## Related questions with answers

A particle moves 3.0 m along a circle of radius 1.5 m. (a) Through what angle does it rotate? (b) If the particle makes this trip in 1.0 s at a constant speed, what is its angular velocity? (c) What is its acceleration?

Solutions

Verified### Information:

$\begin{aligned} s&=3\ \text{m}\\ r&=1.5\ \text{m}\\ t&=1\ \text{s}\\ \end{aligned}$

### Needed:

$\begin{aligned} \theta&=?\\ \omega&=?\\ a&=?\\ \end{aligned}$

$\begin{align*} p&=2 \pi R \\ &=2\pi \left(1.5\right) \\ &=9.425 \text{ m} \\ \implies \Delta \theta &=\left(3.0\right)\left(\dfrac{2\pi}{9.425}\right) \\ &=2.0 \text{ rad} \\ \implies \omega &=\dfrac{d\theta}{dt} \\ &=\dfrac{\Delta \theta}{\Delta t} \\ &=\dfrac{2.0}{1.0} \\ &=2.0 \text{ } \dfrac{\text{rad}}{\text{s}} \\ \implies \alpha&=\dfrac{d\omega}{dt} \\ &=\dfrac{d}{dt}\left(2.0\right) \\ &=0 \\ \implies a_t &= \alpha R \\ &=0 \\ \implies a_c&=\omega^2 R \\ &=\left(2.0\right)^2 \left(1.5\right) \\ &=6.0 \text{ } \dfrac{\text{m}}{\text{s}^2} \end{align*}$

First, we can calculate the covered angle $\Delta \theta$ with the covered distance and the perimeter $p$ (which can be calculated with the radius $R$). Once we have obtained $\Delta \theta$, we proceed to calculate the angular speed $\omega$. Note that $d\theta$ and $dt$ are equal to $\Delta \theta$ and $\Delta t$ respectively, because the rotation occurs at constant speed. Also, notice that there is no tangential acceleration because the angular speed is constant, and therefore the angular acceleration $\alpha$ is zero. Finally, the total acceleration is equal to the centripetal acceleration $a_c$, which can be calculated with $\omega$ and $R$.

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