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 19.4, Problem 19.107P
To determine

(a)

Magnitude of maximum angular velocity of bar AB.

Expert Solution
Check Mark

Answer to Problem 19.107P

The maximum angular velocity of bar AB is 0.45 rad/s.

Explanation of Solution

Given:

Mass of attached sphere is 2kg.

Spring constant is 3.6 kN/m.

Vertical deflection is δmsin(ωft).

Maximum amplitude of deflection is 3 mm.

Frequency of rotation is 15 rad/s.

Concept used:

Draw the FBD for bar AB in equilibrium state.

Vector Mechanics for Engineers: Dynamics, Chapter 19.4, Problem 19.107P , additional homework tip  1

After giving small deflection to bar AB, Draw FBD of bar AB

Vector Mechanics for Engineers: Dynamics, Chapter 19.4, Problem 19.107P , additional homework tip  2

Draw the equivalent kinetic diagram for bar AB as.

Vector Mechanics for Engineers: Dynamics, Chapter 19.4, Problem 19.107P , additional homework tip  3

Write the expression for displacement of point C in terms of angular displacement.

xc=Lθ ...... (1)

Here, xc is the displacement of point C, L is the length of AC and θ is small angular displacement.

Write the expression for displacement of point B in terms of angular displacement.

xB=2Lθ ...... (2)

Here, xB is the displacement of point B and 2L is the length of AB.

Substitute xc for Lθ in equation (2).

xB=2xc ...... (3)

Take moment about point A for equilibrium position.

MA=mg(2L)kΔL

For static equilibrium substitute 0 for MA in above expression,

0=2mgLkΔL

Simplify the above expression.

2mg=kΔ ...... (4)

Here, Δ is the deflection in spring and g is the acceleration due to gravity.

Take moment about point A after giving the angular displacement.

MA=(M)kinetic

Substitute k(xcΔδ)(L)+mg(2L) for MA and ma(2L) for (M)kinetic in above expression.

k(xcΔδ)(L)+mg(2L)=ma(2L)k(xcδ)LkΔL+2mgL=ma2L

Substitute 2mg for kΔ in above expression.

k(xcδ)L=2maLk(xcδ)=2ma

Substitute x¨B for a in above expression.

2mx¨B=k(xcδ)2mx¨B=kxckδ

Substitute xB2 for xc in above expression.

2mx¨B=k(xB)L2kδ4mx¨B=k(xB)L2kδ

Substitute δmsin(ωft) for δ in above expression.

4mx¨B=k(xB)L2k(δmsin(ωft))

Rearrange the above expression as.

4mx¨B+k(xB)=2kδmsin(ωft) ....... (5)

Here, x¨B is the acceleration of point B, t is time and δm is the Maximum amplitude of deflection.

Write the standard equation of motion as.

meqx¨+keqx=F ...... (6)

Compare the coefficient of x¨ in equation (5) and (6).

meq=4m ...... (7)

Compare the coefficient of x in equation (5) and (6).

keq=k ...... (8)

Write the expression for maximum amplitude of force.

Pm=2kδm ...... (9)

Here, Pm is the maximum amplitude of force.

Write the expression for natural frequency of system.

ωn=keqmeq ...... (10)

Here, ωn is the natural frequency of system, keq is the equivalent stiffness of system and meq is the equivalent mass of the system.

Write the expression for the amplitude of vibration of sphere B.

xBmax=(Pmk)(11(ωf2ωn2)) ...... (11)

Here, xBmax is the maximum amplitude of vibration at point B.

Write the expression for magnitude of maximum angular velocity of bar AB.

ωmax=xBmaxωf2L ...... (12)

Here, ωmax is the maximum angular velocity of sphere.

Calculation:

Substitute 2 kg for m in equation (7).

meq=(4)(2)=8 kg

Substitute 3.6 kN/m for k in equation (8).

keq=3.6 kN/m

Substitute 3.6 kN/m for k and 3 mm for δm in equation (9).

Pm=2(3.6 kN/m(1000 N1 kN))(3 mm(1 m1000mm))=21.6N

Substitute 3.6 kN/m for k and 8 kg for m in equation (10).

ωn=(3.6 kN/m(1000 N1 kN))8=21.213 rad/s

Substitute 21.6 N for Pm, 3.6 kN/m for k, 21.213 rad/s for ωn and 15 rad/s for ωf in equation (11).

xBmax=21.6(3.6 kN/m(1000 N1 kN))(1(115221.2132))=0.012 m

Substitute 0.012 m for xBmax, 15 rad/s for ωf and 0.2 m for L in equation (12).

ωmax=(0.012)(15)2(0.2)=0.45 rad/s

The maximum angular velocity of bar AB is 0.45 rad/s.

Conclusion:

Thus, the maximum angular velocity of bar AB is 0.45 rad/s.

To determine

(b)

Magnitude of maximum acceleration of sphere B.

Expert Solution
Check Mark

Answer to Problem 19.107P

The maximum acceleration of sphere is 2.7 m/s2.

Explanation of Solution

Concept used:

Write the expression for maximum acceleration of sphere.

aBmax=xBmaxωf2 ...... (13)

Here, aBmax is the maximum acceleration of sphere.

Calculation:

Substitute 0.012m for xBmax and 15 rad/s for ωf in equation (13).

aBmax=0.012(15)2=2.7 m/s2

The maximum acceleration of sphere is 2.7 m/s2.

Conclusion:

Thus, the maximum acceleration of sphere is 2.7 m/s2.

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Chapter 19 Solutions

Vector Mechanics for Engineers: Dynamics

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