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Question

# Suppose a piece of dust has fallen on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)

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\begin{align*} \omega&= 500 \text{ } \dfrac{\text{rev}}{\text{min}} \cdot \dfrac{2\pi \text{ rad}}{1 \text{ rev}} \cdot \dfrac{1 \text{ min}}{60 \text{ s}} \\ &=52.36 \text{ } \dfrac{\text{rad}}{\text{s}} \\ t&=3 \text{ min} \cdot \dfrac{60 \text{ s}}{1 \text{ min}} \\ &=180 \text{ s} \\ \Delta \theta &= \omega t \\ &=\left(52.36\right) \left(180\right) \\ &=9424.8 \text{ rad} \\ \implies s&= \Delta \theta \: r \\ &= \left(9424.8\right)\left(0.043\right) \\ &=405.27 \text{ m} \end{align*}

First, we need to convert the units of the time $t$ and the angular velocity $\omega$. Then, we use the formula for angular displacement to obtain $\Delta \theta$. In this occasion, the body rotates at a constant rate, therefore the angular acceleration is zero. Finally, we continue to calculate the traveled distance $s$.

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