Day: July 25, 2023

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202307251108 Dynamics Figures (Elementary) Q10

On a wedge of slope \tan\theta with friction, a trolley has a composite mass (M+2m)\,\mathrm{kg} by counting M the mass of body and 2m the masses of its two wheels, each having moment of inertia I=\frac{1}{2}mr^2 for r the radius. The front wheel \text{}_{F} and the rear wheel \text{}_{R} are separated by a distance d.

Parallel to the plane,

F_{\textrm{net}}=W\sin\theta -f_{F}-f_{R} = (M+2m)a

perpendicular to the plane,

\begin{aligned} & \qquad \begin{cases} N_{F}+N_{R} =W\cos\theta \\ xN_F+(x+d)N_R =(x+\frac{d}{2}) W\cos\theta \\ \end{cases} \\ & \Longrightarrow N_F=N_R=\frac{1}{2}(W\cos\theta )\\ \end{aligned}

Note the coefficient of friction (COF) for the static \mu=\mu_\textrm{s} and the kinetic \mu=\mu_\textrm{k}. For each case,

v(t)\begin{cases} v=0 \\ v>0\textrm{ and }\frac{\mathrm{d}v}{\mathrm{d}t}=0 \\ v>0\textrm{ and }\frac{\mathrm{d}v}{\mathrm{d}t}>0 \\ \end{cases}

discuss it.

Roughwork.

For the system, the kinetic energy is

\begin{aligned} T & = \frac{1}{2}(M+2m)v^2+\frac{1}{2}I\omega^2\times 2 \\ & = \frac{1}{2}(M+2m)\dot{x}^2+\bigg(\frac{1}{2}mr^2\bigg)\bigg(\frac{v}{r}\bigg)^2 \\ & = \frac{1}{2}(M+2m)\dot{x}^2+\frac{1}{2}m\dot{x}^2 \\ & = \frac{1}{2}(M+3m)\dot{x}^2 \\ \end{aligned}

and the potential energy

V=-(M+2m)gx\sin\theta.

such that \mathcal{L}=T-V the Lagrangian is

\mathcal{L}(x,\dot{x})\displaystyle{=\frac{1}{2}(M+3m)\dot{x}^2+(M+2m)gx\sin\theta}

Recall Euler-Lagrange equation:

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

Ignoring frictional dissipation (viz. nonsense), one will see

\ddot{x}=\displaystyle{\bigg(\frac{M+2m}{M+3m}\bigg)g\sin\theta}.

With a scalar dissipation function D(x,\dot{x}), the 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}=-\frac{\partial D}{\partial \dot{x}}}}

where D=\mu_\textrm{k}Nv shall answer.

(to be continued)