Find the average acceleration of the mass during the time interval

[quicknote]

I'm suppossed to find the average acceleration of the mass during the time interval from -t to t.
The formula I'm using is [tex]a(-t,t)=\frac{(v(t)-v(-t))}{(t-(-t))} [/tex]

I've already derived v(t) and v(-t) but I'm not sure how to simplify them after subtracting them.

[tex]v(t)=-R\omega sin(\omega t)*\hat{x}+R\omega cos(\omega t)\hat{y}[/tex]
[tex]v(-t)=-R\omega sin(\omega (-t))\hat{x}+R\omega cos(\omega (-t))\hat{y}[/tex]

I got as far as grouping the x and y terms...after that I don't know how else to simplify

[tex][(-R \omega \hat{x}) ((sin(\omega t)+sin(\omega (-t))] + [(R \omega \hat{y}) ((cos \omega t) -cos (\omega (-t))] [/tex]

Any help is appreciated.
thanks!

 

[Gale]

from what you have, i get that the velocities all cancel out. [tex] sin(\omega t) + sin (-\omega t) = sin(\omega t) -sin(\omega t)=0[/tex] because [tex] sin(-\omega t)= -sin(\omega t)[/tex]

ok, well just the sine's cancel...

the cosines simplify [tex] R\omega cos(\omega t) - R\omega cos(-\omega t)= R\omega(cos(\omega t) - cos(-\omega t))= R\omega( cos(\omega t) + cos(\omega t))= 2R\omega cos(\omega t)[/tex]

and that's all in the y hat direction. so all your acceleration is vertical.

 

[quicknote]

Thanks Gale!

I didn't know this rule: sin(-x) = -sin (x)

 

[Gale]

Thanks Gale!

I didn't know this rule: sin(-x) = -sin (x)

hah, don't feel so bad... i didn't either, but i drew it out. when you look at the picture, obviously they're equal.