VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
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 MECH...,DYNAMICS(LOOSE)-W/ACCESS, 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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Two disks A and B, of mass m 1 kg each and of radius 10 cm, are placed on a horizontal table. The disk A is launched in translation with a speed of 10 m/s along the y axis, B is at rest. The coefficient of kinetic friction between the disks is 0.5. The line of impact is at an angle of 60° with the x-axis. The moment of inertia of each disk around its center of mass is I = 1₂ = 0.005 kg.m². The coefficient of restitution between the two disks is e=0.6. a) Determine the velocities of the centers of the two disks, just after the impact. b) Calculate the angular velocities of the disks, just after the impact. c) Calculate the energy loss during the impact.
5. Fig. 3 shows the overhead view of a uniform thin rod of length I and mass M that can rotate freely about a fixed axis on a horizontal surface. The rod is at rest initially with moment of inertia MI². A bullet of mass m is fired towards one end of the rod with initial velocity v. After piercing the rod, the bullet's velocity is reduced to v. The angular velocity of the rod after being shot through is ( ) 3mv А. 2ML 7mv В. 4ML overhead view mv C. MI Fig 3 D. 5mv 3MI
A thin rectangular plate of weight 15 lb rotates about its vertical diagonal AB with an angular velocity V. Knowing that the z axis is perpendicular to the plate and that V is constant and equal to 5 rad/s, determine the angular momentum of the plate about its mass center G. B 12 in. 9 in. ANSWER : - (0.408 slug · ft²/s)i + (0.1398 slug · ft²/s)j.

Chapter 18 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

Ch. 18.1 - Prob. 18.11PCh. 18.1 - Prob. 18.12PCh. 18.1 - Prob. 18.13PCh. 18.1 - Prob. 18.14PCh. 18.1 - Prob. 18.15PCh. 18.1 - For the assembly of Prob. 18.15, determine (a) the...Ch. 18.1 - Prob. 18.17PCh. 18.1 - Determine the angular momentum of the shaft of...Ch. 18.1 - Prob. 18.19PCh. 18.1 - Prob. 18.20PCh. 18.1 - Prob. 18.21PCh. 18.1 - Prob. 18.22PCh. 18.1 - Prob. 18.23PCh. 18.1 - Prob. 18.24PCh. 18.1 - Prob. 18.25PCh. 18.1 - Prob. 18.26PCh. 18.1 - Prob. 18.27PCh. 18.1 - Prob. 18.28PCh. 18.1 - Prob. 18.29PCh. 18.1 - Prob. 18.30PCh. 18.1 - Prob. 18.31PCh. 18.1 - Prob. 18.32PCh. 18.1 - Prob. 18.33PCh. 18.1 - Prob. 18.34PCh. 18.1 - Prob. 18.35PCh. 18.1 - Prob. 18.36PCh. 18.1 - Prob. 18.37PCh. 18.1 - Prob. 18.38PCh. 18.1 - Prob. 18.39PCh. 18.1 - Prob. 18.40PCh. 18.1 - Prob. 18.41PCh. 18.1 - Prob. 18.42PCh. 18.1 - Determine the kinetic energy of the disk of Prob....Ch. 18.1 - Prob. 18.44PCh. 18.1 - Prob. 18.45PCh. 18.1 - Prob. 18.46PCh. 18.1 - Prob. 18.47PCh. 18.1 - Prob. 18.48PCh. 18.1 - Prob. 18.49PCh. 18.1 - Prob. 18.50PCh. 18.1 - Prob. 18.51PCh. 18.1 - Prob. 18.52PCh. 18.1 - Determine the kinetic energy of the space probe of...Ch. 18.1 - Prob. 18.54PCh. 18.2 - Determine the rate of change H.G of the angular...Ch. 18.2 - Prob. 18.56PCh. 18.2 - Determine the rate of change H.G of the angular...Ch. 18.2 - Prob. 18.58PCh. 18.2 - Prob. 18.59PCh. 18.2 - Prob. 18.60PCh. 18.2 - Prob. 18.61PCh. 18.2 - Prob. 18.62PCh. 18.2 - Prob. 18.63PCh. 18.2 - Prob. 18.64PCh. 18.2 - A slender, uniform rod AB of mass m and a vertical...Ch. 18.2 - A thin, homogeneous triangular plate of weight 10...Ch. 18.2 - Prob. 18.67PCh. 18.2 - Prob. 18.68PCh. 18.2 - Prob. 18.69PCh. 18.2 - Prob. 18.70PCh. 18.2 - Prob. 18.71PCh. 18.2 - Prob. 18.72PCh. 18.2 - Prob. 18.73PCh. 18.2 - Prob. 18.74PCh. 18.2 - Prob. 18.75PCh. 18.2 - Prob. 18.76PCh. 18.2 - Prob. 18.77PCh. 18.2 - Prob. 18.78PCh. 18.2 - Prob. 18.79PCh. 18.2 - Prob. 18.80PCh. 18.2 - Prob. 18.81PCh. 18.2 - Prob. 18.82PCh. 18.2 - Prob. 18.83PCh. 18.2 - Prob. 18.84PCh. 18.2 - Prob. 18.85PCh. 18.2 - Prob. 18.86PCh. 18.2 - Prob. 18.87PCh. 18.2 - Prob. 18.88PCh. 18.2 - Prob. 18.89PCh. 18.2 - The slender rod AB is attached by a clevis to arm...Ch. 18.2 - The slender rod AB is attached by a clevis to arm...Ch. 18.2 - Prob. 18.92PCh. 18.2 - The 10-oz disk shown spins at the rate 1=750 rpm,...Ch. 18.2 - Prob. 18.94PCh. 18.2 - Prob. 18.95PCh. 18.2 - Prob. 18.96PCh. 18.2 - Prob. 18.97PCh. 18.2 - Prob. 18.98PCh. 18.2 - Prob. 18.99PCh. 18.2 - Prob. 18.100PCh. 18.2 - Prob. 18.101PCh. 18.2 - Prob. 18.102PCh. 18.2 - Prob. 18.103PCh. 18.2 - A 2.5-kg homogeneous disk of radius 80 mm rotates...Ch. 18.2 - For the disk of Prob. 18.99, determine (a) the...Ch. 18.2 - Prob. 18.106PCh. 18.3 - Prob. 18.107PCh. 18.3 - A uniform thin disk with a 6-in. diameter is...Ch. 18.3 - Prob. 18.109PCh. 18.3 - Prob. 18.110PCh. 18.3 - Prob. 18.111PCh. 18.3 - A solid cone of height 9 in. with a circular base...Ch. 18.3 - Prob. 18.113PCh. 18.3 - Prob. 18.114PCh. 18.3 - Prob. 18.115PCh. 18.3 - Prob. 18.116PCh. 18.3 - Prob. 18.117PCh. 18.3 - Prob. 18.118PCh. 18.3 - Show that for an axisymmetric body under no force,...Ch. 18.3 - Prob. 18.120PCh. 18.3 - Prob. 18.121PCh. 18.3 - Prob. 18.122PCh. 18.3 - Prob. 18.123PCh. 18.3 - Prob. 18.124PCh. 18.3 - Prob. 18.125PCh. 18.3 - Prob. 18.126PCh. 18.3 - Prob. 18.127PCh. 18.3 - Prob. 18.128PCh. 18.3 - An 800-lb geostationary satellite is spinning with...Ch. 18.3 - Solve Prob. 18.129, assuming that the meteorite...Ch. 18.3 - Prob. 18.131PCh. 18.3 - Prob. 18.132PCh. 18.3 - Prob. 18.133PCh. 18.3 - Prob. 18.134PCh. 18.3 - Prob. 18.135PCh. 18.3 - Prob. 18.136PCh. 18.3 - Prob. 18.137PCh. 18.3 - Prob. 18.138PCh. 18.3 - Prob. 18.139PCh. 18.3 - Prob. 18.140PCh. 18.3 - Prob. 18.141PCh. 18.3 - Prob. 18.142PCh. 18.3 - Prob. 18.143PCh. 18.3 - Prob. 18.144PCh. 18.3 - Prob. 18.145PCh. 18.3 - Prob. 18.146PCh. 18 - Prob. 18.147RPCh. 18 - Prob. 18.148RPCh. 18 - A rod of uniform cross-section is used to form the...Ch. 18 - A uniform rod of mass m and length 5a is bent into...Ch. 18 - Prob. 18.151RPCh. 18 - Prob. 18.152RPCh. 18 - A homogeneous disk of weight W=6 lb rotates at the...Ch. 18 - Prob. 18.154RPCh. 18 - Prob. 18.155RPCh. 18 - Prob. 18.156RPCh. 18 - Prob. 18.157RPCh. 18 - Prob. 18.158RP
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Dynamics - Lesson 1: Introduction and Constant Acceleration Equations; Author: Jeff Hanson;https://www.youtube.com/watch?v=7aMiZ3b0Ieg;License: Standard YouTube License, CC-BY