Physics
posted by waqas .
A particular Ferris wheel (a rigid wheel rotating in a vertical plane about a horizontal axis) at a local carnival has a radius of 20.0 m and it completes 1 revolution in 9.84 seconds.
(a) What is the speed (m/s) of a point on the edge of the wheel?
Using the coordinate system shown, nd:
the (b) x component of the acceleration of point A at the top of
the wheel.
the (c) y component of the acceleration of point A at the top of the wheel
the (d) x component of the acceleration of point B at the bottom of the wheel.
the (e) y component of the acceleration of point B at the bottom of the wheel
the (f) x component of the acceleration of point C on the edge of the wheel.
the (g) y component of the acceleration of point C on the edge of the wheel

(a) Divide 2*pi*R by 9.84 s for the speed.
The speed is suspiciously high, and probably unsafe.
(b) and (c) At the top of the wheel, the centripetal acceleration is down (y). Its magnitude is V^2/R everywhere along the outer edge
(d) and (e) At the bottom of the wheel, the centripetal acceleration is up (+y)
(f) and (g) It depends upon which edge. Location C is not shown.