Question b)
17 minutes ⇒ 1 revolution = 2π
1 minute ⇒ 2π/17
Angle of rotation is 2π/17 per minute
Question c)
The period of the rotation is 2π/17
The wheel starts rotating from the height of 5 meters from the ground and the maximum height is 125 meter from the ground.
The amplitude is 125 - 5 = 120 ÷ 2 = 60
The midline of the rotation is at 125 - 5= 120 ÷2 = 60 + 5 = 65 (this is the location of the centre of the wheel). This value is a shift of 65 from the zero midlines (which in this case would be the ground)
Question C
The movement of the wheel can be described as starting from 0° then reaching its peak at 180° and come back to its original position as it stops and it makes a complete circular turn 360°.
If we sketch this on a graph, we will obtain a curve of -cos(x)
We need to apply the period, the amplitude and the shift produced by this rotation into the equation [tex]y = -cos(x)[/tex]
The period is [tex] \frac{2 \pi }{7} [/tex] ⇒ [tex]f(t)=-cos( \frac{2 \pi }{7}t) [/tex]
The amplitude of 60 ⇒ [tex]f(t)=-60cos( \frac{2 \pi }{7}t) [/tex]
The shift of the midline by 65 units upwards ⇒ [tex]f(t) = -60cos( \frac{2 \pi }{7}t)+65 [/tex]
The final equation is
[tex]f(t) = -60cos( \frac{2 \pi }{7}t)+65 [/tex]
and the graph is shown in the third picture below
