Circular Motion and friction problem.

[khuong1994]

Homework Statement

One end of a string is attached to a mass of 7.5kg which is at rest on a horizontal table, coeffecient of friction being = 1/3. String passes through a small hole in the table and supports at its other end a mass of 2.5kg which is revolving in a horizontal of radius 20cm. Find the number of revolutions made per minute if the mass on the table is on the point of slipping.

The Attempt at a Solution

I basically did Fc= 7.5*20* w^2
and then
Fc=T+1/3*7.5*9.8 and since T would be equal to 2.5*g i subbed that into get

Fc=2.5*9.8+1/3*7.5*9.8

From there equated the two Fc to get

7.5*20* w^2 = 2.5*9.8+1/3*7.5*9.8
And then i just found w and changed into revolutions per minute.

Is this method correct?Oh yeah, hi guys as you can see I'm a first time poster and I assume I'll be using this forum, particularly this section, numerous more times in the near future.

 

[cupid.callin]

Fc=T+1/3*7.5*9.8 and since T would be equal to 2.5*g i subbed that into get

How did you got this?