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? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Asked by Abhisek | 1 year ago |  94

1 Answer

Solution :-

Mass of the stone = 0.25 kg

Radius, r = 1.5 m

Number of revolution in a second, 

n= \( \dfrac{40}{60}\)= (\( \dfrac{2}{3}\)) rev/sec

The angular velocity, ω = 2πn = 2 x 3.14 x (\( \dfrac{2}{3}\))

The tension on the string provides the necessary centripetal force

T = mω2r

T = 0.25 x 1.5 x [2 x 3.14 x (\( \dfrac{2}{3}\))]2

= 6.57 N

Maximum tension on the string, Tmax= 200 N

Tmax\( \dfrac{ mv^2_{max}}{r}\)

v2max \( \dfrac{(T_{max} x r)}{m}\)

\( \dfrac{ ( 200 \times 1.5)}{0.25}\) = 1200

vmax = 34. 6 m/s

Therefore, the maximum speed of the stone is 34.64 m/s

Answered by Pragya Singh | 1 year ago

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