Answer
(a) the highest point of the Ferris wheel:
${\bf{a}} = - 15{\bf{T}} + 53.3{\bf{N}} = \left( {15, - 53.3} \right)$ $m/{\min ^2}$
(b) the two points level with the center of the wheel:
${{\bf{a}}_1} = - 15{{\bf{T}}_1} + 53.3{{\bf{N}}_1} = \left( {53.3,15} \right)$ $m/{\min ^2}$
${{\bf{a}}_2} = - 15{{\bf{T}}_2} + 53.3{{\bf{N}}_2} = \left( { - 53.3, - 15} \right)$ $m/{\min ^2}$
Work Step by Step
In Example 6, we are given the following:
- the wheel rotates counterclockwise with a speed of $40$ m/min
- the wheel is slowing at a rate of $15$ $m/{\min ^2}$, that is ${a_{\bf{T}}} = v' = - 15$
- the curvature of the wheel is $\kappa = \frac{1}{R} = \frac{1}{{30}}$, so the normal component of the acceleration is
${a_{\bf{N}}} = \kappa {v^2} = \frac{{{v^2}}}{R} = \frac{{{{40}^2}}}{{30}} \simeq 53.3$
(a) At the highest point of the Ferris wheel we have ${\bf{T}} = \left( { - 1,0} \right)$ and ${\bf{N}} = \left( {0, - 1} \right)$. Therefore,
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
${\bf{a}} = - 15{\bf{T}} + 53.3{\bf{N}} = \left( {15, - 53.3} \right)$ $m/{\min ^2}$
(b) There are two points level with the center of the wheel, that is, at
- the left side: ${{\bf{T}}_1} = \left( {0, - 1} \right)$ and ${{\bf{N}}_1} = \left( {1,0} \right)$
- the right side: ${{\bf{T}}_2} = \left( {0,1} \right)$ and ${{\bf{N}}_2} = \left( { - 1,0} \right)$
Thus,
${{\bf{a}}_1} = - 15{{\bf{T}}_1} + 53.3{{\bf{N}}_1} = \left( {53.3,15} \right)$ $m/{\min ^2}$
${{\bf{a}}_2} = - 15{{\bf{T}}_2} + 53.3{{\bf{N}}_2} = \left( { - 53.3, - 15} \right)$ $m/{\min ^2}$
