Posted by **waqas** on Saturday, August 6, 2011 at 7:45am.

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

- Physics -
**drwls**, Saturday, August 6, 2011 at 8:00am
(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.

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