Respuesta :
Answer:
9.68 m/s2
Explanation:
If it completes 1 revolution, or 2π for every 36 seconds, then its angular speed ω is 2π/36 = 0.1745 rad/s
So the centripetal acceleration would be
[tex]a_c = \omega^2 r = 0.1745^2*9.9 = 0.3 m/s^2[/tex]
The tangential acceleration is:
[tex]a_t = \alpha r = 0.19*9.9 = 1.881 m/s^2[/tex]
When the passenger is on top of the wheel, the tangential acceleration is horizontal, while centripetal acceleration is vertical, acting upward, gravitational acceleration g is 9.8 m/s2 acting downward. The total vertical acceleration is 0.3 - 9.8 = -9.5 m/s2
So the magnitude of the total acceleration consisting of horizontal and vertical components is:
[tex]a = \sqrt{a_v^2 + a_c^2} = \sqrt{9.5^2 + 1.881^2} = \sqrt{90.25 + 3.538161} = \sqrt{93.788161} = 9.68m/s^2[/tex]
Acceleration can be defined as rate of change in the movement. The total magnitude of acceleration of the passenger's at that time is [tex]\bold{ 9.68\ m/s^2 }[/tex] .
The angular speed ω is 2π/36 = 0.1745 rad/s
So the centripetal acceleration will be,
[tex]\bold {a_c = \omega ^2r }\\\\\bold {a_c = 0.1745^2 \times 9.9 }\\\\\bold {a_c = 0.3\ m/s^2}[/tex]
The tangential acceleration,
[tex]\bold {a_t = \alpha r}\\\bold {a_t =0.19\times 9.9 }\\\bold {a_t =1.88\ m/s^2}[/tex]
The total vertical acceleration is
[tex]\bold {a_v = a_c g}\\\bold {a_v = 0.3 \times 9.8 }\\\bold {a_v =-9.5\ m/s^2}[/tex]
So, magnitude of the total acceleration of the passenger's including horizontal and vertical components is:
[tex]\bold {a = \sqrt{a^2_v + a^2_c} }\\\\\bold {a = \sqrt{-9.5^2 + 1.88^2}}\\\\\bold {a = \sqrt{90.25 +3.538 }}\\\\\bold {a = 9.68\ m/s^2}[/tex]
Therefore, the total magnitude of acceleration of the passenger's at that time is [tex]\bold{ 9.68\ m/s^2 }[/tex] .
To know more about acceleration,
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