Respuesta :
Answer:
a. [tex]v=3.03\frac{m}{s}[/tex]
b. [tex]a_c=0.71\frac{m}{s^2}[/tex]
c. [tex]\frac{W_{top}}{W}=0.93[/tex]
d. [tex]\frac{W_{bot}}{W}=1.07[/tex]
Explanation:
a. The speed is the distance travelled divided by the time:
[tex]v=\frac{2\pi r}{T}\\v=\frac{2\pi(13m)}{27s}\\v=3.03\frac{m}{s}[/tex]
b. The magnitud of the centripetal acceleration is given by:
[tex]a_c=\frac{v^2}{r}\\a_c=\frac{(3.03\frac{m}{s})^2}{13m}\\a_c=0.71\frac{m}{s^2}[/tex]
c. According to Newton's second law, at the top we have:
[tex]\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[/tex]
d. According to Newton's second law, at the bottom we have:
[tex]\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[/tex]
The acceleration  ,the ratio of your weight at the top weight and the bottom is mathematically given as
- a=(2pT)^2*r
- wtop/FG = 0.9
- wbottom/FG= 1.101
The acceleration, the ratio of your weight at the top weight and the bottom
Generally the equation for the acceleration  is mathematically given as
a = rv^2
Therefore
a = rv^2
a=(2pT)^2*r
b) Using the ratio of your weight at the top of the ride to your weight while standing on the ground
wtop/FG = 1 - v^2/rg
[tex]wtop/FG = 1 - [r*2*pi/T]^2/rg\\\\wtop/FG = 1 - {[17*2*pi/26]^2/17*9.8}[/tex]
wtop/FG = 0.9
c) Using the ratio of your weight at the bottom of the ride to your weight while standing on the ground
 wbottom/FG = 1 + v^2/rg
[tex]wbottom/FG= 1 + [r*2*pi/T]^2/rg\\\\ wbottom/FG= 1 + {[17*2*pi/26]^2/17*9.8}[/tex]
 wbottom/FG= 1.101
For more information on  Weight
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