Energy Methods

Example 1.1

The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of kG. If it is displaced a slight amount from its equilibrium position and released, determine the natural period of vibration. 
 

Figure 1.1 - Body of arbitrary shape

$$\begin{align} & T+V=\frac{1}{2}\left[ mk_{G}^{2}+m{{d}^{2}} \right]{{\theta }^{2}}+mg\left( d \right)\left( 1-\cos \theta \right) \\ & \left( mk_{G}^{2}+m{{d}^{2}} \right)\dot{\theta }\ddot{\theta }+gd\left( \sin \theta \right)\dot{\theta }=0 \\ & \sin \theta \approx \theta \\ & \ddot{\theta }+\frac{gd}{\left( k_{G}^{2}+{{d}^{2}} \right)}\theta =0 \\ & T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\left( k_{G}^{2}+{{d}^{2}} \right)}{gd}} \\ \end{align}$$

Example 1.2

A square plate has a mass m and is suspended at its corner from a pin O. Determine the natural period of vibration if it is displaced a small amount and released.

Figure 1.2 - Square plate

$$\begin{align} & T+V=\frac{1}{2}\left[ +m{{\left( \frac{a}{\sqrt{2}} \right)}^{2}} \right]{{{\dot{\theta }}}^{2}}+mg\left( \frac{a}{\sqrt{2}} \right)\left( 1-\cos \theta \right) \\ & \frac{2}{3}m{{a}^{2}}\dot{\theta }\ddot{\theta }+mg\left( \frac{a}{\sqrt{2}} \right)\left( \sin \theta \right)\dot{\theta }=0 \\ & \sin \theta =\theta \\ & \ddot{\theta }+\frac{3g}{2\sqrt{2}a}\theta =0 \\ & T=\frac{2\pi }{\omega }=\frac{2\pi }{1.0299}\left( \sqrt{\frac{a}{g}} \right)=6.10\sqrt{\frac{a}{g}} \\ \end{align}$$

Example 1.3

The disk, having a weight of 15 lb is pinned at its center O and supports the block A that has a weight of 3 lb. If the belt which passes over the disk does not slip at its contacting surface, determine the natural period of vibration of the system.

Figure 1.3 - Disk pinned at point O, supporting the block A

$$\begin{align} & s=0.75\theta , \\ & \dot{s}=0.75\dot{\theta }, \\ & T+V=\frac{1}{2}\left[ \frac{1}{2}\left( \frac{15}{32.2} \right){{\left( 0.75 \right)}^{2}} \right]{{{\dot{\theta }}}^{2}}+\frac{1}{2}\left( \frac{3}{32.2} \right){{\left( 0.75\dot{\theta } \right)}^{2}} \\ & +\frac{1}{2}\left( 80 \right){{\left( {{s}_{eq}}+0.75 \right)}^{2}}-3\left( 0.75\theta \right) \\ & 0=0.1834\dot{\theta }\ddot{\theta }+80\left( {{s}_{eq}}+0.75\theta \right)\left( 0.75\dot{\theta } \right)-2.25\dot{\theta } \\ & {{F}_{eq}}=80{{s}_{eq}}=3 \\ & {{s}_{eq}}=0.0375\text{ft} \\ & 0.1834\ddot{\theta }+45\theta =0, \\ & \ddot{\theta }+245.36\theta =0, \\ & T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{245.3}}=0.401s \\ \end{align}$$

The system can be solved without applying the energy method. For equilibrium we have:

$$\begin{align} & {{T}_{st}}=3\text{ lb} \\ & \sum{{{M}_{O}}}={{I}_{O}}\alpha +ma\left( 0.75 \right) \\ & a=0.75\alpha \\ & -{{T}_{st}}0.75-80\cdot {{0.75}^{2}}\theta +3\cdot 0.75=\left[ \frac{1}{2}\left( \frac{15}{32.2} \right)\cdot {{0.75}^{2}} \right]\ddot{\theta }+\left( \frac{3}{32.2} \right)\cdot {{0.75}^{2}}\ddot{\theta } \\ & -2.25-45\theta +2.25=0.131\ddot{\theta }+0.05241\ddot{\theta } \\ & \ddot{\theta }+245.3\theta =0 \\ & T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{245.3}}=0.401\text{ s} \\ \end{align}$$

 

So as you can see by applying both methods we got same result.

Example 1.4

The safe has mass m and is uniformly supported by four springs, each having a stiffness k. Determine the natural period of vertical vibration by applying energy method.

