A small block of mass `4kg` is attached to a cord passing through a hole in a horizontal frictionless surface. The block is originally revolving in a circle of radius `0.5 m` about the hole with a tangential velocity of `4m//s`. The cord is then pulled slowly from below, shorting the radius of the circle in which the block revolves. The breaking strength of the cord is `600 N`. What will be radius of the circle when the cord breaks?
A small block of mass `4kg` is attached to a cord passing through a hole in a horizontal frictionless surface. The block is originally revolving in a circle of radius `0.5 m` about the hole with a tangential velocity of `4m//s`. The cord is then pulled slowly from below, shorting the radius of the circle in which the block revolves. The breaking strength of the cord is `600 N`. What will be radius of the circle when the cord breaks?
Text Solution
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The tension of the rope is the only net force on the block and it does not exert any torque about the axis of rotation. Hence the angular momentum of the block about the axis should remain conserved.
`implies mvr=`constant
Let `r_(1)=0.5 m` and `v_(1)=4m//s`
Let `r_(2),v_(2),T_(2)` be the radius velocity of and tensio, respectively. when the string breaks
`impliesT_(2)=600N`
`mv_(1)r_(1)=mv_(2)r_(2)=` and `T_(2)=mv_(2)^(2)//r_(2)`
`implies mv_(1)r_(1)=msqrt((r_(2)T_(2))/m)r_(2)`
`r_(2)=((v_(1)^(2)r_(1)^(2))/(T_(2))^(1/3)=(4xx16xx0.25)/600)^(1/3)=(16/600)^(1/3)=3.0m`
`implies mvr=`constant
Let `r_(1)=0.5 m` and `v_(1)=4m//s`
Let `r_(2),v_(2),T_(2)` be the radius velocity of and tensio, respectively. when the string breaks
`impliesT_(2)=600N`
`mv_(1)r_(1)=mv_(2)r_(2)=` and `T_(2)=mv_(2)^(2)//r_(2)`
`implies mv_(1)r_(1)=msqrt((r_(2)T_(2))/m)r_(2)`
`r_(2)=((v_(1)^(2)r_(1)^(2))/(T_(2))^(1/3)=(4xx16xx0.25)/600)^(1/3)=(16/600)^(1/3)=3.0m`
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