I tried doing this, but I can't get the exact same answer as the one given(adsbygoogle = window.adsbygoogle || []).push({});

http://img246.echo.cx/my.php?image=untitled0nj.png

I use relative acceleration

[tex]a_a = a_b + \dot \omega \cross r + \omega \cross (\omega \cross r) + 2 \omega \cross V_{rel} + a_{rel}[/tex]

and as

[tex]V_{rel} = 0 [/tex]

[tex]\dot \omega = 0[/tex]

[tex]a_{rel} = 0[/tex]

[tex]a_b = (\doubledot r - r\omega^2) \widehat e_r + (2\dot r+r\dot\omega) \widehat e_\theta[/tex]

so [tex]a_b = b\omega^2 \widehat e_r = b\omega^2 \widehat i[/tex]

and what I ended up for the answer [tex]a_a[/tex] is [tex]b \omega^2 \widehat i + \omega pr \widehat j - p^2r \widehat k[/tex]

fits funny, cause I got the right answer for [tex] a_b[/tex] using the same formula....

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# Homework Help: Relative Motion and acceleration

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