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# A discus thrower starts from rest and begins to rotate with a constant angular acceleration of $2.2 \mathrm{rad} / \mathrm{s}^2$. How many revolutions does it take for the discus thrower's angular speed to reach $6.3 \mathrm{rad} / \mathrm{s}$ ?

Solution

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Using

$\begin{equation*} \omega^2=\omega_0^2 + 2\alpha (\theta - \theta_0) \end{equation*}$

together with $\omega_0=0$ and $\theta_0=0$ we have

$\begin{equation*} \omega^2= 2\alpha \theta \Rightarrow \theta=\frac{\omega^2}{2\alpha} \end{equation*}$

Thus

$\begin{equation*} \theta =\frac{(6.3\mathrm{~rad/s} )^2}{2\times 2.2\mathrm{~rad/s^2}}\approx \color{#c34632}{9\mathrm{~rad}} \end{equation*}$

In order to get number of revolutions we have to divide $\theta$ by $2\pi$ radians:

$\begin{equation*} \text{no. of revolutions} =\frac{\theta}{2\pi\mathrm{~rad}}\approx \boxed{1.4} \end{equation*}$

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