A 3.2 kg block sits on a frictionless table that has a hole cut in the middle of it. The 3.2 kg mass is connected to a string that goes through the hole in the table and attaches to a 4.8 kg mass. The 3.2 kg mass moves in a circle of radius 0.2 m

Question

A 3.2 kg block sits on a frictionless table that has a hole cut in the middle of it. The 3.2 kg mass is connected to a string that goes through the hole in the table and attaches to a 4.8 kg mass. The 3.2 kg mass moves in a circle of radius 0.2 m
How fast does it have to be moving so thaf the 4.8kg mass does not fall? Also what is the tension in the string?

Damon · Answer

Force up = tension in string = 9.8*4.8 = 47 N so 47 = m v^2/r 47 = 3.2 * v^2 /.2 v = 1.71