# Angular velocity of a ferris wheel

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?
Answer in units of 1=s.

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

## Answers and Replies

Doc Al
Mentor
kimikims said:
1) At what angular velocity will this occur?
Answer in units of 1=s.

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
Looks OK to me. What are the units of your answer?
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!

Fredrik
Staff Emeritus
Science Advisor
Gold Member
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?
Answer in units of 1=s.

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 ?