Advanced Mechanics and Dynamical Systems

Virtual displacement. An imaginary, infinitesimally small displacement of a point of a system, consistent with the constraints, made instantaneously (time is not allowed to change) and without regard to the forces that cause actual motion.

Virtual work. The work done by a force during a virtual displacement: $\delta W = \vec F\cdot\delta\vec r$.

Principle of virtual work (coplanar forces on a particle). A particle acted on by coplanar forces $\vec F_1,\vec F_2,\dots$ is in equilibrium if and only if the total virtual work is zero for every small virtual displacement $\delta\vec r$:

$\sum \vec F_i\cdot\delta\vec r = 0 \;\Rightarrow\; \delta x\sum X_i+\delta y\sum Y_i=0 \;\Rightarrow\; \sum X_i=0,\ \sum Y_i=0 .$

Hexagon of rods. Let each rod have length $l$ and weight $W$. Rod $AB$ is fixed horizontally and the hexagon hangs below it. Let $BC$ and $AF$ make an angle $\theta$ with the horizontal (in the regular hexagon $\theta=60^\circ$). By symmetry $CD$ and $FE$ also make angle $\theta$, and $DE$ stays horizontal. The depths below $AB$ are:

• centres of $BC$ and $AF$: $\tfrac12 l\sin\theta$ each
• centres of $CD$ and $FE$: $l\sin\theta+\tfrac12 l\sin\theta=\tfrac32 l\sin\theta$ each
• rod $DE$: $2l\sin\theta$

Sum of the depths of the centres of gravity of the five free rods:

$l\sin\theta+3l\sin\theta+2l\sin\theta=6l\sin\theta .$

The length of the string joining the mid-points of $AB$ and $DE$ is $y=2l\sin\theta$.

Give a small virtual change $\delta\theta$. The weights do work $W\,\delta(6l\sin\theta)$ and the tension $T$ (which resists the lengthening of the string) does work $-T\,\delta(2l\sin\theta)$. By the principle of virtual work

$W\cdot 6l\cos\theta\,\delta\theta-T\cdot 2l\cos\theta\,\delta\theta=0 \;\Rightarrow\; T=3W .$

Thus, with $W$ the weight of each rod, the virtual-work equation gives a tension of $3W$ in the string. The value $W/\sqrt3$ quoted in the question does not follow from this set-up (it would need different rod weights or a different string).

OR: Resultant of forces. The resultant of a system of forces is the single force that has the same effect on a rigid body as the whole system.

Take $P$ along the $y$-axis ($x=0$), $Q$ along the $x$-axis ($y=0$), both in the positive sense, and $R$ along the line $x\cos\theta-y\sin\theta=p$, whose direction is $(\sin\theta,\cos\theta)$. Then

$X=Q+R\sin\theta,\qquad Y=P+R\cos\theta .$

Magnitude of the resultant:

$\mathcal R=\sqrt{X^2+Y^2}=\sqrt{P^2+Q^2+R^2+2PR\cos\theta+2QR\sin\theta}.$

Moment about the origin: $P$ and $Q$ pass through the origin, and $R$ lies on a line at perpendicular distance $p$ from it, so $G=Rp$. The line of action of the resultant is $xY-yX=G$:

$(P+R\cos\theta)\,x-(Q+R\sin\theta)\,y=Rp .$

1. Definition of common catenary

A common catenary is the curve in which a uniform, flexible, inextensible string or chain hangs freely under its own weight when suspended from two fixed points. The lowest point of the curve is called the vertex, and the tangent at the vertex is horizontal.

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2. Equation of the uniform catenary in intrinsic form

Let

• (C) be the vertex,
• (P) be any point on the chain,
• (s = \text{arc } CP),
• (\psi) be the angle which the tangent at (P) makes with the horizontal,
• (w) be the weight per unit length of the chain,
• (T_0) be the tension at the vertex (C).

Consider the equilibrium of the portion (CP) of the chain. The forces acting on it are:

1. Tension (T) at (P), along the tangent,
2. Tension (T_0) at (C), horizontal,
3. Weight (ws) of the portion (CP), acting vertically downward.

Resolving horizontally:

[ T\cos\psi = T_0 ]

Resolving vertically:

[ T\sin\psi = ws ]

Dividing the vertical equation by the horizontal equation,

[ \tan\psi = \frac{ws}{T_0} ]

Put

[ c = \frac{T_0}{w} ]

Then

[ \tan\psi = \frac{s}{c} ]

or

[ \boxed{s = c\tan\psi} ]

This is the intrinsic equation of the common catenary.

Also, from this equation, the Cartesian form can be obtained as

[ \boxed{y + c = c\cosh\left(\frac{x}{c}\right)} ]

where the origin is taken at the vertex.

