Question
Certain stars are believed to be rotating at about $1 \mathrm{rot} / \mathrm{s}$. If such a star has a diameter of 40 $\mathrm{km}$, what is the linear speed of a point on the equator of the star?

Answer

$
\begin{aligned}
& \text { Data : } \mathrm{d}=40 \mathrm{~km}, /=1 \mathrm{rot} / \mathrm{s} \\
& \therefore \mathrm{r}=\frac{d}{2}=\frac{40 \mathrm{~km}}{2}=20 \mathrm{~km}=2 \times 10^4 \mathrm{~m}
\end{aligned}
$
Linear speed, $v=\omega \mathrm{r}=(2 \pi f) r$
$
\begin{aligned}
& =(2 \times 3.142 \times 1)\left(2 \times 10^4\right) \\
& =6.284 \times 2 \times 10^4 \\
& \left.=1.257 \times 10^5 \mathrm{~m} / \mathrm{s} \text { (or } 125.6 \mathrm{~km} / \mathrm{s}\right)
\end{aligned}
$

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