Question
Obtain $\omega=\omega_0+\alpha t$ from first principles.
✓
Answer
The angular acceleration is uniform, hence
$
\frac{ d \omega}{ d t}=\alpha=\text { constant } \quad\quad\text{(i)}
$
Integrating this equation,
$
\begin{aligned}
\omega & =\int \alpha d t+c \\
& =\alpha t+c \quad \text { (as } \alpha \text { is constant) }
\end{aligned}
$
At $t=O, \omega=\omega_0$ (given)
From (i) we get at $t=O, \omega=c=\omega_0$
Thus, $\omega=\alpha t+\omega_0$ as required.
With the definition of $\omega= d \theta / d t$ we may integrate Eq. (6.36) to get Eq. (6.37). This derivation and the derivation of Eq. (6.38) is left as an exercise.
$
\frac{ d \omega}{ d t}=\alpha=\text { constant } \quad\quad\text{(i)}
$
Integrating this equation,
$
\begin{aligned}
\omega & =\int \alpha d t+c \\
& =\alpha t+c \quad \text { (as } \alpha \text { is constant) }
\end{aligned}
$
At $t=O, \omega=\omega_0$ (given)
From (i) we get at $t=O, \omega=c=\omega_0$
Thus, $\omega=\alpha t+\omega_0$ as required.
With the definition of $\omega= d \theta / d t$ we may integrate Eq. (6.36) to get Eq. (6.37). This derivation and the derivation of Eq. (6.38) is left as an exercise.
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