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3. Numerical problem

Given:

[ AB = 462 \text{ ft} ]

Since (A) and (B) lie in the same horizontal line, the vertex (C) lies midway between them. Hence the horizontal distance from the vertex to either support is

[ x = \frac{462}{2} = 231 \text{ ft} ]

Let the tangent at (A) make an angle (\alpha) with the horizontal. Given

[ \alpha = \tan^{-1}\left(\frac{3}{4}\right) ]

so

[ \tan\alpha = \frac{3}{4} ]

For the catenary,

[ y + c = c\cosh\left(\frac{x}{c}\right) ]

Differentiating,

[ \frac{dy}{dx} = \sinh\left(\frac{x}{c}\right) ]

At (A),

[ \left|\frac{dy}{dx}\right| = \tan\alpha = \frac{3}{4} ]

Therefore,

[ \sinh\left(\frac{231}{c}\right) = \frac{3}{4} ]

Let

[ u = \frac{231}{c} ]

Then

[ \sinh u = \frac{3}{4} ]

Now,

[ \sinh u = \frac{e^u - e^{-u}}{2} = \frac{3}{4} ]

[ e^u - e^{-u} = \frac{3}{2} ]

Let (z = e^u). Then

[ z - \frac{1}{z} = \frac{3}{2} ]

[ 2z^2 - 3z - 2 = 0 ]

[ z = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} ]

Taking the positive value,

[ z = 2 ]

Hence

[ e^u = 2 ]

[ u = \ln 2 ]

So

[ \frac{231}{c} = \ln 2 ]

[ c = \frac{231}{\ln 2} ]

The length of the chain from the vertex to one support is

[ s = c\sinh\left(\frac{231}{c}\right) ]

But

[ \sinh\left(\frac{231}{c}\right) = \frac{3}{4} ]

Therefore,

[ s = c \cdot \frac{3}{4} ]

Total length of the chain is

[ L = 2s = 2 \cdot c \cdot \frac{3}{4} = \frac{3c}{2} ]

Substituting (c = \dfrac{231}{\ln 2}),

[ L = \frac{3}{2} \cdot \frac{231}{\ln 2} ]

[ \boxed{L = \frac{693}{2\ln 2} \text{ ft}} ]

Numerically,

[ \ln 2 \approx 0.693147 ]

[ L \approx \frac{693}{2(0.693147)} \approx 499.9 \text{ ft} ]

[ \boxed{L \approx 499.9 \text{ ft}} ]

Hence the length of the chain is approximately 499.9 ft.

Definition:
The centre of gravity of a body is the point through which the resultant weight of the body acts, whatever be the orientation of the body. For a uniform plane lamina, the centre of gravity coincides with its centroid.

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Centre of gravity of a quadrant of an ellipse

Let the ellipse be

[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ]

Take the quadrant in the first quadrant, bounded by (x=0), (y=0), and the ellipse. Its area is

[ A=\frac14(\pi ab)=\frac{\pi ab}{4}. ]

1. (x)-coordinate of centre of gravity, (\bar{x})

Consider a vertical strip at distance (x) from the (y)-axis, of width (dx).
For this strip,

[ y=b\sqrt{1-\frac{x^2}{a^2}}. ]

Hence the area of the strip is

[ dA=y,dx=b\sqrt{1-\frac{x^2}{a^2}},dx. ]

The moment of this strip about the (y)-axis is (x,dA). Therefore,

[ \bar{x}=\frac{1}{A}\int_0^a x,dA ]

[ =\frac{1}{A}\int_0^a x,b\sqrt{1-\frac{x^2}{a^2}},dx. ]

Put

[ x=a\sin\theta,\qquad dx=a\cos\theta,d\theta. ]

When (x=0), (\theta=0); when (x=a), (\theta=\frac{\pi}{2}). Then

[ \int_0^a x,b\sqrt{1-\frac{x^2}{a^2}},dx =ab\int_0^{\pi/2} a\sin\theta\cos^2\theta,d\theta ]

[ =a^2b\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi/2} =\frac{a^2b}{3}. ]

Thus

[ \bar{x}=\frac{a^2b/3}{\pi ab/4} =\frac{4a}{3\pi}. ]

2. (y)-coordinate of centre of gravity, (\bar{y})

Similarly, taking a horizontal strip at distance (y) from the (x)-axis,

[ x=a\sqrt{1-\frac{y^2}{b^2}}, ]

so

[ \bar{y}=\frac{1}{A}\int_0^b y,a\sqrt{1-\frac{y^2}{b^2}},dy. ]

By the same integration,

[ \int_0^b y,a\sqrt{1-\frac{y^2}{b^2}},dy =\frac{ab^2}{3}. ]

Hence

[ \bar{y}=\frac{ab^2/3}{\pi ab/4} =\frac{4b}{3\pi}. ]

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Result

The centre of gravity of the quadrant of the ellipse is

[ \boxed{\left(\bar{x},\bar{y}\right)=\left(\frac{4a}{3\pi},\frac{4b}{3\pi}\right)} ]

measured from the centre of the ellipse along the two axes of the quadrant.

