Chapter 19.3, Problem 19.89P

To determine

(a)

The frequency of small oscillation.

Answer to Problem 19.89P

The frequency of small oscillation is $\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}$.

Explanation of Solution

Given:

Length of inverted pendulum is $l$.

Mass of inverted pendulum is $m$.

Spring constant is $k$.

Concept used:

Write the expression for the potential energy for the maximum displacement position.

$$

$V_1=\frac{1}{2}k{x}_m^2-mg{y}_m$

Here, $V_1$ is the potential energy at maximum displacement position, $k$ is the spring constant, $x_m$ is the deflection in spring, $m$ is the mass of inverted pendulum, $g$ is acceleration due to gravity and $y_m$ is the maximum displacement.

Substitute $a\sin {\theta }_m$ for $x_m$ and $\frac{l}{2}\left(1-\cos {\theta }_m\right)$ for $y_m$ in the above expression.

$V_1=\frac{1}{2}k{a}^2{\sin }^2{\theta }_m-\frac{1}{2}mgl\left(1-\cos {\theta }_m\right)$

As ${\theta }_m$ is very small then $1-\cos {\theta }_m\approx \frac{{\theta }_m^2}{2}$.

Substitute $\frac{{\theta }_m^2}{2}$ for $\left(1-\cos {\theta }_m\right)$ and ${\theta }_m^2$ for ${\sin }^2{\theta }_m$ in above expression.

$V_1=\frac{{\theta }_m^2}{2}\left(k{a}^2-\frac{1}{2}mgl\right)$ ...... (1)

Here, $l$ is the length of inverted pendulum.

Kinetic energy of inverted pendulum at maximum displacement is $0$.

Write the expression for the kinetic energy at maximum velocity position.

$T_2=\frac{1}{2}m{v}^2+I{\dot{\theta }}^2$

Here, $T_2$ is the kinetic energy of system at maximum velocity position, $v$ is the linear velocity of system, $I$ is the moment of inertia of rod and $\dot{\theta }$ is the angular velocity of rod.

Substitute $\frac{m{l}^2}{12}$ for $I$ and $\frac{l\dot{\theta }}{2}$ for $v$ in above expression.

$T_2=\frac{1}{2}m{\left(\frac{l\dot{\theta }}{2}\right)}^2+\frac{1}{2}\left(\frac{m{l}^2}{12}\right){\dot{\theta }}^2$

Substitute $-{\omega }_n{\theta }_m$ for $\dot{\theta }$ in the above expression and simplify the above expression.

$T_2=\frac{1}{6}m{l}^2{\omega }_n^2{\theta }_m^2$ ...... (2)

Here, ${\omega }_n$ is the natural frequency of system.

The potential energy of the system at maximum velocity position is 0.

Write the expression for the conservation of energy.

$T_1+V_1=T_2+V_2$ ...... (3)

Write the expression for the natural frequency of oscillation.

$f_n=\frac{{\omega }_n}{2\pi }$ ...... (4)

Here, $f_n$ is the natural frequency of system.

Calculation:

Substitute $\frac{{\theta }_m^2}{2}\left(k{a}^2-\frac{1}{2}mgl\right)$ for $V_1$, $0$ for $T_2$, $\frac{1}{6}m{l}^2{\omega }_n^2{\theta }_m^2$ for $T_2$ and $0$ for $V_2$ in equation (3).

$\begin{array}{l}\frac{{\theta }_m^2}{2}\left(k{a}^2-\frac{1}{2}mgl\right)=\frac{1}{6}m{l}^2{\omega }_n^2{\theta }_m^2\\ \left(k{a}^2-\frac{1}{2}mgl\right)=\frac{1}{3}m{l}^2{\omega }_n^2\\ \left(6k{a}^2-3mgl\right)=2m{l}^2{\omega }_n^2\end{array}$

Simplify the above expression for ${\omega }_n$.

$\begin{array}{l}{\omega }_n^2=\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}\\ {\omega }_n=\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}\end{array}$

Substitute $\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}$ for ${\omega }_n$ in equation (4).

$\begin{array}{l}{f}_n=\frac{\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}}{2\pi }\\ f_n=\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}\end{array}$

The frequency of small oscillation is $\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}$.

Conclusion:

Thus, the frequency of small oscillation is $\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}$.

To determine

(b)

The smallest value of $a$ for which smallest oscillation occurs.

Answer to Problem 19.89P

The smallest value of $a$ for which smallest oscillation occurs is $\sqrt{\frac{mgl}{2k}}$.

Explanation of Solution

Concept used:

For the oscillation to occur, the value of natural frequency should be real.

$f_n\ge 0$ ...... (5)

Calculation:

Substitute $\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}$ for $f_n$ in equation (5).

$\begin{array}{c}\frac{1}{2\pi }\sqrt{\frac{\left(6k{a}^2-3mgl\right)}{2m{l}^2}}\ge 0\\ \left(6k{a}^2-3mgl\right)\ge 0\\ a^2\ge \frac{mgl}{2k}\\ a\ge \sqrt{\frac{mgl}{2k}}\end{array}$

The smallest value of $a$ for which smallest oscillation occurs is $\sqrt{\frac{mgl}{2k}}$.

Conclusion:

Thus, the smallest value of $a$ for which smallest oscillation occurs is $\sqrt{\frac{mgl}{2k}}$.