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Question

A discus thrower starts from rest and begins to rotate with a constant angular acceleration of 2.2rad/s22.2 \mathrm{rad} / \mathrm{s}^2. How many revolutions does it take for the discus thrower's angular speed to reach 6.3rad/s6.3 \mathrm{rad} / \mathrm{s} ?

Solution

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Using

ω2=ω02+2α(θθ0)\begin{equation*} \omega^2=\omega_0^2 + 2\alpha (\theta - \theta_0) \end{equation*}

together with ω0=0\omega_0=0 and θ0=0\theta_0=0 we have

ω2=2αθθ=ω22α\begin{equation*} \omega^2= 2\alpha \theta \Rightarrow \theta=\frac{\omega^2}{2\alpha} \end{equation*}

Thus

θ=(6.3 rad/s)22×2.2 rad/s29 rad\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 θ\theta by 2π2\pi radians:

no. of revolutions=θ2π rad1.4\begin{equation*} \text{no. of revolutions} =\frac{\theta}{2\pi\mathrm{~rad}}\approx \boxed{1.4} \end{equation*}

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