Equate i component from the equation (2).

0.00182vy=8.8075vy=8.80750.00182vy=4840ft/s

Equate k component from the equation (2).

0.02184vy=145.58ωz

Substitute –4840 ft/s for vy.

0.02184(4840)=145.58ωzωz=0.02184(4840)145.58ωz=0.726rad/s

Calculate the kinetic energy of motion of the probe relative to its mass center (T) using the formula:

T=12(Ixωx2+Iyωy2+Izωz2)=12m(kx2ωx2+ky2ωy2+kz2ωz2)

Substitute 93.17lbs2/ft for m, 1.375 ft for kx, 1.425 ft for ky, 1.25 ft for kz, 0.05 rad/s for ωx and -0.12 rad/s for ωy, and 0.726rad/s for ωz.

T=12×93.17[(1.3752×0.052)+(1.4252×0.122)+(1.252×(0.7262))]=12×93.17(0.00473+0.02924+0.824)=39.9lbft

Thus, the kinetic energy (T) of the space probe is 39.9lbft_.

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Chapter 18 Solutions

<LCPO> VECTOR MECH,STAT+DYNAMICS

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