Целую неделю "убил" на эту задачу. Вердикт таков: результаты уважаемых
lel0lel и
fred1996 будут совпадать в пределе малых деформаций (точнее когда интегральная относительная деформация мала). А это как раз то что нам нужно, поскольку в задаче требуется применение закона Гука в дифференциальной форме. в случае больших деформаций решение
lel0lel более корректно, поскольку учитывает измение поперечного сечения. Однако, при больших деформациях нужно учитывать и отклонения нелинейные от закона Гука (см. топик.
topic128791.html). Ниже приведу правильное "школьное" решение (уж простите за мой аглицкий).
Let us consider the homogeneous tourniquet of mass
![$m$ $m$](https://dxdy-01.korotkov.co.uk/f/0/e/5/0e51a2dede42189d77627c4d742822c382.png)
and natural length
![$l$ $l$](https://dxdy-03.korotkov.co.uk/f/2/f/2/2f2322dff5bde89c37bcae4116fe20a882.png)
with stiffness
![$k=ES/l$ $k=ES/l$](https://dxdy-03.korotkov.co.uk/f/a/7/b/a7bfa55cfaa2ea284e5af23f550d70ab82.png)
, where
![$E$ $E$](https://dxdy-01.korotkov.co.uk/f/8/4/d/84df98c65d88c6adf15d4645ffa25e4782.png)
is the Young's modulus,
![$S$ $S$](https://dxdy-03.korotkov.co.uk/f/e/2/5/e257acd1ccbe7fcb654708f1a866bfe982.png)
is the cross-sectional area of the tourniquet. One of its end is fixed on a smooth table, and point mass
![$M$ $M$](https://dxdy-04.korotkov.co.uk/f/f/b/9/fb97d38bcc19230b0acd442e17db879c82.png)
is attached to the other end. We assume that the tourniquet uniformly rotates in horizontal plane with an angular velocity of
![$\omega$ $\omega$](https://dxdy-03.korotkov.co.uk/f/a/e/4/ae4fb5973f393577570881fc24fc205482.png)
. Let
![$x$ $x$](https://dxdy-04.korotkov.co.uk/f/3/3/2/332cc365a4987aacce0ead01b8bdcc0b82.png)
be the horizontal coordinate of some cross-section before stretching. In this case, according to the Hooke's law, the absolute value of tension force
![$T$ $T$](https://dxdy-03.korotkov.co.uk/f/2/f/1/2f118ee06d05f3c2d98361d9c30e38ce82.png)
at the cross-section with coordinate
![$x$ $x$](https://dxdy-04.korotkov.co.uk/f/3/3/2/332cc365a4987aacce0ead01b8bdcc0b82.png)
is equal to
![$$
T(x)=E S \frac{{\rm d}u}{{\rm d}x}=k l\frac{{\rm d}u}{{\rm d}x} .
$$ $$
T(x)=E S \frac{{\rm d}u}{{\rm d}x}=k l\frac{{\rm d}u}{{\rm d}x} .
$$](https://dxdy-02.korotkov.co.uk/f/9/3/e/93e0f554a3aceb71fae353f76475897782.png)
where
![$u(x)>0$ $u(x)>0$](https://dxdy-04.korotkov.co.uk/f/f/a/e/faee54fb0cf6ae9fe1d9f23b5105491b82.png)
is the elongation of the tourniquet element of length
![$x$ $x$](https://dxdy-04.korotkov.co.uk/f/3/3/2/332cc365a4987aacce0ead01b8bdcc0b82.png)
. Equation (1) (the linear theory) is valid for small deformations, where
![$u(x)\ll x$ $u(x)\ll x$](https://dxdy-04.korotkov.co.uk/f/7/8/5/78540ad90c8f105a0fbc968ad162bbc582.png)
and
![${\rm d}u \ll {\rm d}x$ ${\rm d}u \ll {\rm d}x$](https://dxdy-02.korotkov.co.uk/f/d/0/7/d07b638e84115141380bc61dbcc1204382.png)
.
