IOM 2016, Теория — Задача 1. «Three Pulleys»

• 1) $T_2/T_1 = e^{\mu\pi} = 2$

  Tension of the «downstream» rope section is greater by the factor of 2.

• 2) $m_2 > \overline{m}_2 = e^{2\mu\pi}m_1 = 4m_1$

• 3) 800 g

• 4) $a = \frac{2(n-1)m_1}{2(n+1)m_1 + M\cdot e^{-\mu\pi}}g = \frac{4(n-1)m_1}{4(n+1)m_1 + M}g$

• 5) $a = \frac{g}{9}$

• 6) $a = \frac{ke^{-\mu\pi} - 1}{ke^{-\mu\pi} + 1}g = \frac{k-2}{k+2}g$

• 7) $a = \frac{g}{3}$

• 8) $a_\tau = \frac{2}{9}g$

• 9) $M \le 4(e^{2\mu\pi} - e^{\mu\pi})m_1 = 8m_1$

  The rope does not slide under the rotating pulley for any m2.

• 10) $\frac{T_2}{T_3} = \frac{14}{11}$

  Tension on the «right» to the rotating pulley is 14/11 of the tension on the «left».

TABLE OF ANSWERS:

Proposed Solution:

1. If the weights are in motion the rope has to slide over the fixed pulleys. Therefore for any rope element of a length $dl$ in a pulley groove the absolute value of friction force equals $dF_{\mathrm{fr}} = \mu dN$, where $dN$ is a normal force exerted by the pulley. The friction force balances a net tangential force of rope tension $dT$. The force pressing the rope element to the pulley is the normal component of the net tension force $Td\varphi = dN$. Here $d\varphi = dl/R$ is the angular size of the rope element). Thus $dT = dF_{\mathrm{fr}} = \mu dN = \mu Td\varphi$ and the tension of the rope sliding over a fixed pulley varies according to an exponential law $T(\varphi) = T_0\cdot e^{\mu\varphi}$.

Suppose that the weight $m_2$ is moving downward and the rope tensions are labeled in the Figure. Then according to the exponential law $T_2 = \alpha T_1$ and $T_4 = \alpha T_3$ where $\alpha \equiv e^{-\mu\pi} = 0{,}5$.

2, 3) Now let us write down equations of motion of the weights. Since the rope is inextensible the absolute value of a weight acceleration is the same. Let the vertical axis be directed upward. Then one obtains for tensions $T_{2,3}$:

Concerning the rotating pulley, the rope does not have to slide under it and they can rotate together. If this is the case the net tension $dT$ for any rope element is still balanced by a friction force, although now it is static friction ($dF_{\mathrm{fr}} < \mu dN$). The net torque exerted on the rotating pulley due to friction is $\sum R\cdot dF_{\mathrm{fr}} = R\sum dT = R(T_2 - T_3)$. Besides, if there is no sliding, a tangential acceleration of the pulley circumference equals a linear acceleration $a$ of the weights and rope, so an angular acceleration of the pulley is $\varepsilon = a/R$. Since the pulley moment of inertia can be approximated by $I = \dfrac{1}{2}MR^2$, the equation of angular motion becomes $\dfrac{1}{2}MR^2\dfrac{a}{R} = R(T_2 - T_3)$, whence $T_2 - T_3 = \dfrac{1}{2}Ma$. Substituting here the tensions found above one obtains the acceleration of weights and rope (providing the rope does not slide under the rotating pulley):

This equation is correct if two conditions are met: $a > 0$ (we assume that the system is moving) and $T_3 > \alpha T_2$ (we assume the rope does not slide under the rotating pulley, otherwise $dT = dF_{\mathrm{fr}} = \mu dN = \mu Td\varphi$ and $T_3 = \alpha T_2$). The considered motion is realised if $m_2 > \overline{m}_2 = \alpha^{-2}m_1 = 4m_1 = 800$ g (otherwise friction force holds the system at rest) and if $m_2 < \dfrac{Mm_1}{\alpha[\alpha^2 M - 4m_1(1 - \alpha)]} = \dfrac{8Mm_1}{M - 8m_1} = 2{,}4$ kg (for a greater $m_2$ the rope slides under the «middle» pulley). One can see that if $m_2$ is a bit more than 800 g the above conditions are met and the system starts moving as soon as it has been released.

4, 5) For $m_2 = n\overline{m}_2 = 1{,}6$ kg both inequalities derived above are satisfied, so $a = \dfrac{2(n-1)m_1}{2(n+1)m_1 + \alpha M}g = \dfrac{4(n-1)m_1}{4(n+1)m_1 + M}g = \dfrac{g}{9}$.

6, 7) When $m_2$ increases to $k\overline{m}_2 = 3{,}2$ kg the second inequality no longer holds. This means that the rope is now sliding under the rotating pulley and $T_3 = \alpha T_2$. The angular acceleration of the pulley is $\varepsilon < a/R$. Using expressions for the tensions derived above one obtains:

8. Equation of angular motion for the rotating pulley now becomes:

Substituting here the expressions for $m_2$ and the acceleration one obtains:

Obviously this acceleration does not exceed $a = \dfrac{g}{3}$, as it should be.

9, 10) Notice that if $M \le 4\alpha^{-2}(1 - \alpha)m_1 = \alpha^{-2}(e^{2\mu\pi} - e^{\mu\pi})m_1 = 8m_1 = 1{,}6$ kg, the condition $T_3 > \alpha T_2$ holds true for any $m_2$, i.e. there is no sliding. Therefore acceleration of the «lightweight» pulley can be evaluated according to the equation derived in 4). Substituting $m_2 = 8m_1$ and $M = 6m_1$ one obtains: $a = \dfrac{2g}{9}$. Then $T_2 = \dfrac{28}{9}m_1 g$ and $T_3 = \dfrac{22}{9}m_1 g$, whence $\dfrac{T_2}{T_3} = \dfrac{14}{11}$.