Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 17.3, Problem 17.120P

For the beam of Prob. 17.119, determine the velocity of the 1-oz bullet for which the maximum angle of rotation of the beam will be 90°.

Expert Solution & Answer
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To determine

The velocity of the bullet for which the maximum angle of rotation of the beam will be 900

Answer to Problem 17.120P

Velocity of a bullet for 900 of the rotation angle of the beam will be 1887 ft/sec.

Explanation of Solution

Given information:

θm=900

Mass of bullet w’ =1 oz=1 ounce =0.06251 b.

Mass of beam w=18 lb.

Velocity of beam v0=750 mi/h =1100ft/sec

Concept used:

Moment of inertia =I=1/12 ml2

Low of conservation of energY

Calculation:

Mass ratio = β = w/w=0.0625/18 =0.00347

We can say that w = βw or

M = βm

Since β is so small mass of the bullet is ignored while calculating the motion of beam after impact.

As per kinetics,

Vector Mechanics for Engineers: Dynamics, Chapter 17.3, Problem 17.120P , additional homework tip  1

Linear component, M = βmv0+0=mv2m

v2=βv0+0=mv2v2=βv0

Taking moment at B,

0+0=Iwmv2L/2Iw=mv2L/2W= mv2L/2I=12/2xmv2L/mL2=6v2/LW=6βv0/L

Motion after impact,

v2=mg L/2T2=1/2Iw22+1/2mv22=1/2(1/12mL2)(6βv0/L)2+1/2m(βv0)2=mL2/24(36(β2v02/L2)+m/2β2v02=3mβ2v02/2+mβ2v02/2=4/2mβ2v02T2=2mβ2v02

New position:

W=0 and Q= Qmax

Vector Mechanics for Engineers: Dynamics, Chapter 17.3, Problem 17.120P , additional homework tip  2

According to figure, v3=mgL/2 cosθm

T3=1/2mv32

Linear components,

Mv2+0=mv3V3=v2=βv0

Using low of conservation of energy,

T3+V3=T3+V3

2mβ2v02mgL/2=1/2m(βv02)mgL/2 cosθm

2m(βv02)mgL/2=1/2mgL/2cosθm=m(βv02)/2

2m(βv02)mgL/2(1cosθm)=m(βv0)2/2

m[2(βv02)gL/2(1cosθm)]=m(βv02)/2

2m/2[2(βv02)gL/2(1cosθm)]=(βv02)

2m(βv02)mgL/2(1cosθm)=(βv0)2

2m(βv02)(βv0)2=mgL/2(1cosθm)

(βv0)2(2m1)=mgL/2(1cosθm)

(βv0)2=mgL/2(1cosθm)/4m2

(βv0)2=1/3gL(1cosθm)βv0

=1/3gL/2(1cosθm)

=1/3(32.2)(4)(1cos90)

βv0=6.55 ft/secv0=6.55/0.00347v0=1887.1 ft/sec

Conclusion:

We can find that bullet velocity for a maximum rotation angle of beam 900 is 1887 ft/sec.

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Chapter 17 Solutions

Vector Mechanics for Engineers: Dynamics

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