Respuesta :
Answer:
a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: [tex]a_{R} = 2.571\,\frac{m}{s^{2}}[/tex], [tex]\angle = 90^{\circ}[/tex].
b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: [tex]a_{R} = 2.571\,\frac{m}{s^{2}}[/tex], [tex]\angle = 270^{\circ}[/tex].
c) The Ferris wheel takes 14.646 seconds to make a revolution.
Explanation:
a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.
In addition, centripetal acceleration is determined by the following expression:
[tex]a_{R} = \frac{v^{2}}{R}[/tex] (Eq. 1)
Where:
[tex]a_{R}[/tex] - Centripetal acceleration, measured in meters per square second.
[tex]v[/tex] - Linear speed, measured in meters per second.
[tex]R[/tex] - Radius of the Ferris wheel, measured in meters.
If we know that [tex]v = 6\,\frac{m}{s}[/tex] and [tex]R = 14\,m[/tex], the magnitude of radial acceleration is:
[tex]a_{R} = \frac{\left(6\,\frac{m}{s} \right)^{2}}{14\,m}[/tex]
[tex]a_{R} = 2.571\,\frac{m}{s^{2}}[/tex]
The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: [tex]a_{R} = 2.571\,\frac{m}{s^{2}}[/tex], [tex]\angle = 90^{\circ}[/tex].
b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.
Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: [tex]a_{R} = 2.571\,\frac{m}{s^{2}}[/tex], [tex]\angle = 270^{\circ}[/tex].
c) At first we need to calculate the angular velocity of the Ferris wheel ([tex]\omega[/tex]), measured in radians per second, by using the following expression:
[tex]\omega = \frac{v}{R}[/tex] (Eq. 2)
If we know that [tex]v = 6\,\frac{m}{s}[/tex] and [tex]R = 14\,m[/tex], then the angular velocity of the Ferris wheel is:
[tex]\omega = \frac{6\,\frac{m}{s} }{14\,m}[/tex]
[tex]\omega = 0.429\,\frac{rad}{s}[/tex]
Now we proceed to obtain the period of the Ferris wheel ([tex]T[/tex]), measured in seconds, which is the time needed by that wheel to make on revolution:
[tex]T = \frac{2\pi}{\omega}[/tex] (Eq. 3)
Where [tex]\omega[/tex] is the angular velocity of the Ferris wheel, measured in radians per second.
If we get that [tex]\omega = 0.429\,\frac{rad}{s}[/tex], then:
[tex]T = \frac{2\pi}{0.429\,\frac{rad}{s} }[/tex]
[tex]T = 14.646\,s[/tex]
The Ferris wheel takes 14.646 seconds to make a revolution.
The acceleration experienced by the person, at different locations and the time taken to make one revolution is required in the Ferris wheel.
At highest and lowest point the magnitude of acceleration is [tex]2.5714\ \text{m/s}^2[/tex] and the direction is towards the center.
The time taken is [tex]14.66\ \text{s}[/tex]
r = Radius = 14 m
v = Velocity = 6 m/s
Acceleration is given by
[tex]a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{6^2}{14}\\\Rightarrow a=2.5714\ \text{m/s}^2[/tex]
The magnitude of acceleration at the lowest and highest point of the Ferris wheel will be same the direction will also be the same i.e. towards the center.
Time is given by
[tex]t=\dfrac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi 14}{6}\\\Rightarrow t=14.66\ \text{s}[/tex]
Learn more about rotational motion:
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