An equation for instantaneous angular acceleration is given as:

$$ \alpha \equiv \lim_{\Delta t\to0}\frac{\Delta \omega}{\Delta t} = \frac{d\omega}{dt} $$

The text I am reading says writing this equation in differential form

$$ dw = \alpha dt $$

and integrating from $t_1$ = 0 to $t_f$ = $t$ gives

$$ \omega_f = \omega_i + \alpha t $$

I am not exactly sure how the authors came to this. Any help would be much appreciated.

  • $\begingroup$ To be even more specific I understand the $\alpha t$ term, but I cannot figure out where the $\omega_f$ & $\omega_i$ came from. $\endgroup$ – albertjorlando Nov 27 '13 at 4:58

Start with \begin{align} \alpha(t) = \frac{d\omega}{dt}(t). \end{align} Integrate both sides from $t_i$ to $t_f$; \begin{align} \int_{t_i}^{t_f}dt\,\alpha(t) = \int_{t_i}^{t_f}dt\, \frac{d\omega}{dt}(t) = \omega(t_f) - \omega (t_i) \end{align} The second equality on the right follows from the fundamental theorem of calculus which basically says that if you integrate the derivative of a function, you recover the original function.

Now, if $\alpha(t)$ is constant in time, namely it has no dependence on $t$, then it can be pulled outside of the integral on the left, and we obtain \begin{align} \alpha\cdot(t_f-t_i) = \omega(t_f) -\omega(t_i) \end{align} Which is the desired identity.

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  • $\begingroup$ @ao2130 Sure thing. Welcome to physics.SE. $\endgroup$ – joshphysics Nov 27 '13 at 5:08

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