Uniform circular motion on a table

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Homework Help Overview

The problem involves a small block in uniform circular motion on a frictionless tabletop, connected by a string to a larger block suspended below. The context includes analyzing the forces acting on both blocks and determining the conditions for the motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the blocks and the conditions for the larger block's motion. Questions arise regarding the interpretation of the problem statement and the assumptions about the masses involved.

Discussion Status

Some participants have provided insights into the equations of motion and the relationships between the masses. There is an ongoing exploration of the implications of canceling masses in the equations, with suggestions to reconsider this approach.

Contextual Notes

Participants note the specific condition that the mass of the hanging block is greater than that of the block on the table, which influences the analysis of the forces involved.

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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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