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.

Answer to Problem 4.67P

The equation of motion for the system shown in part (a) is $m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}-2K_1{L}_1^2\right)\theta =0$.

The equation of motion for the system shown in part (b) is $m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}+2K_1{L}_1^2\right)\theta =0$.

The system is neutrally stable.

Explanation of Solution

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

Figure-(1)

Here, the distance of mass from pivot point is $L$, the distance of the spring from pivot point is $L_1$, the angular displacement is $\theta$, 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).

$m{L}^2\ddot{\theta }=\left(mg\sin \varphi \right)L-2k\left(\delta +L_1\theta \right)L_1$..... (I)

Here, the angular acceleration is $\ddot{\theta }$, the angle of displacement of pendulum is $\varphi$, the static deflection in the spring is $\delta$, the acceleration due to gravity is $g$.

Substitute $\sin \left(45°+\theta \right)$ for $\sin \varphi$ in Equation (I).

$\begin{array}{l}m{L}^2\ddot{\theta }=\left(mg\sin \left(45°+\theta \right)\right)L-2k\left(\delta +L_1\theta \right)L_1\\ m{L}^2\ddot{\theta }=mgL\left(\sin 45°\cos \theta +\cos 45°\sin \theta \right)-2k{L}_1\delta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }=mgL\frac{\sqrt{2}}{2}+mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1\delta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }=mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta +mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \end{array}$        ....... (II)

Since $\theta$ is small, Substitute $0$ for $\theta$ in Equation (II).

$\begin{array}{c}m{L}^2\left(0\right)=mgL\frac{\sqrt{2}}{2}\left(0\right)-2k{L}_1^2\left(0\right)+mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \\ 0=mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \end{array}$        ....... (III)

Substitute $0$ for $mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta$ in Equation (II).

$\begin{array}{l}m{L}^2\ddot{\theta }=mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta +\left(0\right)\\ m{L}^2\ddot{\theta }=mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}-2k{L}_1^2\right)\theta =0\end{array}$        ....... (IV)

Take Laplace for Equation (IV).

$\begin{array}{l}m{L}^{2}L\left[\ddot{\theta }\right]+\left(mgL\frac{\sqrt{2}}{2}-2k{L}_1^2\right)L\left[\theta \right]=0\\ m{L}^2{s}^2\theta \left(s\right)+\left(mgL\frac{\sqrt{2}}{2}-2k{L}_1^2\right)\theta \left(s\right)=0\\ \left[m{L}^2{s}^2+\left(mgL\frac{\sqrt{2}}{2}-2k{L}_1^2\right)\right]\theta \left(s\right)=0\\ m{L}^2{s}^2+mgL\frac{\sqrt{2}}{2}-2k{L}_1^2=0\end{array}$        ....... (V)

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

Figure-(2)

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

$m{L}^2\ddot{\theta }=\left(mg\sin \varphi \right)L-2k\left(\delta +L_1\theta \right)L_1$..... (VI)

Substitute $\sin \left(135°+\theta \right)$ for $\sin \varphi$ in Equation (VI).

$\begin{array}{l}m{L}^2\ddot{\theta }=\left(mg\sin \left(135°+\theta \right)\right)L-2k\left(\delta +L_1\theta \right)L_1\\ m{L}^2\ddot{\theta }=mgL\left(\sin 135°\cos \theta +\cos 135°\sin \theta \right)-2k{L}_1\delta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }=mgL\frac{\sqrt{2}}{2}-mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1\delta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }=-mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta +mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \end{array}$        ....... (VII)

Since $\theta$ is small, Substitute $0$ for $\theta$ in Equation (VII).

$\begin{array}{c}m{L}^2\left(0\right)=-mgL\frac{\sqrt{2}}{2}\left(0\right)-2k{L}_1^2\left(0\right)+mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \\ 0=mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta \end{array}$        ....... (VIII)

Substitute $0$ for $mgL\frac{\sqrt{2}}{2}-2k{L}_1\delta$ in Equation (VII).

$\begin{array}{l}m{L}^2\ddot{\theta }=-mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta +\left(0\right)\\ m{L}^2\ddot{\theta }=-mgL\frac{\sqrt{2}}{2}\theta -2k{L}_1^2\theta \\ m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}+2k{L}_1^2\right)\theta =0\end{array}$..... (IX)

Take Laplace for Equation (IX).

$\begin{array}{l}m{L}^{2}L\left[\ddot{\theta }\right]+\left(mgL\frac{\sqrt{2}}{2}+2k{L}_1^2\right)L\left[\theta \right]=0\\ m{L}^2{s}^2\theta \left(s\right)+\left(mgL\frac{\sqrt{2}}{2}+2k{L}_1^2\right)\theta \left(s\right)=0\\ \left[m{L}^2{s}^2+\left(mgL\frac{\sqrt{2}}{2}+2k{L}_1^2\right)\right]\theta \left(s\right)=0\\ m{L}^2{s}^2+\left(mgL\frac{\sqrt{2}}{2}+2k{L}_1^2\right)=0\end{array}$        ....... (X)

Write the expression for neutral stable condition.

$mgL\frac{\sqrt{2}}{2}+2k_1{L}_1^2>0$        ....... (XI)

Write the expression for stable condition.

$mgL\frac{\sqrt{2}}{2}-2k_1{L}_1^2\ge 0$        ....... (XII)

Write the expression for unstable condition.

$mgL\frac{\sqrt{2}}{2}-2k_1{L}_1^2<0$        ....... (XIII)

Conclusion:

The equation of motion for the system shown by the Equation (V) for system shown in part (a) is $m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}-2K_1{L}_1^2\right)\theta =0$.

The equation of motion for the system shown by the Equation (X) for the system shown in part (b) is $m{L}^2\ddot{\theta }+\left(mgL\frac{\sqrt{2}}{2}+2K_1{L}_1^2\right)\theta =0$.

The system is neutrally stable.