- #1
bookworm031
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- Homework Statement
- Determine, using Euler-Lagrange's equation, the acceleration for B when the weights are moving vertically.
- Relevant Equations
- ##\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{y}}\bigg) = \frac{\partial L}{\partial y}##, ##L = T - V##
##m_{A} = 3 kg##
##m_{B} = 2 kg##
##y_{A} + y_{B} = c \Leftrightarrow y_{A} = c - y_{B}##, where c is a constant.
##\Rightarrow \dot{y_{A}} = -\dot{y_{B}}##
The Lagrangian:
$$L = T - V$$
##T =\frac{1}{2}m_{A}\dot{y_{B}}^{2} + \frac{1}{2}m_{B}\dot{y_{B}}^{2}##
##V = m_{A}g(c - y_{B}) + m_{B}gy_{B}##
##\Leftrightarrow L = \frac{1}{2}m_{A}\dot{y_{B}}^{2} + \frac{1}{2}m_{B}\dot{y_{B}}^{2} - (m_{A}g(c - y_{B}) + m_{B}gy_{B})##
Applying Euler-Lagrange's equation:
##\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{y}}\bigg) = \ddot{y_{B}}(m_{B} + m_{A})##
##\frac{\partial L}{\partial y} = g(m_{A} - m_{B})##
Solving for ##\ddot{y_{B}}##:
##\ddot{y_{B}} = \frac{g(m_{A} - m_{B})}{(m_{B} + m_{A})} = 1.9604 \frac{m}{s^{2}}##
The answer is supposed to be ##1.78 \frac{m}{s^{2}}##. What am I doing wrong? I'm completely lost.
Thanks!
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