__Question 5.21__:
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?

__Solution__:Mass of the stone,

*m*= 0.25 kg

*r*= 1.5 m

Number of revolution per second, n = 40 / 60 = 2 / 3 rps

Angular velocity, ω = v / r = 2πn

The centripetal force for the stone is provided by the tension

*T*, in the string, i.e.,

T = F

_{Centripetal}

= mv

^{2}/ r = mrω = mr(2πn)

^{2}

= 0.25 X 1.5 X (2 X 3.14 X (2/3) )

^{2}

= 6.57 N

Maximum tension in the string,

*T*

_{max}= 200 N

T

_{max}= mv

_{}

^{}

^{2}

_{max}/ r

_{}

∴ v

_{max}= (T

_{max}X r / m)

^{1/2}

= (200 X 1.5 / 0.25)

^{1/2}

= (1200)

^{1/2}= 34.64 m/s

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

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