Chapter 5, Problem 53P

A Ferris wheel 22.0 m in diameter rotates once every 12.5 s (see Fig. 5-9). What is the ratio of a person's apparent weight to her real weight at (a) the top, and (b) the bottom?

To determine

The ratio of apparent weight to the real weight at the top of a Ferris wheel.

Answer to Problem 53P

Solution:

0.717.

Explanation of Solution

Given:

The diameter of the Ferris wheel, $d=22\text{m}$

The radiusof the Ferris wheel, $r=11\text{m}$

Period, $T=12.5\text{s}$

Formula Used:

Angular velocity can be obtained by:

  $\omega =\frac{2\pi }{T}$

Where, T is the period.

Acceleration can be obtained by:

  $a={\omega }^{2}r$

Calculation:

The angular velocitycan be calculated as follows:

  $\begin{array}{c}\omega =\frac{2\pi }{T}\\ =\frac{2\pi }{12.5}\\ =0.5027\text{rad/s}\end{array}$

The acceleration can be calculated as follows:

  $a={\omega }^{2}r$

  $\begin{array}{c}a={0.5027}^2\times 11\\ =2.78{\text{m/s}}^{\text{2}}\end{array}$

Apparent weight at the top of the Ferris wheel

  $W_a=m\left(g-a\right)$

  $\begin{array}{l}{W}_a=m\left(9.81-2.78\right)\\ W_a=7.03\text{m}\end{array}$

The ratio of apparent weight to the real weight at the top of a Ferris wheel

  $\frac{W_a}{W_r}=\frac{mg-ma}{mg}$

  $\begin{array}{l}\frac{W_a}{W_r}=\frac{g-a}{g}\\ \frac{W_a}{W_r}=\frac{9.81-2.78}{9.81}\end{array}$

  $\frac{W_a}{W_r}=0.717$ .

Conclusion:

Therefore, the ratio of apparent weight to the real weight at the top of a Ferris wheel is 0.717.

To determine

The ratio of apparent weight to the real weight at the bottom of a Ferris wheel.

Answer to Problem 53P

Solution:

1.283

Explanation of Solution

Given:

The diameter of the Ferris wheel, $d=22\text{m}$

The radius of the Ferris wheel, $r=11\text{m}$

Period, $T=12.5\text{s}$

Formula Used:

Angular velocity can be obtained by:

  $\omega =\frac{2\pi }{T}$

Where, T is the period.

Acceleration can be obtained by:

  $a={\omega }^{2}r$

Calculation:

The angular velocity can be calculated as follows:

  $\begin{array}{c}\omega =\frac{2\pi }{T}\\ =\frac{2\pi }{12.5}\\ =0.5027\text{rad/s}\end{array}$

The acceleration can be calculated as follows:

  $a={\omega }^{2}r$

  $\begin{array}{c}a={0.5027}^2\times 11\\ =2.78{\text{m/s}}^{\text{2}}\end{array}$

Apparent weight at the top of the Ferris wheel

  $W_a=m\left(g+a\right)$

  $W_a=m\left(9.81+2.78\right)=7.0\text{3}\text{m}$

The ratio of apparent weight to the real weight at the top of a Ferris wheel

  $\frac{W_a}{W_r}=\frac{mg+ma}{mg}$

  $\begin{array}{l}\frac{W_a}{W_r}=\frac{g+a}{g}\\ \frac{W_a}{W_r}=\frac{9.81+2.78}{9.81}\end{array}$

  $\frac{W_a}{W_r}=1.283$

Conclusion:

Therefore, the ratio of apparent weight to the real weight at the bottom of a Ferris wheel is 1.283.