Write the kinematical equations for circular motion in analogy with linear motion.
Q 29
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For circular motion of a particle with constant angular acceleration $\alpha$, average angular speed, $\omega_{\mathrm{av}}=\frac{\omega_0+\omega}{2}$
$
\alpha=\frac{\omega-\omega_0}{t} \text { and } \theta-\theta_0=\omega_{\text {av }} \cdot t
$
where $\omega_0$ and $\omega$ are the initial and final angular speeds, $t$ is the time, $\omega_{a v}$ the average angular speed and $\theta_0$ and $\theta$ the initial and final angular positions of the particle.
Then, the angular kinematical equations for the circular motion are (in analogy with linear kinematical equations for constant linear acceleration)
$
\begin{aligned}
& \omega=\omega_0+\alpha t \\
& \theta-\theta_0=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \omega^2=\omega_0^2+2 \alpha\left(\theta-\theta_0\right)
\end{aligned}
$
$
\alpha=\frac{\omega-\omega_0}{t} \text { and } \theta-\theta_0=\omega_{\text {av }} \cdot t
$
where $\omega_0$ and $\omega$ are the initial and final angular speeds, $t$ is the time, $\omega_{a v}$ the average angular speed and $\theta_0$ and $\theta$ the initial and final angular positions of the particle.
Then, the angular kinematical equations for the circular motion are (in analogy with linear kinematical equations for constant linear acceleration)
$
\begin{aligned}
& \omega=\omega_0+\alpha t \\
& \theta-\theta_0=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \omega^2=\omega_0^2+2 \alpha\left(\theta-\theta_0\right)
\end{aligned}
$
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