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

Explanation

Given `m=0.5 kg`
`r=2 m`
`g=9.8ms^(-2)`
The maximum tension that the string can withstand
`F=4.8 kg wt =4.8 xx 9.8 N`
Let the maximum number of revolutions per second=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`