On the merry-go-round, each horse moves up and down five times in one complete revolution. The up- and - down motion of each horse can be modeled by the function h(t) = 25cos(5ϴ), where h is the horse's displacement in centimetre from its centre and ϴ is the rotation angle of the merry-go-round. Assume the rides begins when ϴ=0 degrees for a given horse.
A) At what rotation angles will the horse be displaced 15 cm above its original position in one revolution?
B) At what rotation angles will the horse be displaced 20 cm below its original position in one revolution?
C) How long will it take for one complete revolution if the carousel rotates at a speed of 24degrees/s?
I see now.....they start at the high point when theta = 0 originally so they will be at 25 above center and never reach 15cm above that! d'Oh!
A: 15cm = 25cos(5t) t= theta
15/25 = cos 5t
Arcos (15/25) = 5t t= .185 R =10.62 degrees
B -20 = 25 cos(5t)
-20/25 = cos 5t
Arcos (-20/25) = 5t t = .4996 R 28.62 degrees
C 360 degrees/ (24 degrees/sec) = 15 seconds
Ooops sorry....I gave onlt the FIRST position of those displacements.....back in a few
A 5 times every rotation is every 72 degrees
10.62 + 72 = 82.62 154.62 226.62 298.62 370.62
BUT, they also will be displaced 15 cm on the way DOWN for another 5 times each rotation
arcos(-15/25) = 5t = .44 R 25.37 degrees 97.37 169.37 2.41.37 313.37 degrees
(I think !)
A) At what rotation angles will the horse be displaced 15 cm above its original position in one revolution?
The horse is at its max height [25cm] when the ride starts [at θ = 0] ......thus, it never rises15 cm above its original poisition.......
B) At what rotation angles will the horse be displaced 20 cm below its original position in one revolution?
The height at those angles will be 5cm....so we have
5 = 25cos(5θ)
Have a look at the graph here [in degrees].......https://www.desmos.com/calculator/pusq0nejlq
The graph shows that this height occurs 10 times in one revolution
C) How long will it take for one complete revolution if the carousel rotates at a speed of 24degrees/s?
360 / 24 = 15 seconds
NOW I am more confused h = 25 cos(5theta) is it's position from its CENTER......
Is that what you assumed , Chris??
I see now.....they start at the high point when theta = 0 originally so they will be at 25 above center and never reach 15cm above that! d'Oh!
I think, as the question is written......you're answers are correct.....AND you are correct again when you say the Q is confusing !