Moment of inertia of a uniform rod. Take the rod along the $x$-axis from $-a$ to $a$ with the axis through its mid-point, perpendicular to the rod. Mass per unit length is $M/2a$. For an element $dx$ at distance $x$:

$I=\int_{-a}^{a}\frac{M}{2a}\,x^2\,dx=\frac{M}{2a}\cdot\frac{2a^3}{3}=\frac{Ma^2}{3}.$

OR: A particle falling from infinity through a hole to the centre of the earth.

Treat the earth as a uniform sphere of mass $M$ and radius $R$ (with $g=GM/R^2$), with a narrow hole along a diameter; neglect air resistance and rotation. Let $r$ be the distance of the particle from the centre.

1. Motion outside the earth ($r\ge R$). The force is $mGM/r^2=mgR^2/r^2$ towards the centre. The particle starts from rest at infinity, so by energy conservation

$\tfrac12 v^2-\frac{GM}{r}=0\;\Rightarrow\; v^2=\frac{2GM}{r}=\frac{2gR^2}{r}.$

At the surface $v=\sqrt{2gR}\approx 11.2\ \text{km s}^{-1}$ (the escape speed). The fall from infinity takes an infinite time.

2. Motion inside the hole ($r\le R$). The attraction at distance $r$ is due only to the mass inside radius $r$, so it is proportional to $r$:

$\ddot r=-\frac{g}{R}\,r .$

This is simple harmonic motion with $\omega=\sqrt{g/R}$. Energy conservation from the surface, where $v^2=2gR$:

$v^2=2gR+\frac{g}{R}\left(R^2-r^2\right)=\frac{g}{R}\left(3R^2-r^2\right).$

The motion is $r=\sqrt3\,R\sin\omega t$, with $t$ measured from the instant the particle is at the centre. The amplitude $\sqrt3R$ is larger than $R$, so the particle leaves the hole before reaching the end of the SHM path.

3. Speed at the centre.

$v_{\max}^2=3gR\;\Rightarrow\; v_{\max}=\sqrt{3gR}\approx 13.7\ \text{km s}^{-1}.$

4. Time through the hole. The surface is reached when $\sin\omega t=1/\sqrt3$, so the time from the surface to the centre is

$t_1=\sqrt{\frac{R}{g}}\,\sin^{-1}\frac{1}{\sqrt3}\approx 8.3\ \text{min}.$

The passage from one surface to the other takes $2t_1\approx16.6$ minutes.

Conclusion. The particle accelerates from rest at infinity, enters the hole with speed $\sqrt{2gR}$, reaches its maximum speed $\sqrt{3gR}$ at the centre, and leaves the other side with speed $\sqrt{2gR}$. Its total energy is zero, so it then decelerates and just reaches infinity with zero speed. It does not oscillate and never returns.

Tangential and normal components of acceleration. For a particle moving on a plane curve with speed $v$, where $\phi$ is the angle the tangent makes with a fixed line and $\rho=ds/d\phi$ is the radius of curvature, the acceleration is resolved along the tangent and the principal normal:

• Tangential component: $a_t=\dfrac{dv}{dt}=v\dfrac{dv}{ds}$ (rate of change of speed, along the tangent).
• Normal component: $a_n=\dfrac{v^2}{\rho}=v^2\dfrac{d\phi}{ds}=v\dfrac{d\phi}{dt}$ (rate of change of direction, towards the centre of curvature).

Proof. Given:

$v\frac{dv}{ds}=f\ \text{(constant)},\qquad \frac{v}{a_n}=k\ \text{(constant)}.$

From the second condition $a_n=v/k$, and since $a_n=v^2\,d\phi/ds$,

$v\frac{d\phi}{ds}=\frac1k .$

Dividing the first relation by this one:

$\frac{dv}{d\phi}=kf \;\Rightarrow\; v=kf\,\phi+c_1 .$

Then $\dfrac{ds}{d\phi}=kv=k\left(kf\,\phi+c_1\right)$, and integrating,

$s=\frac{k^2f}{2}\,\phi^2+kc_1\,\phi+c_2 ,$

which is the intrinsic equation, a polynomial in $\phi$ with $s$ and $\phi$ the only variables.

Form $s=A\phi+B$. The linear form follows when the ratio that is constant is $v^2/a_n$. Then $a_n=v^2/A$ gives

$v^2\frac{d\phi}{ds}=\frac{v^2}{A}\;\Rightarrow\;\frac{ds}{d\phi}=A\;\Rightarrow\; s=A\phi+B,$

since $v^2/a_n=\rho=ds/d\phi$ is then constant. With the ratio $v/a_n$ constant, as in the literal statement, the path has $s$ quadratic in $\phi$ as derived above.