Arc length

jisbon

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
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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
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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

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
2018 Award
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

jisbon

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

Homework Helper
Gold Member
2018 Award
Arc length has no units
It has dimension length, so the SI unit is metres.

• Orodruin

"Arc length"

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