Question

A ferris wheel 22.0 m in diameter rotates once every 12.5 seconds. WHat is the ratio of a person's apparent weight to her real weight at (a) the top, and (b) the bottom?

Answer 1

Concepts and reason

The concepts used to solve this problem are centripetal acceleration and Newton’s second law of motion.

Initially, the angular velocity is found to determine the centripetal force.

Finally, use Newton’s second law of motion to determine the ratio of apparent weight to real weight of the person at the bottom and top.

Fundamentals

The centripetal acceleration is the one acting on a body, moving along a circular path and is directed towards the center of the curvature of the path.

The expression for centripetal acceleration is as follows:

${a_c} = \frac{{{v^2}}}{r}$

Here, ${a_c}$ is the centripetal acceleration of the body, $v$ is its velocity and $r$ is the radius of the circular path.

According to Newton’s second law of motion, the acceleration of a body produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.

The expression for Newton’s second law of motion is,

$F = ma$

Here $F$ is the net force, $m$ is the mass of the body and $a$ is its acceleration.

The force which results in centripetal acceleration is the centripetal force.

The expression for centripetal force is,

${F_c} = m\frac{{{v^2}}}{r}$

Here, ${F_c}$ is the centripetal force.

The expression for angular velocity is,

$\omega = \frac{{2\pi }}{T}$

Here, $\omega$ is the angular velocity and $T$ is the time period.

The expression for angular velocity in terms of velocity and radius is,

$\omega = \frac{v}{r}$

Here, $v$ is the velocity of the body and $r$ is the radius of the circular path.

The expression for radius is,

$r = \frac{d}{2}$

Here, $d$ is the diameter.

The apparent weight of the person is the normal force acting on her and the real weight is the gravitational force.

The expression for apparent weight at the bottom of the circular path is,

${N_b} = mg + {F_c}$

Here, ${N_b}$ is the apparent weight at the bottom and $g$ is the acceleration due to gravity.

The expression for apparent weight at the top of the circular path is,

${N_t} = mg - {F_c}$

Here, ${N_t}$ is the apparent weight at the top.

The expression for real weight is,

${F_g} = mg$

Here, ${F_g}$ is the real weight.

(a)

The expression for angular velocity is,

$\omega = \frac{{2\pi }}{T}$

Substitute $12.5\,{\rm{s}}$ for $T$ and $3.14$ for $\pi$ .

$\begin{array}{c}\\\omega = \frac{{2\pi }}{T}\\\\ = \frac{{\left( 2 \right)\left( {3.14} \right)}}{{12.5\,{\rm{s}}}}\\\\ = 0.5024\,{\rm{rad/s}}\\\end{array}$

The expression for radius is,

$r = \frac{d}{2}$

Substitute $22.0\,{\rm{m}}$ for $d$ .

$\begin{array}{c}\\r = \frac{d}{2}\\\\ = \frac{{22.0\,{\rm{m}}}}{2}\\\\ = 11\,{\rm{m}}\\\end{array}$

The expression for the apparent weight of the person at the top,

${N_t} = mg - {F_c}$

The expression for real weight is,

${F_g} = mg$

The ratio between apparent weight and real weight is,

$\begin{array}{c}\\\frac{{{N_t}}}{{{F_g}}} = \frac{{mg - {F_c}}}{{mg}}\\\\ = 1 - \frac{{{F_c}}}{{mg}}\\\end{array}$

Substitute $m\frac{{{v^2}}}{r}$ for ${F_c}$ .

$\begin{array}{c}\\\frac{{{N_t}}}{{{F_g}}} = 1 - \frac{{m\left( {\frac{{{v^2}}}{r}} \right)}}{{mg}}\\\\ = 1 - \frac{{\left( {\frac{{{v^2}}}{r}} \right)}}{g}\\\end{array}$

Substitute $r\omega$ for $v$ .

$\frac{{{N_t}}}{{{F_g}}} = 1 - \frac{{{\omega ^2}r}}{g}$

Substitute $0.5024\,{\rm{rad/s}}$ for $\omega$ , $11\,{\rm{m}}$ for $r$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ .

$\begin{array}{c}\\\frac{{{N_t}}}{{{F_g}}} = 1 - \frac{{{{\left( {0.5024\,{\rm{rad/s}}} \right)}^2}(11\,{\rm{m)}}}}{{(9.8\,m/{s^2})}}\\\\ = 0.717\\\end{array}$

(b)

The expression for the apparent weight of the person at the bottom,

${N_b} = mg + {F_c}$

The expression for real weight is,

${F_g} = mg$

The ratio between apparent weight and real weight is,

$\begin{array}{c}\\\frac{{{N_b}}}{{{F_g}}} = \frac{{mg + {F_c}}}{{mg}}\\\\ = 1 + \frac{{{F_c}}}{{mg}}\\\end{array}$

Substitute $m\frac{{{v^2}}}{r}$ for ${F_c}$ .

$\begin{array}{c}\\\frac{{{N_b}}}{{{F_g}}} = 1 + \frac{{m\left( {\frac{{{v^2}}}{r}} \right)}}{{mg}}\\\\ = 1 + \frac{{\left( {\frac{{{v^2}}}{r}} \right)}}{g}\\\end{array}$

Substitute $r\omega$ for $v$ .

$\frac{{{N_b}}}{{{F_g}}} = 1 + \frac{{{\omega ^2}r}}{g}$

Substitute $0.5024\,{\rm{rad/s}}$ for $\omega$ , $11\,{\rm{m}}$ for $r$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ .

$\begin{array}{c}\\\frac{{{N_b}}}{{{F_g}}} = 1 + \frac{{{{\left( {0.5024\,{\rm{rad/s}}} \right)}^2}(11\,{\rm{m)}}}}{{(9.8\,m/{s^2})}}\\\\ = 1.283\\\end{array}$

Ans: Part a

The ratio of a person’s apparent weight to real weight at the top is ${\bf{0}}{\bf{.717}}$ .

Part b

The ratio of a person’s apparent weight to real weight at the bottom is ${\bf{1}}{\bf{.283}}$ .