202211021054 Dynamics Figures (Elementary) Q9

Referring to the figure below, the pulleys are assumed to be weightless and frictionless and the ropes massless and inextensible.

Prove that m_1=2m_2 if the system is in equilibrium. Find the equation of motion for each mass, i. if m_1>2m_2; and ii. if m_1<2m_2.


Roughwork.

In equilibrium,

\textrm{Net }\mathbf{F}=m\mathbf{a}=\mathbf{0}.

Thus

\begin{cases} W_1-2T =m_1(0) \\ W_2-T =m_2(0) \end{cases} \Longrightarrow\quad \begin{cases} m_1g =2T \\ m_2g =T \\ \end{cases}

\therefore m_1=2m_2.

Mass m_1 will move down and m_2 up if m_1>2m_2, and the reverse if m_1<2m_2. When m_1 makes displacement of \Delta h the vertical, m_2 makes 2\Delta h.

Without gain of speciality, one may obtain the equations of motion by either (a) Newton’s second law, or (b) conservation of mechanical energy, or (c) Euler-Lagrange method.

Take downward positive. When m_1>2m_2, in one’s mind one can draw two free-body diagrams giving two equations:

\begin{cases} W_1(=m_1g)-2T=m_1a_1\\ W_2(=m_2g)-T=-m_2a_2 \end{cases}

Hence,

\begin{cases} 2T = m_1(g-a_1) \\ T = m_2(g+a_2) \end{cases}

note that 2a_1=a_2. And so

\begin{aligned} a_1 & = \bigg( 1-\frac{6m_2}{m_1+4m_2}\bigg)\cdot g \\ a_2 & = \bigg( \frac{3m_1}{m_1+4m_2}-1\bigg)\cdot g \\ \end{aligned}

Note the inequalities 0\leqslant a_1\leqslant a_2<g. One can thus find tension in magnitude T, yet this is left the reader.

The Lagrangian \mathcal{L}=T-V of the system is

\mathcal{L}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+m_1g\Delta h-2m_2g\Delta h,

or,

\mathcal{L}=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2(\dot{2x})^2+m_1gx-2m_2gx.

where v_1=v_1(t) and v_2=v_2(t) vary with time t as are subjected to acceleration, and v_2=2v_1.

Euler-Lagrange equation reads

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial \mathcal{L}}{\partial\dot{x}}\bigg)-\frac{\partial\mathcal{L}}{\partial x}=0}.

Computing term by term,

\begin{aligned} \frac{\partial \mathcal{L}}{\partial x} & = m_1g-2m_2g \\ \frac{\partial\mathcal{L}}{\partial \dot{x}} & = m_1\dot{x}+4m_2\dot{x} \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial\dot{x}}\bigg) & = m_1\ddot{x}+4m_2\ddot{x} \\ \end{aligned}

and the result follows.

(to be continued)