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 Homework Statement:

If a ##1.5## kg mass is rotating at uniform speed on a table and another ##2## kg mass is connected with a string which goes through the hole on the table at the center of rotation of the ##1.5## kg mass making the ##2## kg mass hanging then what should be the speed of rotation of the ##1.5## kg mass to keep the ##2## kg mass stationary?
Coefficient of sliding friction of the ##1.5## kg mass is ##μ_k=0.2## and the string is inextensible.
 Relevant Equations:
 Newtonian Mechanics
**My Attempt:** Here, I considered for 2 kg mass the resultant motion is zero, which means it's acceleration (a) would be zero. So, if we consider the Tension force from the 1.5 kg mass to keep 2 kg mass from falling is L, then 2×9.8L=ma=2× 0 ⇒ L=19.6 N. But where does the tension force from 1.5 kg comes from? I thought it's from the centrifugal (or centripetal) force. But should I take into account friction now? If the friction is f_{k} then can I write this ##\frac{mv^2}{r}f_kL=0## in the next step to find out speed of the 1.5 kg mass?
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