Answer:
51.96 feet is the value with which the seat on the rim is rising in the vertical direction
Explanation:
Angular velocity = ω= 2 rad/s
radius of ferris wheel = r = 30 feet
We need to find that , how fast is a seat on rim rising (in vertical direction) when it is 15 feet above horizontal line through center of wheel, meaning how quickly a point on rim of wheel is rising vertically at point where this point is 15 ft above horizontal line through center of wheel.
To write the expression :
y = r sin (ω t)
y= 30 sin (2t)
y = 60 sin(t) cos(t)
y= 60
![(cos(t)^(2)- sin (t)^(2) )](https://img.qammunity.org/2021/formulas/mathematics/college/qz6uyad0bnsds71fjynzskhlb30bwc809t.png)
y = 60 cos(2t) _____________________(Equation 1)
The position of 15 feet up is given by angle of 2t
Sin (2t) =
=
![(1)/(2)](https://img.qammunity.org/2021/formulas/physics/middle-school/ukxexrkoplrwscaxd96qbbkphc5fo6w2ur.png)
Then,
2t = 30 degrees
cos (2t) = cos (30) =
![(√(3))/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/8jgno9u9e5q3mu4wko1bmv6qvl6dicfd9h.png)
put in equation 1 , we get:
y = 60cos(30)
y = 60 (0.866)
y= 51.96 feet
which is the value with which the seat on the rim is rising in the vertical direction when it is 15 feet above horizontal line through center of wheel.