Question

A stone of mass 0.3kg tied to the end of a string in a horizontal plane is whirled round in a circle of radius 1m with the frequency of 40rpm. (i) What is the tension in the string ? (ii) What is the maximum speed with which the stone can be whirled round if the string can widthstand a maximum tension of 200N.

Answer 1

Given :
mass (m) = 0.3kg
radius (r)=1m
no. of rotation per second (n)= 40/60 rpm
                                                 =2/3 rps
Angular velocity (ω)=v/r=2πn= 2πx2/3=4π/3 rad /s
Tmax=200N
formula : Centripetal force for stone is provided by tension T.
T = F centripetal
 = mv²/r=mrω²
=0.3x 1 x(4π/3)²
=5.258N.
For maximum velocity , tension is maximum.
vmax=√tmaxx r /m
Vmax=√200x1/0.3
         =√2000/3
vmax=25.82m/s
Therefore tension force = 5.258N
maximum velocity= 25.82m/s

Answer 2

Tension T in the string is equal to the Centripetal force required for the  stone to rotate in a circle.

T = m r ω² = 0.3 kg * 1 m * (2π*40/60)² rad²/sec²
   = 1.6π²/3 Newtons

Maximum Tension = 200 N

Maximum ω = √(T/m r) = √[200/ (0.3*1) ] = 20√5/√3  rad/sec
                    = 20√5 /(2π√3)  rev/sec
                    = 600 √5 /(√3π)  rpm