A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to one end tothe string 2malong and is ratated in a horizontal circle caculate the greatest number of revolutions that the mass can make without breaking the string

A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to

1.	A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to one end tothe string 2malong and is ratated in a horizontal circle caculate the greatest number of revolutions that the mass can make without breaking the string
Answer» Solution :Given`m=0.5 kg` `r=2 m` `g=9.8ms^(-2)` Themaximumtensionthat the stringcanwithstand `F=4.8 kg wt=4.8xx 9.8 N` LET themaximum NUMBER of revolutions persecond=v Now F = `mr epsilon^(2)= mr (2 pi v)^(2) = 4PI^(2) mrv^(2)` or `v^(2)=(F )/(4pi^(2)mr)` `=(4.8 xx 9.8)/( 4xx 9.8 xx 0.5 xx 2) =1.215` or `v = SQRT(1.215 )= 1.102 rps` `=1.102 xx 60 =66 .13 rpm`