A stone of mass 0.25; kg tied to the end of a | vedclass

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Question

Mass of the stone, $m=0.25 \,kg$

Radius of the circle, $r=1.5 \,m$

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3} \,rps$

Angul

Vedclass · Accepted Answer

$34.64$

Vedclass · Answer

Mass of the stone, $m=0.25 \,kg$

Radius of the circle, $r=1.5 \,m$

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3} \,rps$

Angular velocity, $\omega=\frac{v}{r}=2 \pi n$

The centripetal force for the stone is provided by the tension $T$, in the string, i.e., $T=F_{	ext {Centripetal }}$

$=\frac{m v^{2}}{r}=m r \omega^{2}=m r(2 \pi n)^{2}$

$=0.25 	imes 1.5 	imes\left(2 	imes 3.14 	imes \frac{2}{3}ight)^{2}$

$=6.57\, N$

Maximum tension in the string, $T_{\max }=200 \,N$

$T_{\max }=\frac{m v_{\max }^{2}}{r}$

$	herefore v_{\max }=\sqrt{\frac{T_{\max } 	imes r}{m}}$

$=\sqrt{\frac{200 	imes 1.5}{0.25}}$

$=\sqrt{1200}=34.64 \,m / s$

Therefore, the maximum speed of the stone is $34.64 \,m / s$

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