Uniform circular motion on a table

AI Thread Summary
A small block on a frictionless table is in uniform circular motion, connected to a larger suspended block via a string. The relationship between the gravitational force on the larger block and the centripetal force acting on the smaller block is established through the equation Mg = Ma. The correct approach involves recognizing that the masses cannot be canceled due to the specified relationship between them. The key equation becomes m_1 a = (m_2 v^2)/r, where m_1 is the mass of the larger block and m_2 is the mass of the smaller block. The goal is to solve for the velocity v at which the larger block remains stationary.
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Homework Statement


A small block with mass m rests on a frictionless horizontal tabletop a distance r from a hole in the center of the table. A string tied to the small block passes down through the hole, and a larger block with mass M is suspended from the free end of the string. The small block is set into uniform circular motion with radius r and speed v.


Homework Equations


a=v2/R
F=ma

The Attempt at a Solution


I really have know idea...but here goes nothing
Mg=Ma
Mg=Mv2/r
g=v2/r
v=sqrt(g/r)
 

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What is the question..? That is just a statement.
I assume they are asking for the velocity at which larger block is neither rising nor falling?
 
yes, sorry, that's what theyre looking for
 
You have got the right idea with the equation, by letting ma = \frac{mv^2}{r}

however this is where the error is. You canceled the masses. You can't do this because the question specifically states that the mass hanging on the bottom of the string is larger than that on the table.

So your equation is now: m_1 a = \frac{m_2 v^2}{r} where m_1 > m_2

Solve for v.
 
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