Figure 1.4 - Safe supported by four springs

First of all we know that that sum of potential and kinetic energy is constant.

$$T+V=const.$$

Now we will calculate kinetic and potential energy and determine the value of the natural period.

$$\begin{align} & T=\frac{1}{2}m{{{\dot{y}}}^{2}} \\ & V=mgy+\frac{1}{2}4k{{\left( \Delta s-y \right)}^{2}} \\ & T+V=\frac{1}{2}m{{{\dot{y}}}^{2}}+mgy+\frac{1}{2}4k{{\left( \Delta s-y \right)}^{2}} \\ & T+V=\frac{1}{2}m{{{\dot{y}}}^{2}}+mgy+\frac{1}{2}4k\left( {{\left( \Delta s \right)}^{2}}-2\Delta sy+{{y}^{2}} \right) \\ & m\dot{y}\ddot{y}+mg\dot{y}-4k\left( \Delta sy \right)\dot{y}=0 \\ & m\ddot{y}+mg-4ky-4k\Delta s=0 \\ & \text{Since }\Delta s=\frac{mg}{4k} \\ & m\ddot{y}+4ky=0 \\ & \ddot{y}+\frac{4k}{m}y=0 \\ & \omega =\sqrt{\frac{4k}{m}}, \\ & T=\frac{2\pi }{\omega }=\pi \sqrt{\frac{m}{k}} \\ \end{align}$$

Example 1.5

Determine the differential equation of motion of the 15 kg spool. Assume that it does not slip at the surface of contact as it oscillates. The radius of gyration of the spool about its center of mass is 125 mm. The springs are originally un-stretched.

Figure 1.5 - Spool connected with springs

Since the spool rolls without slipping, there is no friction, the stretching of both spring can be approximated as x1=0.1θ, and x2=0.2θ when the spool is being displaced by a small angular displacement θ. 
 Applying the energy principle method we get:

$$\begin{align} & U=T+V \\ & {{V}_{P}}=\frac{1}{2}kx_{1}^{2}+\frac{1}{2}kx_{2}^{2}=\frac{1}{2}200\cdot {{0.1}^{2}}{{\theta }^{2}}+\frac{1}{2}200\cdot {{0.2}^{2}}{{\theta }^{2}}=5{{\theta }^{2}} \\ & T=\frac{1}{2}{{I}_{A}}{{\omega }^{2}} \\ & {{I}_{A}}=15\cdot {{0.125}^{2}}+15\cdot {{0.1}^{2}}=0.3845375 \\ & T=\frac{1}{2}{{I}_{A}}{{\omega }^{2}}=\frac{1}{2}0.3845375{{{\dot{\theta }}}^{2}}=0.1921875{{{\dot{\theta }}}^{2}} \\ & U=T+V=0.1921875{{{\dot{\theta }}}^{2}}+5{{\theta }^{2}} \\ & 0.3845375\dot{\theta }\ddot{\theta }+10\theta \dot{\theta }=0 \\ & \dot{\theta }\left( 0.3845375\ddot{\theta }+10\theta \right)=0 \\ & \dot{\theta }\ne 0 \\ & 0.3845375\ddot{\theta }+10\theta =0/:0.3845375 \\ & \ddot{\theta }+26\theta =0 \\ & \\ \end{align}$$

Example 1.6

Determine the natural period of vibration of the disk having a mass m and radius r. assume the disk doesn’t slip on the surface of contact as it oscillates. 
 

Figure 1.6 - Disk connected with spring

$$\begin{align} & T+V=const. \\ & s=\left( 2r \right)\theta \\ & T+V=\frac{1}{2}\left[ \frac{1}{2}m{{r}^{2}}+m{{r}^{2}} \right]{{{\dot{\theta }}}^{2}}+\frac{1}{2}k{{\left( 2r\theta \right)}^{2}} \\ & 0=\frac{3}{2}m{{r}^{2}}\dot{\theta }\ddot{\theta }+4k{{r}^{2}}\theta \dot{\theta } \\ & \ddot{\theta }+\frac{8k}{3m}\theta =0 \\ & \tau =\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{8k}{3m}}}=3.85\sqrt{\frac{m}{k}} \\ \end{align}$$

Example 1.7

If the wheel is given a small angular displacement of [theta] and released from rest, it is observed that it oscillates with a natural period of T. Determine the wheels radius of gyration about its center of mass G. The wheel’s radius of gyration about its center of mass G. The wheel has a mass of m and rolls on the rails without slipping.

Figure 1.7 -Wheel

With reference to the datum established in previous figure the gravitational potential energy of the wheel is:

$$V=-W{{y}_{G}}=-mgR\cos \theta$$

The kinetic energy of the wheel:

$$\begin{align} & T=\frac{1}{2}mv_{G}^{2}+\frac{1}{2}{{I}_{G}}{{\omega }^{2}} \\ & T=\frac{1}{2}m{{\left( \dot{\theta }R \right)}^{2}}+\frac{1}{2}\left( mk_{G}^{2} \right){{\left[ \left( \frac{R}{r} \right)\dot{\theta } \right]}^{2}} \\ & T=\frac{1}{2}m{{R}^{2}}\left( \frac{{{r}^{2}}+k_{G}^{2}}{{{r}^{2}}} \right){{{\dot{\theta }}}^{2}} \\ \end{align}$$

