describe the motion of a point on a wheel that has a centre 4m off the ground, has radius of 15 cm, makes a full rotation every 10 seconds and starts at its highest point. use y=asink(t-b)+c model or cosine function.

Question

anonymous · Answer

c=4 and a=.15  ...  is that part clear?

anonymous · Answer

a=15 not .15..

anonymous · Answer

false.
:)
continuing...
$$f=\frac{ 1 }{10}$$
$$\omega = 2  \pi f$$
$$w= \frac{ \pi  }{ 5 }$$

anonymous · Answer

agree so far?

anonymous · Answer

yes i got that part

anonymous · Answer

so the phase...

anonymous · Answer

we want it to be at the highest point when t=0...
sin =1 when theta = pi/2

pi/5 *(t-b) = pi/2
pi/5(-b) = pi/2
b= 5/2

anonymous · Answer

$$y = .15 \sin (\frac{ \pi  }{ 5} (t+\frac{ 5 }{2})) +4$$

anonymous · Answer

would it be appropriate to keep(t-b) it at pi/5 *(t) + pi/2 ?

anonymous · Answer

you can... that's not the form they ask you to use though...

anonymous · Answer

whoops,  should say b=-5/2 up there... typo

anonymous · Answer

oh okay. 
& also for cosine model it would just be $$y=.15\cos(\frac{ \pi }{ 5 }t) +4$$ 
right??

anonymous · Answer

yes

anonymous · Answer

why would it be -5/2  and not 5/2 ? :S

anonymous · Answer

why is b= -5/2?  is that what you're asking?

anonymous · Answer

OH WAIT.. nvm.. you were referring to your reply before the equation. nvm. lol

anonymous · Answer

thanks!

anonymous · Answer

sure:)