Unit for the slope of a circular motion relationship

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SUMMARY

The discussion centers on determining the unit of the slope in the formula T=(√((4π^2 m)/Mg)) √l, where T represents the period of circular motion. The slope, represented by m, simplifies to √g, where g is the acceleration due to gravity (9.81 m/s²). The participant initially calculated the unit as s²/m but later acknowledged the correct unit for the slope is √g, which leads to a unit of s/√m when considering the relationship between the variables.

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  • Understanding of circular motion dynamics
  • Familiarity with the formula for the period of a pendulum
  • Basic knowledge of dimensional analysis
  • Concept of gravitational acceleration (g)
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Epsillon
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Homework Statement


so given the formula T=(√((4π^2 m)/Mg)) √l+0
which is y=mx+b m being (√((4π^2 m)/Mg)) and x being √l

what is the unit of the slope??





The Attempt at a Solution


So the kgs cross out in the mass and we are just left with the units of g which is m/s^2
since g is on the denominator i got the unit s^2/m

HOWEVER my problem is that when the slope is multiplied by the x shoudnt it give you the units for y?

The way I did it it doesent.
 
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Hi Epsillon! :smile:
Epsillon said:
… and we are just left with the units of g which is m/s^2
since g is on the denominator i got the unit s^2/m

erm … it's √g. :redface:
 

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