What Speed and Radius Conditions Keep the Hanging Mass Stationary?

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Homework Help Overview

The problem involves a mass on a frictionless table connected to a hanging mass by a cord, requiring the determination of conditions for the hanging mass to remain stationary. The subject area includes dynamics and circular motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between speed and radius, with one suggesting that multiple answers may exist due to their interdependence. There is also mention of using dimensional analysis to argue that insufficient information is provided to separate the variables.

Discussion Status

Some participants have offered insights regarding the limitations of the information provided, indicating that additional parameters would be necessary to solve for both speed and radius independently. There is acknowledgment of the need for further clarification on the problem setup.

Contextual Notes

Participants note that the absence of mass dimensions in the requested values complicates the analysis, suggesting that either the radius or speed must be specified to progress further.

toesockshoe
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Homework Statement


A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table; the hole has no frictional effect on the string. Find the condition (the speed of the mass and the radius of its circular motion) with which it must spin for M to remain at rest

Homework Equations



f=ma
a_c=v^2/r

The Attempt at a Solution



is the answer v=sqrt(Mgr/m)? i feel like you need to separate r and v but I can't find a way to express both variable individually with the givens... i think speed and radius depend on each other so there can be multiple answers. is this correct?

how I solved it:

system m
x components:

Ftension = ma
Ftension = mv^2/r

system M
Fgravity - F tension (cordinate system with y-axis going downard) = ma
a is 0
so Fgravity = FtensionTENSION FORCES NEED TO BE SAME FOR BOTH SYSTEMS

so Fgravity can be rewritted as Mg
so Mg = mv^2/r
and thus v = sqrt(Mgr/m) right?[/B]
 
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Dimensional analysis can be used to prove there's not enough information. There's no mass dimension in either of the requested values, so the only use of the given masses is to take the ratio. That leaves you with only an acceleration (g) from which to obtain both a distance and a speed. No chance.
 
haruspex said:
Dimensional analysis can be used to prove there's not enough information. There's no mass dimension in either of the requested values, so the only use of the given masses is to take the ratio. That leaves you with only an acceleration (g) from which to obtain both a distance and a speed. No chance.
so is the answer i have the best i can do with the given information? i think either r or v would have to be given for me to have separate equations for both r and v... right?
 
toesockshoe said:
so is the answer i have the best i can do with the given information? i think either r or v would have to be given for me to have separate equations for both r and v... right?
Yes.
 

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