Answer:
422.36 N
Explanation:
given,
time of rotation = 4.30 s
T = 4.30 s
Assuming the diameter of the ring equal to 16 m
radius, R = 8 m
[tex]v = \dfrac{2\pi R}{T}[/tex]
[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]
v = 11.69 m/s
now, Force does the ring push on her at the top
[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]
[tex] N + m g = \dfrac{mv^2}{R}[/tex]
[tex] N = \dfrac{mv^2}{R}- m g[/tex]
[tex] N = m(\dfrac{v^2}{R}- g)[/tex]
[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]
N = 422.36 N
The force exerted by the ring to push her is equal to 422.36 N.
The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
It is Given that
Time rotation T= 4.30 s
Mass m= 58 kg
Now the Velocity will be calculated as
[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]
Now by balancing the forces
[tex]N=\dfrac{mv^2}{R} -mg[/tex]
[tex]N=m(\dfrac{v^2}{R} -g)[/tex]
[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]
[tex]N=422.36 \ Newton[/tex]
Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
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