Answer to Question #157327 in Physics for Duc

Question

A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v=2.4m/s in a circle of radius R1=0.8m. The string is then pulled slowly through the hole so that the radius is reduced to R=0.48m. What is the speed, V2, of the mass now?

Answer 1

$mv_1r_1=mv_2r_2\to$

$v_2=v_1r_1/r_2=2.4\cdot 0.8/0.48=4(m/s)$ . Answer