Respuesta :
Answer:
a. 0.0628 m
b. 0.000263 m/s2
c.99.997%
d.100.003%
Explanation:
Suppose the radius is 15 m and each loop takes 25 minutes (25*60 = 1500 seconds) to complete. The circumference of the Ferris wheel would be
C = 2πR = 2π*15 = 94.25 m
The your speed is the length of each loop per unit of time to complete each loop
v = C/t = 94.25 / 1500 = 0.0628 m/s
b. Since the circular motion is constant, the only acceleration from the motion is the centripetal acceleration, which can be calculated as the following:
[tex]a_c = v^2/r = 0.0628^2/15 = 0.000263 m/s^2[/tex]
c. Let g = 9.8 m/s^2. At top of the ride, the centripetal acceleration would be pushing you up, while gravity pushes you down, so the weight you experience (compare to standing on ground would be):
[tex]\frac{g-a_c}{g} = \frac{9.8 - 0.000263}{9.8} = 0.99997 = 99.997 \%[/tex]
d. At the bottom of the ride, both centripetal acceleration and gravity would be pushing you downward, making you feel like
[tex]\frac{g+a_c}{g} = \frac{9.8 + 0.000263}{9.8} = 1.00003 = 100.003 \%[/tex]
Answer:
a) The speed is 3.77 m/s
b) The acceleration is 0.948 m/s²
c) The ratio of your weight at the top to your weight standing on the ground is 0.903
d) The ratio of your weight at the bottom of the ride to your weight while standing on the ground is 1.097
Explanation:
a) The speed of the wheel is equal to:
[tex]v=rw[/tex]
Where
r = radius of the circle = 15 m
w = angular velocity
[tex]w=\frac{2\pi }{T}[/tex]
[tex]v=\frac{2\pi r}{T} =\frac{2\pi *15}{25} =3.77m/s[/tex]
b) The acceleration is:
[tex]a=\frac{v^{2} }{r} =\frac{3.77^{2} }{15} =0.948m/s^{2}[/tex]
c) Applying the Newton´s second law:
[tex]W_{true} -N=ma\\W_{true}-N=\frac{mv^{2} }{r} \\N=W_{app} =W_{true}-\frac{mv^{2} }{r}[/tex]
Dividing the expression by Wtrue:
[tex]\frac{W_{app}}{W_{true} } =1-\frac{mv^{2} }{rW_{true}} \\\frac{W_{app}}{W_{true} }=1-\frac{v^{2} }{rg} =1-\frac{3.77^{2} }{15*9.8} =0.903[/tex]
d) Applying the Newton´s second law:
[tex]N-W_{true}=ma\\N-W_{true}=\frac{mv^{2} }{r} \\N=W_{app} =W_{true}+\frac{mv^{2} }{r}[/tex]
Dividing the expression by Wtrue:
[tex]\frac{W_{app}}{W_{true} } =1+\frac{mv^{2} }{rW_{true}} \\\frac{W_{app}}{W_{true} }=1+\frac{v^{2} }{rg} =1+\frac{3.77^{2} }{15*9.8} =1.097[/tex]
