Правка раздела «Solution»
en/2.7.27.md
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| ### Statement | |||
| $2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released? | |||
|  | |||
| ### Solution | |||
|  | |||
| Newton's second law for a rotational movement | |||
| $$ | |||
| M = I\varepsilon | |||
| $$ | |||
| $$ | |||
| m_1gR - m_2 g R = I\varepsilon\tag{1} | |||
| $$ | |||
| Since there is no slippage: | |||
| $$ | |||
| \varepsilon = \frac{a}{R} | |||
| $$ | |||
| Substituting into $(1)$ | |||
| $$ | |||
| \boxed{(m_1 - m_2) g R^2 = Ia}\tag{2} | |||
| $$ | |||
| Conservation of angular momentum | |||
| $$ | |||
| I\omega = m_1vR + m_2vR\tag{3} | |||
| $$ | |||
| Let's express velocity through angular velocity | |||
| $$ | |||
| v=\omega R\tag{4} | |||
| $$ | |||
| After substituting $(4)$ into $(3)$ | |||
| $$ | |||
| \boxed{I = (m_1+m_2)R^2} \tag{5} | |||
| $$ | |||
| Let's equate the expressions $(2)$ and $(5)$ | |||
| $$ | |||
| (m_1 - m_2) g R^2 = (m_1+m_2)aR^2 | |||
| $$ | |||
| After mathematical transformations | |||
| $$ | |||
| a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6} | |||
| $$ | |||
| Describe the forces acting on the vertical axis | |||
| $$ | |||
| N = (m_1+m_2)g - a (m_1-m_2)\tag{7} | |||
| $$ | |||
| @@ -69,7 +69,7 @@Solution | |||
| Substituting the acceleration $(6)$ into expression $(7)$ | |||
| $$ | |||
| − | N = g\cdot\left((m_1+m_2) | ||
| + | N = g\cdot\left((m_1+m_2)- \frac{(m_1-m_2)^2}{m_1+m_2}\right) | ||
| $$ | |||
| From here we find the support reaction force: | |||
| $$ | |||
| \boxed{N = \frac{4m_1m_2g}{m_1+m_2}} | |||
| $$ | |||
| #### Answer | |||
| $$ | |||
| N = 4m_1m_2g/(m_1 + m_2) | |||
| $$ | |||
| Mustafa Bakhodirov | |||
| ещё строк без изменений 12 | |||
| ### Statement | ### Statement | ||
| $2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released? | $2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released? | ||
|  |  | ||
| ### Solution | ### Solution | ||
|  |  | ||
| Newton's second law for a rotational movement | Newton's second law for a rotational movement | ||
| $$ | $$ | ||
| M = I\varepsilon | M = I\varepsilon | ||
| $$ | $$ | ||
| $$ | $$ | ||
| m_1gR - m_2 g R = I\varepsilon\tag{1} | m_1gR - m_2 g R = I\varepsilon\tag{1} | ||
| $$ | $$ | ||
| Since there is no slippage: | Since there is no slippage: | ||
| $$ | $$ | ||
| \varepsilon = \frac{a}{R} | \varepsilon = \frac{a}{R} | ||
| $$ | $$ | ||
| Substituting into $(1)$ | Substituting into $(1)$ | ||
| $$ | $$ | ||
| \boxed{(m_1 - m_2) g R^2 = Ia}\tag{2} | \boxed{(m_1 - m_2) g R^2 = Ia}\tag{2} | ||
| $$ | $$ | ||
| Conservation of angular momentum | Conservation of angular momentum | ||
| $$ | $$ | ||
| I\omega = m_1vR + m_2vR\tag{3} | I\omega = m_1vR + m_2vR\tag{3} | ||
| $$ | $$ | ||
| Let's express velocity through angular velocity | Let's express velocity through angular velocity | ||
| $$ | $$ | ||
| v=\omega R\tag{4} | v=\omega R\tag{4} | ||
| $$ | $$ | ||
| After substituting $(4)$ into $(3)$ | After substituting $(4)$ into $(3)$ | ||
| $$ | $$ | ||
| \boxed{I = (m_1+m_2)R^2} \tag{5} | \boxed{I = (m_1+m_2)R^2} \tag{5} | ||
| $$ | $$ | ||
| Let's equate the expressions $(2)$ and $(5)$ | Let's equate the expressions $(2)$ and $(5)$ | ||
| $$ | $$ | ||
| (m_1 - m_2) g R^2 = (m_1+m_2)aR^2 | (m_1 - m_2) g R^2 = (m_1+m_2)aR^2 | ||
| $$ | $$ | ||
| After mathematical transformations | After mathematical transformations | ||
| $$ | $$ | ||
| a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6} | a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6} | ||
| $$ | $$ | ||
| Describe the forces acting on the vertical axis | Describe the forces acting on the vertical axis | ||
| $$ | $$ | ||
| N = (m_1+m_2)g - a (m_1-m_2)\tag{7} | N = (m_1+m_2)g - a (m_1-m_2)\tag{7} | ||
| $$ | $$ | ||
| @@ -69,7 +69,7 @@Solution | |||
| Substituting the acceleration $(6)$ into expression $(7)$ | Substituting the acceleration $(6)$ into expression $(7)$ | ||
| $$ | $$ | ||
| N = g\cdot\left((m_1+m_2) |
N = g\cdot\left((m_1+m_2)- \frac{(m_1-m_2)^2}{m_1+m_2}\right) | ||
| $$ | $$ | ||
| From here we find the support reaction force: | From here we find the support reaction force: | ||
| $$ | $$ | ||
| \boxed{N = \frac{4m_1m_2g}{m_1+m_2}} | \boxed{N = \frac{4m_1m_2g}{m_1+m_2}} | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| N = 4m_1m_2g/(m_1 + m_2) | N = 4m_1m_2g/(m_1 + m_2) | ||
| $$ | $$ | ||
| Mustafa Bakhodirov | Mustafa Bakhodirov | ||
| ещё строк без изменений 12 | |||