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 1

Mass of the stone, *m* = 0.25 kg

Radius of the circle, *r* = 1.5 m

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

Angular velocity , `omega= v/r = 2pin`

The centripetal force for the stone is provided by the tension *T*, in the string, i.e.,

`T = F_"Centripetal"`

= `(mv^2)/r = mromega^2 = mr(2pin)^2`

`= 0.25 xx 1.5 xx (2xx3.14xx2/3)^2`

= 6.57 N

Maximum tension in the string, *T*_{max} = 200 N

`T_"max"` = `(mv_" max"^2)/m`

`:.v_"max" = sqrt((T_max xx r)/m)`

` = sqrt((200xx1.5)/0.25)`

= `sqrt(1200)` = 34.64 m/s

Therefore the maximum speed of the stone is 34 .64 m/s

#### Solution 2

Her m = 0.25 kg, r = 1.5 m

`n = 40 "rpm" = 40/60 rps = 2/3 "rps"`

Now `T = mromega^2 = mr(2pin)^2 = 4pi^2 mrn^2`

`T = 4xx22/7xx22x7xx 0.25xx1.5xx(2/3)^3 = 6.6 N`

if T_"max" = 200 N then from

`T_"max" = mv_"max"^2 => v_"max"^2 = (T_"max" xx r)/m`

or `v_"max"^2 = (200xx1.5)/0.25 = 1200 => v_max = sqrt1200 = 34.6 ms^(-1)`