Question

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 horizontal plane. What is the tension in the string?

Answer 1

Mass of stone $=0.25Kg$

Radius of circular path $=1.5m$

Angular frequency (f)$=40rev/min$

$=\frac{40}{60}rev/sec$

$=\frac{2}{3}rev/s$

We know, in uniform circular motion , centripital force is provided by tension in string.

e.g Tension in string = mω²r

Angular speed (ω) = 2πf

So,

$Tension=m\times 4\pi r²f²$

$=0.25\times 1.5\times 4\times (\frac{22}{7}{)}^2\times (\frac{2}{3}{)}^2$

$=6.6N$

Given,

Maximum tension which can be withstand by the string .

$Tmax=200N$

$Tmax=mVmax²/r$

$Vmax=\surd Tmax\times r/m$

$Vmax=\surd 1200m/s=34.6m/s$