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 19.3, Problem 19.87P
To determine

The period of small oscillations of the system.

Expert Solution & Answer
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Answer to Problem 19.87P

The period of small oscillations of the system: τn=2π12r2+2l23gl

Explanation of Solution

Given information:

The mass of gear C is m, and the mass of gear A is 4m.

Calculations:

Vector Mechanics For Engineers, Chapter 19.3, Problem 19.87P

For the figures shown above:

Kinematics:2rθA=rθC2θA=θC2θ˙A=θ˙CLetθA=θm2θm=(θC)m2θ˙m=(θ˙C)m

For Position 1:Kinetic energy: T1=12I¯Aθ.m2+12I¯C(2θ.m)2+12I¯ABθ.m2+12I¯CD(2θ.m)2+12mAB(l2θ.m)2+12mCD(l22θ.m)2where,I¯A=12(4m)(2r)2=8 mr2I¯C=12(m)(r)2=12 mr2I¯AB=112ml2I¯CD=112ml2substituting,T1=12m[8r2+(r22)4+l212+l23+l24+l2]θ.m2T1=12m[10r2+53l2]θ.m2Potential energy: V1=0

For Position 2:Kinetic energy: T2=0Potential energy: V2=mgl2(1cosθm)+mgl2(1cos2θm)For small angles (1cosθm=2sin2θm2θm22)(1cos2θm=2sin2θm2θm2)V2=12mgl(θm22+2θm2)=12mgl5θm22

Now, from the law of conservation of energy:T1+V1=T2+V2with, (θ˙m2=ωn2θm2)12m[10r2+53l2]ωn2θm2+0=0+12mgl5θm22ωn2=52gl10r2+53l2ωn2=3gl12r2+2l2Thus, the period of oscillation:τn=2πωnτn=2π12r2+2l23gl

Conclusion:

The period of small oscillations of the system is τn=2π12r2+2l23gl

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Chapter 19 Solutions

Vector Mechanics For Engineers

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