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anonymous
 4 years ago
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(tb)+c model or cosine function.
anonymous
 4 years ago
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(tb)+c model or cosine function.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c=4 and a=.15 ... is that part clear?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0false. :) continuing... \[f=\frac{ 1 }{10}\] \[\omega = 2 \pi f\] \[w= \frac{ \pi }{ 5 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we want it to be at the highest point when t=0... sin =1 when theta = pi/2 pi/5 *(tb) = pi/2 pi/5(b) = pi/2 b= 5/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y = .15 \sin (\frac{ \pi }{ 5} (t+\frac{ 5 }{2})) +4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it be appropriate to keep(tb) it at pi/5 *(t) + pi/2 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can... that's not the form they ask you to use though...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whoops, should say b=5/2 up there... typo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay. & also for cosine model it would just be \[y=.15\cos(\frac{ \pi }{ 5 }t) +4\] right??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why would it be 5/2 and not 5/2 ? :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why is b= 5/2? is that what you're asking?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH WAIT.. nvm.. you were referring to your reply before the equation. nvm. lol
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