A stone of mass 100 g attached to a string of length 50 cm is whirled in a vertical circle by giving velocity at lowest point as 7 m/s. Find the velocity at the highest point. [Acceleration due to gravity = 9.8 m/s2 ]

#### Solution

**Given:-**

m = 100 g = 0.1 kg,

r = 50 cm = 0.5m,

g = 9.8 m/s^{2},

v_{2} = 7 m/s

**To find:-** Velocity at the highest point (v_{H})

**Formula:-** `v_"H"= sqrt((2("TE"_("H")))/"m"-4"gr")`

**Calculation:-**

The total energy at the bottom,

E_{bot} ( = KE + PE = `1/2` 2.45 J )

T.E.(H)= K.E. at lowest point = (1/2)mv_{L}^{2}

∴ T.E_{(H)} = `1/2` x 0.1 x 7^{2}

=2.45 J

The total energy at the top,

E_{top }= KE + PE = `1/2"mv"_1^2 + "mg(2r)"`

E_{top }= `1/2(0.1)v_1^2 + (0.1)(9.8)( 2 xx 0.5 )`

E_{top }= 0.05`v_1^2` + 0.98

By the Principle of conservation of energy,

E_{top} = E_{bot}

`0.05v_1^2` + 0.98 = 2.45

`v_1^2 = (2.45 - 0.98 )/(0.05) = 147/5 = 29.4`

The required velocity v_{1 }= `sqrt(29.4)` = 5.422 m/s.