Question
Mathematically establish the third equqtion of rotational motion $\omega^2-\omega^2_0=2\alpha\theta$
✓
Answer
We know that angular acceleration is defined as $\alpha=\frac{\text{d}\omega}{\text{dt}}.$
$\therefore\alpha=\frac{\text{d}\omega}{\text{d}\theta}.\frac{\text{d}\theta}{\text{dt}}=\omega\frac{\text{d}\omega}{\text{d}\theta}$
$[\because\omega=\frac{\text{d}\theta}{\text{dt}}]$
or $\alpha.\text{d}.\theta=\omega\text{d}\omega$
On integrating, we have
$\int\limits^\theta_0\alpha\text{d}\theta=\int\limits^{\omega}_{\omega_0}\omega\text{d}\omega$
$\therefore[\alpha\theta]^\theta_0=\Big[\frac{\omega^2}{2}\Big]^{\omega}_{\omega_{0}}$
$\text{or }\alpha(\theta-0)=\frac{\omega^2}{2}-\frac{\omega^2_0}{2}$
$\Rightarrow\omega^2-\omega^2_0=2\alpha\theta, $ which is the requisite relation.
$\therefore\alpha=\frac{\text{d}\omega}{\text{d}\theta}.\frac{\text{d}\theta}{\text{dt}}=\omega\frac{\text{d}\omega}{\text{d}\theta}$
$[\because\omega=\frac{\text{d}\theta}{\text{dt}}]$
or $\alpha.\text{d}.\theta=\omega\text{d}\omega$
On integrating, we have
$\int\limits^\theta_0\alpha\text{d}\theta=\int\limits^{\omega}_{\omega_0}\omega\text{d}\omega$
$\therefore[\alpha\theta]^\theta_0=\Big[\frac{\omega^2}{2}\Big]^{\omega}_{\omega_{0}}$
$\text{or }\alpha(\theta-0)=\frac{\omega^2}{2}-\frac{\omega^2_0}{2}$
$\Rightarrow\omega^2-\omega^2_0=2\alpha\theta, $ which is the requisite relation.
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