# Arc length

Homework Statement:
An object is undergoing circular motion in horizontal plane at fixed radius## r = 0.12m##
Calculate arc length the object swept through the first 2 seconds.
Relevant Equations:
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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

Delta2

tnich
Homework Helper
##s =\int_{0}^{2} \sqrt{\frac{2+2t}{0.12}}##
What are the units of arc length? What are the units of ##\int \omega dt##?

Delta2 and jisbon
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
Staff Emeritus
Homework Helper
Gold Member
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.

Delta2
Or simply go with integrating the tangential velocity from your first expression and never care about angular velocity.
Oh yep, that's another alternative :)

haruspex