Equation for θ Deduce an equation satisfied by θ alone and find the speeds of [latex] P_{1} \text { and } P_{2} [/latex] when the string becomes vertical.

From the linear momentum equation (10.38), [latex] 4 \dot{x}+a \dot{\theta} \cos \theta=0 [/latex], (10.38) [latex] \dot{x}=-\frac{1}{4} a \dot{\theta} \cos \theta [/latex] and, if we now eliminate [latex] \dot{x} [/latex] from the energy equation (10.39), we obtain, after simplification [latex] 4 \dot{x}^{2}+a^{2} \dot{\theta}^{2}+2 a \dot{x} \dot{\theta} \cos \theta=g a(2 \cos \theta-1).[/latex] (10.39) [latex] \dot{\theta}^{2}=\frac{4 g}{a}\left(\frac{2 \cos \theta-1}{4-\cos ^{2} \theta}\right), [/latex] (10.40) which is an equation for θ alone. It follows from this equation that, when the string becomes vertical (that is, when θ = 0), [latex]\dot{\theta}^{2}[/latex] = 4g/3a. Hence, at this instant, [latex] \dot{\theta}=-(4 g / 3 a)^{1 / 2} [/latex] and (from the momentum conservation equation) [latex] \dot{x}=+(g / 12 a)^{1 / 2} [/latex]. Hence, the speed of [latex] P_{1} \text { is }(a g / 12)^{1 / 2} [/latex] and the speed of [latex] P_{2} \text { is }|\dot{x}+a \dot{\theta}|=(3 a g / 4)^{1 / 2}.[/latex]

Question 10.Q.1: Equation for θ Deduce an equation satisfied by θ alone and f...

Classical Mechanics

Equation for θ

Deduce an equation satisfied by θ alone and find the speeds of $P_{1} \text { and } P_{2}$ when the string becomes vertical.

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From the linear momentum equation (10.38),

$4 \dot{x}+a \dot{\theta} \cos \theta=0$ , (10.38)

$\dot{x}=-\frac{1}{4} a \dot{\theta} \cos \theta$

and, if we now eliminate $\dot{x}$ from the energy equation (10.39), we obtain, after simplification

$4 \dot{x}^{2}+a^{2} \dot{\theta}^{2}+2 a \dot{x} \dot{\theta} \cos \theta=g a(2 \cos \theta-1).$ (10.39)

$\dot{\theta}^{2}=\frac{4 g}{a}\left(\frac{2 \cos \theta-1}{4-\cos ^{2} \theta}\right),$ (10.40)

which is an equation for θ alone.
It follows from this equation that, when the string becomes vertical (that is, when θ = 0), $\dot{\theta}^{2}$ = 4g/3a. Hence, at this instant, $\dot{\theta}=-(4 g / 3 a)^{1 / 2}$ and (from the momentum conservation equation) $\dot{x}=+(g / 12 a)^{1 / 2}$ . Hence, the speed of $P_{1} \text { is }(a g / 12)^{1 / 2}$ and the speed of $P_{2} \text { is }|\dot{x}+a \dot{\theta}|=(3 a g / 4)^{1 / 2}.$