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A weightless thread can bear tension upto 37 N . A stone of mass 500 g is tied to it and revolved in a circular path of radius 4 m in a vertical plane. If $g=10 \mathrm{~ms}^{-2}$, then the maximum angular velocity of the stone will be
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Verified Answer
The correct answer is:
$4 \mathrm{rad} \mathrm{s}^{-1}$
Maximum tension in the thread is given by or
$T_{\max }=m g+\frac{m v^2}{r}$
or $\quad T_{\max }=m g+m r \omega^2$
$(\because v=r \omega)$
or $\quad \omega^2=\frac{T_{\max }-m g}{m r}$
Given,
$\begin{aligned}
& T_{\max }=37 \mathrm{~N}, m=500 \mathrm{~g}=0.5 \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}, \\
& r=4 \mathrm{~m} \\
& \therefore \quad \omega^2=\frac{37-0.5 \times 10}{0.5 \times 4}=\frac{37-5}{2} \\
& \text { or } \quad \omega^2=16 \text { or } \omega=4 \mathrm{rad} \mathrm{s}^{-1}
\end{aligned}$
$T_{\max }=m g+\frac{m v^2}{r}$
or $\quad T_{\max }=m g+m r \omega^2$
$(\because v=r \omega)$
or $\quad \omega^2=\frac{T_{\max }-m g}{m r}$
Given,
$\begin{aligned}
& T_{\max }=37 \mathrm{~N}, m=500 \mathrm{~g}=0.5 \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}, \\
& r=4 \mathrm{~m} \\
& \therefore \quad \omega^2=\frac{37-0.5 \times 10}{0.5 \times 4}=\frac{37-5}{2} \\
& \text { or } \quad \omega^2=16 \text { or } \omega=4 \mathrm{rad} \mathrm{s}^{-1}
\end{aligned}$
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