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Arc length

  • Thread starter jisbon
  • Start date
199
16
Homework Statement
An object is undergoing circular motion in horizontal plane at fixed radius## r = 0.12m##
Radial acceleration is ##2+2t ##m/s
Calculate arc length the object swept through the first 2 seconds.
Homework Equations
-
From what I understand,

##a_{r} = v_{tan}^2 /r##
##a_{r} = (r\omega)^2 /r##
##a_{r} = r\omega^2##
##\omega^2 = \frac{a_{r}}{r}##
##\omega^2 = \frac{2+2t}{0.12}##
##\omega = \sqrt{\frac{2+2t}{0.12}}##
##s =\int_{0}^{2} \sqrt{\frac{2+2t}{0.12}}##
After integrating, I still can't seem to get the correct answer which is 1.37m
Are my concepts wrong or..?
Thanks
 
199
16
A
What are the units of arc length? What are the units of ##\int \omega dt##?
Arc length has no units, ##\omega## has a SI units : rad s−1
I realized that when I integrate ##\omega## I will get angular displacement instead of arc length. So after I get the angular displacement I can just multiply it by radius to get the arc length?

EDIT: Solved~! Thanks for the reminder
 
Last edited:

Orodruin

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So after I get the angular displacement I can just multiply it by radius to get the arc length?
Or simply go with integrating the tangential velocity from your first expression and never care about angular velocity.
 
199
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Or simply go with integrating the tangential velocity from your first expression and never care about angular velocity.
Oh yep, that's another alternative :)
 

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