The energy function of the wheel is:

$$\begin{align} & T+V=\operatorname{constant} \\ & \frac{1}{2}m{{R}^{2}}\left( \frac{{{r}^{2}}+k_{G}^{2}}{{{r}^{2}}} \right){{{\dot{\theta }}}^{2}}-mgR\cos \theta =\operatorname{constant} \\ & m{{R}^{2}}\left( \frac{{{r}^{2}}+k_{G}^{2}}{{{r}^{2}}} \right)\dot{\theta }\ddot{\theta }+mgR\sin \theta \dot{\theta }=0 \\ & m{{R}^{2}}\left( \frac{{{r}^{2}}+k_{G}^{2}}{{{r}^{2}}} \right)\ddot{\theta }+mgR\sin \theta =0 \\ & \ddot{\theta }+\frac{g}{R}\left( \frac{{{r}^{2}}}{{{r}^{2}}+k_{G}^{2}} \right)\theta =0 \\ & {{\omega }_{n}}=\sqrt{\frac{g}{R}\left( \frac{{{r}^{2}}}{{{r}^{2}}+k_{G}^{2}} \right)} \\ & \tau =\frac{2\pi }{{{\omega }_{n}}}=2\pi \sqrt{\frac{g}{R}\left( \frac{{{r}^{2}}}{{{r}^{2}}+k_{G}^{2}} \right)} \\ & {{k}_{G}}=\frac{r}{2\pi }\sqrt{\frac{{{\tau }^{2}}g-4{{\pi }^{2}}R}{R}} \\ \end{align}$$

Example 1.8

A torsional spring of stiffness k is attached to a wheel that has a mass of  M. If the wheel is given a small angular displacement of [theta] determine the natural period of oscillation. The wheel has a radius of gyration about the z axis of kz. 

Figure 1.8 - Torsional spring

$$\begin{align} & V={{V}_{e}}=\frac{1}{2}k{{\theta }^{2}} \\ & {{I}_{z}}=M{{\left( {{k}_{z}} \right)}^{2}} \\ & {{T}_{1}}=\frac{1}{2}{{I}_{z}}{{{\dot{\theta }}}^{2}}=\frac{1}{2}Mk_{z}^{2}{{{\dot{\theta }}}^{2}} \\ & {{T}_{1}}+V=\operatorname{constant} \\ & \frac{1}{2}Mk_{z}^{2}{{{\dot{\theta }}}^{2}}+\frac{1}{2}k{{\theta }^{2}}=\operatorname{constant} \\ & Mk_{z}^{2}\dot{\theta }\ddot{\theta }+k\theta \dot{\theta }=0 \\ & Mk_{z}^{2}\ddot{\theta }+k\theta =0 \\ & \ddot{\theta }+\frac{k}{Mk_{z}^{2}}\theta =0 \\ & {{\omega }_{n}}=\sqrt{\frac{k}{Mk_{z}^{2}}} \\ & \tau =\frac{2\pi }{{{\omega }_{n}}}=2\pi \sqrt{\frac{Mk_{z}^{2}}{k}} \\ \end{align}$$

Example 1.9

Determine the frequency of oscillation of the cylinder of mass m when it is pulled own slightly and released. Neglect the mass of the small pulley.

Figure 1.9

Solution: Potential and Kinetic Energy: Referring to the free body diagram of the system at equilibrium position as shown in figure a

$$\begin{align} & \sum{{{F}_{y}}}=0, \\ & 2mg-{{\left( {{F}_{sp}} \right)}_{st}}=0 \\ & {{\left( {{F}_{sp}} \right)}_{st}}=2mg \\ & 2{{s}_{p}}+{{s}_{c}}=l, \\ & 2\Delta {{s}_{p}}-\Delta {{s}_{c}}=0, \\ & \Delta {{s}_{p}}=-\frac{\Delta {{s}_{c}}}{2}=\Delta {{s}_{c}}\uparrow \\ & {{V}_{e}}=\frac{1}{2}k{{\left( {{s}_{0}}+{{s}_{1}} \right)}^{2}}=\frac{1}{2}k{{\left( \frac{2mg}{k}+\frac{y}{2} \right)}^{2}} \\ & {{V}_{g}}=-Wy=-mgy, \\ & T+V=\operatorname{constant} \\ & \frac{1}{2}m{{{\dot{y}}}^{2}}+k{{\left( \frac{2mg}{k}+\frac{y}{2} \right)}^{2}}-mgy=\operatorname{constnat} \\ & m\dot{y}\ddot{y}+k\left( \frac{2mg}{k}+\frac{y}{2} \right)\left( \frac{{\dot{y}}}{2} \right)-mg\dot{y}=0 \\ & \dot{y}\left( m\ddot{y}+\frac{k}{4}y \right)=0 \\ & m\ddot{y}+\frac{k}{4}y=0 \\ & \ddot{y}+\frac{k}{4m}y=0 \\ & {{\omega }_{n}}=\frac{1}{2}\sqrt{\frac{k}{m}} \\ & {{f}_{n}}=\frac{{{\omega }_{n}}}{4\pi }=\frac{1}{4\pi }\sqrt{\frac{k}{m}} \\ \end{align}$$