Respuesta :
Answer:
Explanation:
A ) angular velocity ω = 2π / T
= 2 x 3.14 / 60
= .10467 rad / s
linear velocity v = ω R
= .10467 x 50
= 5.23 m / s
centripetal force = m v² / R
= mg v² / gR
= 834 x 5.23² / 9.8 x 50
= 46.55 N
B )
apparent weight
= mg - centripetal force
= 834 - 46.55
= 787.45 N
C ) apparent weight
= mg + centripetal force
= 834 + 46.55
= 880.55 N.
D )
For apparent weight to be zero
centripetal force = mg
mg = mv² / R
v² = gR
= 9.8 x 50
= 490
v = 22.13 m /s
time period of revolution
= 2π R /v
2 x 3.14 x 50 / 22.13
= 14.19 s
For the Cosmoclock 21 Ferris wheel in Yokohama City, Japan the speed and weight of the passenger,
- Part A: The speed of the passengers when the Ferris wheel is rotating at this rate is 5.23 m/s.
- Part B: The apparent weight of the passenger at the highest point of the Ferris wheel is 787.45.l
- Part C: The apparent weight of the passenger at the lowest point on the Ferris wheel is 880.55
- Part D: The time for one revolution if the passenger's apparent weight at the highest point were zero is 14.19 s.
- Part E: The passenger's apparent weight at the lowest point is 1667.7 N.
What is angular speed of a body?
The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,
[tex]\omega= \dfrac{\Delta \theta}{\Delta t}[/tex]
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m.
Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60 s).
- Part A: The speed of the passengers when the Ferris wheel is rotating at this rate (in m/s)-
Speed of the body is the product of angular velocity and radius. Angular velocity is the ratio of 2π to the time period, and radius is 50 m (100/2). . Thus, the speed is,
[tex]v=\dfrac{2\pi\times50}{60}v=5.23 \rm m/s[/tex]
- Part B: Apparent weight of the passenger at the highest point of the Ferris wheel- (in N)-
The passenger weighs 834 N. Thus the mass of passenger is,
[tex]m=\dfrac{834}{9.81}\\m\cong85\rm kg[/tex]
The apparent weight of the passenger at the highest point of the Ferris wheel is the difference of his weight and the centripetal force. Therefore,
[tex]W_a=W-\dfrac{mv^2}{r}\\W_a=834-\dfrac{85\times5.23^2}{50}\\W_a=787.45\rm N[/tex]
- Part C: The apparent weight at the lowest point on the Ferris wheel
The apparent weight of the passenger at the lowest point of the Ferris wheel is the sum of his weight and the centripetal force. Therefore,
[tex]W_a=W+\dfrac{mv^2}{r}\\W_a=834+\dfrac{85\times5.23^2}{50}\\W_a=880.55\rm N[/tex]
- Part D: Time for one revolution if the passenger's apparent weight at the highest point were zero-
For apparent weight at the highest point to be zero, the weight and centripetal force of the passenger should be equal. Thus,
[tex]W=\dfrac{mv^2}{r}\\845=\dfrac{85\times v^2}{50}\\v=22.07\rm m/s[/tex]
Thus, the time period is,
[tex]t=\dfrac{2\pi (50)}{22.07}\\t=14.23\rm s[/tex]
- Part E: What then would be the passenger's apparent weight at the lowest point? (in N)
It can be given as,
[tex]W=m(g+a)\\W=85(9.81+9.81)W=1667.7\rm N[/tex]
Hence, For the Cosmoclock 21 Ferris wheel in Yokohama City, Japan the speed and weight of the passenger,
- Part A: The speed of the passengers when the Ferris wheel is rotating at this rate is 5.23 m/s.
- Part B: The apparent weight of the passenger at the highest point of the Ferris wheel is 787.45.l
- Part C: The apparent weight of the passenger at the lowest point on the Ferris wheel is 880.55
- Part D: The time for one revolution if the passenger's apparent weight at the highest point were zero is 14.19 s.
- Part E: The passenger's apparent weight at the lowest point is 1667.7 N.
Learn more about the angular speed here;
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