In the reference frame associated with the rotating tourniquet, the equilibrium equation of its infinitesimal element
![${\rm d}x$ ${\rm d}x$](https://dxdy-03.korotkov.co.uk/f/2/d/8/2d83f4fa245ee0e19b46ae02d661318382.png)
has the following form:
![$$
T+{\rm d}T + F_{\rm centrifugal}-T=0,
$$ $$
T+{\rm d}T + F_{\rm centrifugal}-T=0,
$$](https://dxdy-02.korotkov.co.uk/f/9/3/d/93d0cb4da280ed520ec768393631e5fd82.png)
or
![$$
{\rm d}T(x) =-{\rm d}m \omega^2 \left[ x+u(x)\right] \approx -{\rm d}m \omega^2 x,
$$ $$
{\rm d}T(x) =-{\rm d}m \omega^2 \left[ x+u(x)\right] \approx -{\rm d}m \omega^2 x,
$$](https://dxdy-01.korotkov.co.uk/f/c/5/6/c5635070fb60971817331b1dd1e6f73982.png)
where
![${\rm d}m=(m/l){\rm d}x$ ${\rm d}m=(m/l){\rm d}x$](https://dxdy-02.korotkov.co.uk/f/5/3/d/53df0276a6ec588897d70232b953241482.png)
is the mass of the infinitesimal element
![${\rm d}x$ ${\rm d}x$](https://dxdy-03.korotkov.co.uk/f/2/d/8/2d83f4fa245ee0e19b46ae02d661318382.png)
.
Using equations (1) and (2), we get:
![$$
\frac{{{\rm d^2}u}}{{\rm d}x^2}=-\frac{m \omega^2}{kl^2}x.
$$ $$
\frac{{{\rm d^2}u}}{{\rm d}x^2}=-\frac{m \omega^2}{kl^2}x.
$$](https://dxdy-02.korotkov.co.uk/f/d/a/5/da538dfd30d8c63f22008d51681f091382.png)
Hence
![$$
\frac{{\rm d}u}{{\rm d}x}=-\frac{m \omega^2}{2kl^2}x^2+C_1.
$$ $$
\frac{{\rm d}u}{{\rm d}x}=-\frac{m \omega^2}{2kl^2}x^2+C_1.
$$](https://dxdy-03.korotkov.co.uk/f/e/4/5/e453205a3f8a89c5d90f5698f0c0847282.png)
Since
![$T(l+u(l))=M\omega^2(l+u(l))\approx T(l)\approx M\omega^2 l$ $T(l+u(l))=M\omega^2(l+u(l))\approx T(l)\approx M\omega^2 l$](https://dxdy-01.korotkov.co.uk/f/0/6/9/069cd2c637cbdc6ab0267a3edbfe7f4e82.png)
, then using equation (1), we obtain the integration constant:
![$$
C_1=\left( M+\frac{m}{2}\right)\frac{\omega^2}{k}.
$$ $$
C_1=\left( M+\frac{m}{2}\right)\frac{\omega^2}{k}.
$$](https://dxdy-04.korotkov.co.uk/f/7/8/a/78ac5614f923e663c3998db30499adfa82.png)
Therefore, in the case of
![$m\neq 0$ $m\neq 0$](https://dxdy-01.korotkov.co.uk/f/8/d/c/8dc46ff2b6bb335459c416513f008b8c82.png)
the deformation is inhomogeneous (
![${\rm d}u/{\rm d}x \neq \operatorname{const}$ ${\rm d}u/{\rm d}x \neq \operatorname{const}$](https://dxdy-04.korotkov.co.uk/f/b/2/d/b2dfe227ac09ca83014f43c17fbd414982.png)
). Equations (1), (5), and (6) also tell us that the tension force is maximum at the tourniquet attachment point and decreases quadratically with increasing
![$x$ $x$](https://dxdy-04.korotkov.co.uk/f/3/3/2/332cc365a4987aacce0ead01b8bdcc0b82.png)
-coordinate.
Considering again equations (5) and (6), we derive:
![$$
u(x)=-\frac{m \omega^2}{6kl^2}x^3+\left( M+\frac{m}{2}\right)\frac{\omega^2}{k}x+C_2.
$$ $$
u(x)=-\frac{m \omega^2}{6kl^2}x^3+\left( M+\frac{m}{2}\right)\frac{\omega^2}{k}x+C_2.
$$](https://dxdy-02.korotkov.co.uk/f/d/9/7/d971e0f73575a6f0aeccdcf4f292b16882.png)
At
![$u=0$ $u=0$](https://dxdy-02.korotkov.co.uk/f/9/9/8/998a4942cf3e6a4f88122511dc88431882.png)
. Then
![$C_2=0$ $C_2=0$](https://dxdy-01.korotkov.co.uk/f/0/3/3/03344b7e00d7c6b1ded15487279600a382.png)
. Applying equation (7), we find the total elongation:
![$$
\Delta l=u(l)=\left( M+\frac{m}{3}\right)\frac{\omega^2 l}{k}.
$$ $$
\Delta l=u(l)=\left( M+\frac{m}{3}\right)\frac{\omega^2 l}{k}.
$$](https://dxdy-04.korotkov.co.uk/f/7/7/7/777c1035c783918ba208909766f2079a82.png)