Question

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 13 m , and you use your watch to find that each loop around takes 27 s . a.What is your speed?
b.What is the magnitude of your acceleration?
c.What is the ratio of your weight at the top of the ride to yourweight while standing on the ground?
d.What is the ratio of your weight at the bottom of the ride toyour weight while standing on the ground?

Answer 1

Answer:

a. $v=3.03\frac{m}{s}$

b. $a_c=0.71\frac{m}{s^2}$

c. $\frac{W_{top}}{W}=0.93$

d. $\frac{W_{bot}}{W}=1.07$

Explanation:

a. The speed is the distance travelled divided by the time:

$v=\frac{2\pi r}{T}\\v=\frac{2\pi(13m)}{27s}\\v=3.03\frac{m}{s}$

b. The magnitud of the centripetal acceleration is given by:

$a_c=\frac{v^2}{r}\\a_c=\frac{(3.03\frac{m}{s})^2}{13m}\\a_c=0.71\frac{m}{s^2}$

c. According to Newton's second law, at the top we have:

$\sum F_{top}:W-W_{top}=ma_c\\W_{top}=mg-ma_c\\W_{top}=m(g-a_c)\\W_{top}=m(9.8\frac{m}{s^2}-0.71\frac{m}{s^2})\\W_{top}=m(9.09\frac{m}{s^2})\\\frac{W_{top}}{W}=\frac{m(9.09\frac{m}{s^2})}{m(9.8\frac{m}{s^2})}\\\frac{W_{top}}{W}=0.93$

d. According to Newton's second law, at the bottom we have:

$\sum F_{bot}:-W+W_{bot}=ma_c\\W_{bot}=mg+ma_c\\W_{bot}=m(g+a_c)\\W_{bot}=m(9.8\frac{m}{s^2}+0.71\frac{m}{s^2})\\W_{bot}=m(10.51\frac{m}{s^2})\\\frac{W_{bot}}{W}=\frac{m(10.51\frac{m}{s^2})}{m(9.8\frac{m}{s^2})}\\\frac{W_{bot}}{W}=1.07$