Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 19.3, Problem 19.89P

(a)

To determine

The frequency (fn) of small oscillation for a is equal to 0.5 m.

(a)

Expert Solution
Check Mark

Answer to Problem 19.89P

The frequency (fn) of small oscillation for a is equal to 0.5 m is 0.802Hz_.

Explanation of Solution

Given information:

The mass (mS) of the sphere is 2 kg.

The mass (mABC) of the bar ABC is 1.5 kg.

The length (l) of the bar is 0.7 m.

The spring constant (k) is 200N/m.

The value of a is 0.5 m.

Assume the acceleration due to gravity (g) as 9.81m/s2.

Calculation:

Show the position 1 and position 2 of the system as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 19.3, Problem 19.89P

For position 1(Maximum velocity):

Calculate the mass moment of inertia bar ABC (IABC) using the relation:

IABC=112mABCl2

Substitute 1.5 kg for mABC and 0.7 m for l.

IABC=112×1.5×(0.7)2=0.06125kgm2

Write the expression the maximum velocity of sphere (vS)m:

(vS)m=lθ˙m

Write the expression the maximum velocity of rod ABC (vABC):

vABC=l2θ˙m

Write the expression for the kinetic energy (T1):

T1=12mS(vS)m2+12(IABC)θ˙m2+12mABC(vABC)m2

Substitute 2 kg for mS, lθ˙m for (vS)m, 0.06125kgm2 for IABC, 1.5 kg for mABC, l2θ˙m for vABC, and 0.7 m for l.

T1=12×2×(0.72θ˙m2)+12(0.06125)θ˙m2+12×1.5×(0.72)2θ˙m2=[0.49+0.030625+0.091875]θ˙m2=0.6125θ˙m2

Write the expression for the potential energy (V1):

V1=0

For position 2:

Write the expression for the kinetic energy (T2):

T2=0

Write the expression for the displacement in sphere (hS):

hS=(llcosθm)=l(1cosθm)((1cosθm=2sin2θm2))=l2sin2θm2

For small oscillation 2sin2θm2θm22.

hS=lθm22

Write the expression for the displacement of rod ABC (hABC):

hABC=(l2l2cosθm)=l2(1cosθm)((1cosθm=2sin2θm2))=l22sin2θm2

For small oscillation 2sin2θm2θm22.

hABC=l2θm22

Write the expression for the displacement of spring (xm):

xm=asinθm

For small angle sinθmθm.

xm=aθm

Write the expression for the potential energy (V2):

V2=12kxm2mSghSmABCghABC

Substitute aθm for xm, l2θm22 for hABC, and lθm22 for hS.

V2=12k(aθm)2mSg(lθm22)mABCg(l2θm22)

Substitute 200N/m for k, 0.7 m for l, 2 kg for mS, 1.5 kg for mABC, and 9.81m/s2 for g.

V2=12×200×(aθm)22×9.81×(0.7×θm22)1.5×9.81×(0.72θm22)=100a2θm26.867θm22.575θm2=(100a29.442)θm2

Express the term (θ˙m):

θ˙m=ωnθm

Write the expression for the conservation of energy

T1+V1=T2+V2

Substitute 0.6125θ˙m2 for T1, 0 for V1, 0 for T2 and (100a29.442)θm2 for V2.

0.6125θ˙m2+0=0+(100a29.442)θm20.6125θ˙m2=(100a29.442)θm2

Substitute ωnθm for θ˙m.

0.6125ωn2θm2=(100a29.442)θm2ωn2=(100a29.442)0.6125 (1)

Substitute 0.5 m for a in Equation (1).

ωn2=(100(0.5)29.442)0.6125ωn=25.401ωn=5.0399rad/s

Calculate the frequency (τn) of small vibration using the relation:

fn=ωn2π

Substitute 5.0399rad/s for ωn.

fn=5.03992π=0.802Hz

Therefore, the frequency (fn) of small oscillation for a is equal to 0.5 m is 0.802Hz_.

(b)

To determine

The smallest value of a for which oscillation will occur.

(b)

Expert Solution
Check Mark

Answer to Problem 19.89P

The smallest value of a for which oscillation will occur is 0.307m_.

Explanation of Solution

Given information:

The mass (mS) of the sphere is 2 kg.

The mass (mABC) of the bar ABC is 1.5 kg.

The length (l) of the bar is 0.7 m.

The spring constant (k) is 200N/m.

The value of a is 0.5 m.

Assuming the acceleration due to gravity (g) as 9.81m/s2.

Calculation:

Calculate the value of a:

Substitute 0 for ωn in Equation (1).

0=(100a29.442)0.61250=100a29.442a2=9.442100a=0.09442a=0.307m

Therefore, the smallest value of a for which oscillation will occur is 0.307m_.

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Chapter 19 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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