Angular velocity of a ferris wheel

• kimikims
In summary, the passenger's normal weight is 367 N on a ferris wheel with a radius of 14 m and an acceleration of gravity of 9.8 m/s^2. To make the customers at the top of the wheel feel zero g's, the operator must rev it up to an angular velocity of 0.835 1/s. Assuming the same angular velocity, the customer will feel a weight of 735.97 N at the bottom of the wheel.
kimikims
This is a hard problem! Anyone know what my mistakes are??

-----
A passenger on the ferris wheel normally
weighs 367 N. The ferris wheel has a 14 m radius and is
equipped with a powerful motor. The operator revs it up so that the customers at the top of the wheel feel zero g's (they momentarily
lift slightly of their seats). The acceleration of gravity is 9.8 m/s^2

1) At what angular velocity will this occur?

N = 367 N
R = 14m
g = 9.8 m/s^2
N = 0

Fc = mg = m(V^2/R)
gR = V^2

(9.8)(14) = v^2 = 137.2 = 11.71

(Angular Velocity) W= V/R

11.71 / 14 = .835

2) Assume: The rotating angular velocity is
same as in Part 1.
What weight does the customer feel at the
bottom of the wheel? Answer in units of N.

Fc = N - mg = m(V^2/R)

N = mg + m(V^2/R)

V = WR

N = mg + m [(W^2 x R^2)/(R)]

= mg + mW^2R

= (367)(9.8) + (367) (.84)^2 (14)

= 7221.9728

kimikims said:
1) At what angular velocity will this occur?

N = 367 N
R = 14m
g = 9.8 m/s^2
N = 0

Fc = mg = m(V^2/R)
gR = V^2

(9.8)(14) = v^2 = 137.2 = 11.71

(Angular Velocity) W= V/R

11.71 / 14 = .835
2) Assume: The rotating angular velocity is
same as in Part 1.
What weight does the customer feel at the
bottom of the wheel? Answer in units of N.

Fc = N - mg = m(V^2/R)

N = mg + m(V^2/R)

V = WR

N = mg + m [(W^2 x R^2)/(R)]

= mg + mW^2R

= (367)(9.8) + (367) (.84)^2 (14)

= 7221.9728
One problem: 367 N is the passenger's weight, not mass!

Another hint for part 2 is that the magnitude of the centrifugal force is always the same. If you understand that, you should be able to immediately write down the correct answer.

kimikims said:
This is a hard problem! Anyone know what my mistakes are??

-----
A passenger on the ferris wheel normally
weighs 367 N. The ferris wheel has a 14 m radius and is
equipped with a powerful motor. The operator revs it up so that the customers at the top of the wheel feel zero g's (they momentarily
lift slightly of their seats). The acceleration of gravity is 9.8 m/s^2

1) At what angular velocity will this occur?

N = 367 N
R = 14m
g = 9.8 m/s^2
N = 0

Fc = mg = m(V^2/R)
gR = V^2

(9.8)(14) = v^2 = 137.2 = 11.71

(Angular Velocity) W= V/R

11.71 / 14 = .835

2) Assume: The rotating angular velocity is
same as in Part 1.
What weight does the customer feel at the
bottom of the wheel? Answer in units of N.

Fc = N - mg = m(V^2/R)

N = mg + m(V^2/R)

V = WR

N = mg + m [(W^2 x R^2)/(R)]

= mg + mW^2R

= (367)(9.8) + (367) (.84)^2 (14)

= 7221.9728

So for part 2...

it should be uhm

(11.71) (9.8) + (11.71) (.84)^2 (14)

=230.43 N?

How did you get a mass of 11.71 kg? The passenger's normal weight is 367 Newtons...and

weight = mass * g.

kimikims said:
So for part 2...

it should be uhm

(11.71) (9.8) + (11.71) (.84)^2 (14)

=230.43 N?

So would it be...

(37.4) (9.8) + (37.4) (.84)^2 (14)

= 735.97 ?

1. What is angular velocity?

Angular velocity is a measure of how fast an object is rotating or moving in a circular path. It is usually measured in radians per second or degrees per second.

2. How is angular velocity different from linear velocity?

Angular velocity measures the rate at which an object is rotating, while linear velocity measures the rate at which an object is moving in a straight line. Angular velocity is dependent on the object's distance from the center of rotation, while linear velocity is not.

3. How is angular velocity of a ferris wheel calculated?

The angular velocity of a ferris wheel can be calculated by dividing the change in the wheel's angular position by the change in time. This can also be expressed as the ratio of the circumference of the wheel to the time it takes to complete one full rotation.

4. Why does the angular velocity of a ferris wheel change?

The angular velocity of a ferris wheel changes because its speed and direction of rotation are constantly changing. At the top of the wheel, the velocity is highest, while at the bottom it is lowest. This is due to the varying distance of the different parts of the wheel from the center of rotation.

5. How does angular velocity affect the ride experience on a ferris wheel?

The angular velocity of a ferris wheel can affect the ride experience by determining the speed at which the riders are moving and the forces acting on them. A higher angular velocity can result in a more thrilling and faster ride, while a lower angular velocity can result in a slower and more relaxed ride.

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