System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 4, Problem 4.67P
To determine

The equation of motion for the system shown in part (a).

The equation of motion for the system shown in part (b).

The nature of stability of the system.

Expert Solution & Answer
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Answer to Problem 4.67P

The equation of motion for the system shown in part (a) is mL2θ¨+(mgL222K1L12)θ=0.

The equation of motion for the system shown in part (b) is mL2θ¨+(mgL22+2K1L12)θ=0.

The system is neutrally stable.

Explanation of Solution

Figure (1) shows the free body diagram of the system shown in part (a).

System Dynamics, Chapter 4, Problem 4.67P , additional homework tip  1

Figure-(1)

Here, the distance of mass from pivot point is L, the distance of the spring from pivot point is L1, the angular displacement is θ, the torsional spring constant is k the mass of the block is m.

Write the expression for equilibrium condition for the pendulum for Figure-(1).

mL2θ¨=(mgsinϕ)L2k(δ+L1θ)L1..... (I)

Here, the angular acceleration is θ¨, the angle of displacement of pendulum is ϕ, the static deflection in the spring is δ, the acceleration due to gravity is g.

Substitute sin(45°+θ) for sinϕ in Equation (I).

mL2θ¨=(mgsin(45°+θ))L2k(δ+L1θ)L1mL2θ¨=mgL(sin45°cosθ+cos45°sinθ)2kL1δ2kL12θmL2θ¨=mgL22+mgL22θ2kL1δ2kL12θmL2θ¨=mgL22θ2kL12θ+mgL222kL1δ        ....... (II)

Since θ is small, Substitute 0 for θ in Equation (II).

mL2(0)=mgL22(0)2kL12(0)+mgL222kL1δ0=mgL222kL1δ        ....... (III)

Substitute 0 for mgL222kL1δ in Equation (II).

mL2θ¨=mgL22θ2kL12θ+(0)mL2θ¨=mgL22θ2kL12θmL2θ¨+(mgL222kL12)θ=0        ....... (IV)

Take Laplace for Equation (IV).

mL2L[θ¨]+(mgL222kL12)L[θ]=0mL2s2θ(s)+(mgL222kL12)θ(s)=0[mL2s2+(mgL222kL12)]θ(s)=0mL2s2+mgL222kL12=0        ....... (V)

Figure (2) shows the free body diagram of the system shown in part (b).

System Dynamics, Chapter 4, Problem 4.67P , additional homework tip  2

Figure-(2)

Write the expression for equilibrium condition for the pendulum for Figure-(2).

mL2θ¨=(mgsinϕ)L2k(δ+L1θ)L1..... (VI)

Substitute sin(135°+θ) for sinϕ in Equation (VI).

mL2θ¨=(mgsin(135°+θ))L2k(δ+L1θ)L1mL2θ¨=mgL(sin135°cosθ+cos135°sinθ)2kL1δ2kL12θmL2θ¨=mgL22mgL22θ2kL1δ2kL12θmL2θ¨=mgL22θ2kL12θ+mgL222kL1δ        ....... (VII)

Since θ is small, Substitute 0 for θ in Equation (VII).

mL2(0)=mgL22(0)2kL12(0)+mgL222kL1δ0=mgL222kL1δ        ....... (VIII)

Substitute 0 for mgL222kL1δ in Equation (VII).

mL2θ¨=mgL22θ2kL12θ+(0)mL2θ¨=mgL22θ2kL12θmL2θ¨+(mgL22+2kL12)θ=0..... (IX)

Take Laplace for Equation (IX).

mL2L[θ¨]+(mgL22+2kL12)L[θ]=0mL2s2θ(s)+(mgL22+2kL12)θ(s)=0[mL2s2+(mgL22+2kL12)]θ(s)=0mL2s2+(mgL22+2kL12)=0        ....... (X)

Write the expression for neutral stable condition.

mgL22+2k1L12>0        ....... (XI)

Write the expression for stable condition.

mgL222k1L120        ....... (XII)

Write the expression for unstable condition.

mgL222k1L12<0        ....... (XIII)

Conclusion:

The equation of motion for the system shown by the Equation (V) for system shown in part (a) is mL2θ¨+(mgL222K1L12)θ=0.

The equation of motion for the system shown by the Equation (X) for the system shown in part (b) is mL2θ¨+(mgL22+2K1L12)θ=0.

The system is neutrally stable.

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Chapter 4 Solutions

System Dynamics

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