Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.2

Posted on by

Structure and Interpretation of Classical Mechanics

.

…

\displaystyle{\begin{aligned} R \frac{d^2 \theta}{dt^2} &= g \sin \theta - \frac{\beta R \dot \theta^2}{2} \\ \end{aligned}}

DSolve[
    R theta''[t]
        == g Sin[theta[t]]
            - (beta*R (theta'[t])^2)/2,

    theta[t],
    t
] // Simplify // TeXForm

\displaystyle{\begin{aligned}     c_1 + t &= - \int_1^{\theta(t)} \frac{\sqrt{\beta^2 R + R}}{\sqrt{\beta^2 c_1 R e^{- \beta \xi} + c_1 R e^{- \beta \xi} + 2 \beta g \sin \xi - 2 g \cos \xi}} d \xi \\    c_1 + t &=   \int_1^{\theta(t)} \frac{\sqrt{\beta^2 R + R}}{\sqrt{\beta^2 c_1 R e^{- \beta \zeta} + c_1 R e^{- \beta \zeta} + 2 \beta g \sin \zeta - 2 g \cos \zeta}} d\zeta \\    \end{aligned}}

\displaystyle{\begin{aligned}     c_1 + t &= - \int_1^{\theta(t)} \frac{\sqrt{R (\beta^2  + 1)}}{\sqrt{  (\beta^2 + 1)  c_1 R e^{- \beta \xi} + 2 g ( \beta \sin \xi - \cos \xi ) }} d \xi \\    c_1 + t &=   \int_1^{\theta(t)} \frac{\sqrt{R (\beta^2  + 1)}}{\sqrt{  (\beta^2 + 1)   c_1 R e^{- \beta \zeta} + 2 g (\beta g \sin \zeta - \cos \zeta )}} d\zeta \\    \end{aligned}}

Let

\displaystyle{\begin{aligned} u &= \dot \theta^2 \\ \end{aligned}}

Then

\displaystyle{\begin{aligned} \frac{du}{dt} &= 2 \left( \pm \sqrt{u} \right) \ddot \theta \\ \ddot \theta &= \frac{\pm 1}{2} \frac{du}{d \theta} \\ \frac{du}{d \theta} + \beta u &= \pm \frac{2 g}{R} \sin \theta \\ \end{aligned}}

This is a first-order linear differential equation. The integrating factor is

\displaystyle{\begin{aligned} e^{\int \beta d \theta} &= e^{\beta \theta} \\ \end{aligned}}

So

\displaystyle{\begin{aligned} \frac{du}{d \theta} + \beta u &= \pm \frac{2 g}{R} \sin \theta \\ \frac{d}{d \theta} \left( u e^{\beta \theta} \right) &= \pm \frac{2 g e^{\beta \theta} }{R} \sin \theta \\ \end{aligned}}

\displaystyle{\begin{aligned} u e^{\beta \theta} &= \pm \frac{2 g }{R} \int e^{\beta \theta} \sin \theta d \theta \\ u &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + C e^{-\beta \theta} \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + C e^{-\beta \theta} \\ \end{aligned}}

When t=0, \theta=0 and \dot \theta = 0:

\displaystyle{\begin{aligned} 0 &= \pm \frac{- 2 g }{R (\beta^2 + 1)} + C \\ \end{aligned}}

\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + \pm \frac{2 g }{R (\beta^2 + 1)} e^{-\beta \theta} \\ &= \pm \frac{2 g }{R (\beta^2 + 1)} \left( \beta \sin \theta - \cos \theta + e^{-\beta \theta} \right) \\ \end{aligned}}

…

— Me@2023-05-23 11:02:25 AM

.

.

2023.06.03 Saturday (c) All rights reserved by ACHK