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3-2.Motion in Plane
medium
A stone of mass $0.25\; kg$ tied to the end of a string is whirled round in a circle of radius $1.5 \;m$ with a speed of $40\; rev./min$ in a hortzontal plane. What is the tenston in the string? What is the maximum speed in $m/s $ with which the stone can be whirled around if the string can withstand a maximum tension of $200\; N ?$
A$34.64$
B$42.26$
C$26.5$
D$12.26$
Solution
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., $T=F_{\text {Centripetal }}$
$=\frac{m v^{2}}{r}=m r \omega^{2}=m r(2 \pi n)^{2}$
$=0.25 \times 1.5 \times\left(2 \times 3.14 \times \frac{2}{3}\right)^{2}$
$=6.57\, N$
Maximum tension in the string, $T_{\max }=200 \,N$
$T_{\max }=\frac{m v_{\max }^{2}}{r}$
$\therefore v_{\max }=\sqrt{\frac{T_{\max } \times r}{m}}$
$=\sqrt{\frac{200 \times 1.5}{0.25}}$
$=\sqrt{1200}=34.64 \,m / s$
Therefore, the maximum speed of the stone is $34.64 \,m / s$
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., $T=F_{\text {Centripetal }}$
$=\frac{m v^{2}}{r}=m r \omega^{2}=m r(2 \pi n)^{2}$
$=0.25 \times 1.5 \times\left(2 \times 3.14 \times \frac{2}{3}\right)^{2}$
$=6.57\, N$
Maximum tension in the string, $T_{\max }=200 \,N$
$T_{\max }=\frac{m v_{\max }^{2}}{r}$
$\therefore v_{\max }=\sqrt{\frac{T_{\max } \times r}{m}}$
$=\sqrt{\frac{200 \times 1.5}{0.25}}$
$=\sqrt{1200}=34.64 \,m / s$
Therefore, the maximum speed of the stone is $34.64 \,m / s$
Standard 11
Physics
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