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.
$$\begin{align}
& T+V=\frac{1}{2}\left[ mk_{G}^{2}+m{{d}^{2}} \right]{{\theta }^{2}}+mg\left( d \right)\left( 1-\cos \theta \right) \\
& \left( mk_{G}^{2}+m{{d}^{2}} \right)\dot{\theta }\ddot{\theta }+gd\left( \sin \theta \right)\dot{\theta }=0 \\
& \sin \theta \approx \theta \\
& \ddot{\theta }+\frac{gd}{\left( k_{G}^{2}+{{d}^{2}} \right)}\theta =0 \\
& T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\left( k_{G}^{2}+{{d}^{2}} \right)}{gd}} \\
\end{align}$$
 |
Figure 1.1 - Body of arbitrary shape |
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.
$$\begin{align}
& T+V=\frac{1}{2}\left[ +m{{\left( \frac{a}{\sqrt{2}} \right)}^{2}} \right]{{{\dot{\theta }}}^{2}}+mg\left( \frac{a}{\sqrt{2}} \right)\left( 1-\cos \theta \right) \\
& \frac{2}{3}m{{a}^{2}}\dot{\theta }\ddot{\theta }+mg\left( \frac{a}{\sqrt{2}} \right)\left( \sin \theta \right)\dot{\theta }=0 \\
& \sin \theta =\theta \\
& \ddot{\theta }+\frac{3g}{2\sqrt{2}a}\theta =0 \\
& T=\frac{2\pi }{\omega }=\frac{2\pi }{1.0299}\left( \sqrt{\frac{a}{g}} \right)=6.10\sqrt{\frac{a}{g}} \\
\end{align}$$
 |
Figure 1.2 - Square plate |
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.
$$\begin{align}
& s=0.75\theta , \\
& \dot{s}=0.75\dot{\theta }, \\
& T+V=\frac{1}{2}\left[ \frac{1}{2}\left( \frac{15}{32.2} \right){{\left( 0.75 \right)}^{2}} \right]{{{\dot{\theta }}}^{2}}+\frac{1}{2}\left( \frac{3}{32.2} \right){{\left( 0.75\dot{\theta } \right)}^{2}} \\
& +\frac{1}{2}\left( 80 \right){{\left( {{s}_{eq}}+0.75 \right)}^{2}}-3\left( 0.75\theta \right) \\
& 0=0.1834\dot{\theta }\ddot{\theta }+80\left( {{s}_{eq}}+0.75\theta \right)\left( 0.75\dot{\theta } \right)-2.25\dot{\theta } \\
& {{F}_{eq}}=80{{s}_{eq}}=3 \\
& {{s}_{eq}}=0.0375\text{ft} \\
& 0.1834\ddot{\theta }+45\theta =0, \\
& \ddot{\theta }+245.36\theta =0, \\
& T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{245.3}}=0.401s \\
\end{align}$$
 |
Figure 1.3 - Disk pinned at point O, supporting the block A |
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 |
Applying the energy principle method we get:
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.
$$\begin{align}
& T+V=const. \\
& s=\left( 2r \right)\theta \\
& T+V=\frac{1}{2}\left[ \frac{1}{2}m{{r}^{2}}+m{{r}^{2}} \right]{{{\dot{\theta }}}^{2}}+\frac{1}{2}k{{\left( 2r\theta \right)}^{2}} \\
& 0=\frac{3}{2}m{{r}^{2}}\dot{\theta }\ddot{\theta }+4k{{r}^{2}}\theta \dot{\theta } \\
& \ddot{\theta }+\frac{8k}{3m}\theta =0 \\
& \tau =\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{8k}{3m}}}=3.85\sqrt{\frac{m}{k}} \\
\end{align}$$
 |
Figure 1.6 - Disk connected with spring |
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.
$$\begin{align}
& V={{V}_{e}}=\frac{1}{2}k{{\theta }^{2}} \\
& {{I}_{z}}=M{{\left( {{k}_{z}} \right)}^{2}} \\
& {{T}_{1}}=\frac{1}{2}{{I}_{z}}{{{\dot{\theta }}}^{2}}=\frac{1}{2}Mk_{z}^{2}{{{\dot{\theta }}}^{2}} \\
& {{T}_{1}}+V=\operatorname{constant} \\
& \frac{1}{2}Mk_{z}^{2}{{{\dot{\theta }}}^{2}}+\frac{1}{2}k{{\theta }^{2}}=\operatorname{constant} \\
& Mk_{z}^{2}\dot{\theta }\ddot{\theta }+k\theta \dot{\theta }=0 \\
& Mk_{z}^{2}\ddot{\theta }+k\theta =0 \\
& \ddot{\theta }+\frac{k}{Mk_{z}^{2}}\theta =0 \\
& {{\omega }_{n}}=\sqrt{\frac{k}{Mk_{z}^{2}}} \\
& \tau =\frac{2\pi }{{{\omega }_{n}}}=2\pi \sqrt{\frac{Mk_{z}^{2}}{k}} \\
\end{align}$$
 |
Figure 1.8 - Torsional spring |
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
$$\begin{align}
& \sum{{{F}_{y}}}=0, \\
& 2mg-{{\left( {{F}_{sp}} \right)}_{st}}=0 \\
& {{\left( {{F}_{sp}} \right)}_{st}}=2mg \\
& 2{{s}_{p}}+{{s}_{c}}=l, \\
& 2\Delta {{s}_{p}}-\Delta {{s}_{c}}=0, \\
& \Delta {{s}_{p}}=-\frac{\Delta {{s}_{c}}}{2}=\Delta {{s}_{c}}\uparrow \\
& {{V}_{e}}=\frac{1}{2}k{{\left( {{s}_{0}}+{{s}_{1}} \right)}^{2}}=\frac{1}{2}k{{\left( \frac{2mg}{k}+\frac{y}{2} \right)}^{2}} \\
& {{V}_{g}}=-Wy=-mgy, \\
& T+V=\operatorname{constant} \\
& \frac{1}{2}m{{{\dot{y}}}^{2}}+k{{\left( \frac{2mg}{k}+\frac{y}{2} \right)}^{2}}-mgy=\operatorname{constnat} \\
& m\dot{y}\ddot{y}+k\left( \frac{2mg}{k}+\frac{y}{2} \right)\left( \frac{{\dot{y}}}{2} \right)-mg\dot{y}=0 \\
& \dot{y}\left( m\ddot{y}+\frac{k}{4}y \right)=0 \\
& m\ddot{y}+\frac{k}{4}y=0 \\
& \ddot{y}+\frac{k}{4m}y=0 \\
& {{\omega }_{n}}=\frac{1}{2}\sqrt{\frac{k}{m}} \\
& {{f}_{n}}=\frac{{{\omega }_{n}}}{4\pi }=\frac{1}{4\pi }\sqrt{\frac{k}{m}} \\
\end{align}$$
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