Question

A Ferris wheel of radius 12m is turning about a horizontal axis through its center, such that the linear speed of a passenger on the rim is constant and equal to 9m/s.

a) What are the magnitude and direction of the acceleration of the passenger as he passes

through the lowest point in his circular motion?

Answer 1

Explanations & Calculations

• It is the centripetal acceleration that acts on the passenger as he moves in the circle.
• Acceleration is therefore,

$\qquad\qquad \begin{aligned} \small a&=\small \frac{v^2}{r}\\ &=\small \frac{(9\,ms^{-1})^2}{12\,m}\\ &=\small 6.8\,ms^{-2} \end{aligned}$

• This acts on him towards the centre of the circle and as it is the bottom, this directs vertically upwards.