Question

A person riding on a Ferris Wheel of radius 14.0 m at some moment has a tangential speed of 2.00 m/s (rotating clockwise) and is slowing down at a rate of 1.00x10-1 m/s2 . At the middle point on the right, indicated by the red circle, determine the magnitude and exact direction of his acceleration vector.

140m

Answer 1

Radius of the wheel r = 14 m/s

Tangential acceleration is a_T = -0.10 m/s², Centripetal acceleration is given by

$a_{c}= \frac{v^{2}}{r}= \frac{(2 m/s)^{2}}{14 m} = 0.286 m/s^{2}$

Net magnitude of the acceleration is

$a =\sqrt{ a_{c}^{2}+a_{T}^{2}}=\sqrt{ (0.286 m/s^{2})^{2}+(-0.10 m/s^{2})^{2}}= 0.303 m/s^{2}$

Direction will be

$\Phi = \tan^{-1}\left ( \frac{a_{T}}{a_{c}} \right )= \tan^{-1}\left ( \frac{-0.100}{0.286} \right )= -19.3^{0}$

Negative sign indicates clocwise angle wrt radius vector or centripetal acceleration vector.