A certain string $500 \mathrm{~cm}$ long breaks under a tension of $45 \mathrm{~kg}$ wt. An object of mass $100 \mathrm{~g}$ is attached to this string and whirled in a horizontal circle. Find the maximum number of revolutions that the object can make per second without breaking the string, $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ ]
Q 39.2
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Data : $\mathrm{m}=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{r}=500 \mathrm{~cm}=5 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \mathrm{~F}=45 \mathrm{~kg} w \mathrm{t}=45 \times 9.8 \mathrm{~N}$ The breaking tension is equal to the maximum centripetal force that can be applied. $\therefore \mathrm{F}=\mathrm{m} \omega^2 \mathrm{r}$,
But $\omega=2 \pi \mathrm{f}$, where/is the corresponding frequency of revolution.
$
\begin{aligned}
& \therefore F=m(2 \pi f)^2 r=4 \pi^2 m f^2 r \\
& \therefore f=\sqrt{\frac{F}{4 \pi^2 m r}}=\sqrt{\frac{45 \times 9.8}{4 \times(3.142)^2 \times 0.1 \times 5}}
\end{aligned}
$
The maximum number of revolutions per second, $f=4.726 \mathrm{~Hz}$
But $\omega=2 \pi \mathrm{f}$, where/is the corresponding frequency of revolution.
$
\begin{aligned}
& \therefore F=m(2 \pi f)^2 r=4 \pi^2 m f^2 r \\
& \therefore f=\sqrt{\frac{F}{4 \pi^2 m r}}=\sqrt{\frac{45 \times 9.8}{4 \times(3.142)^2 \times 0.1 \times 5}}
\end{aligned}
$
The maximum number of revolutions per second, $f=4.726 \mathrm{~Hz}$
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