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)?

A mass m is attached to a string passing through a small hole in a fri

1.	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» Solution :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.