A mass m is attached to a string passing through a small hole in a frictionless, horizontal surface. The mass is initially orbiting with a velocity `v_(1)` in a circle of radius `r_(1)` The string is then slowly pulled from below, decreasing the radius of the circles to `r_2` (i) What is the speed of the mass when the radius is `r_2` ? (ii) What is the tension in the string ? (iii) What is the work done in moving the mass m from `r_(1)` to `r_(2)`?

Answer

459

Explanation

The torque acting on the rotating mass about the vertical axis is zero. So law of conservation of angular momentum holds good,
`mv_(1)r_(1) = mv_(2)r_(2)`
`v_(2) = (v_(1)r_(1))/r_(2)`
The tension in the string is T
`T = (mv_(2)^(2))/r_(2) =(mv_(1)^(2) r_(1)^(2))/r_(2)^(3)`
The change in K.E. `=1/2 mv_(2)^(2) -1/2mv_(1)^(2)`
`=1/2mv_(1)^(2)[(r_(1)^(2)-r_(2)^(2))/r_(2)^(2)]`
The work done in moving the mass from `r_(1)` to `r_(2)` is equal to change in K.E.