Angular velocity of a ferris wheel

  • Thread starter kimikims
  • Start date
  • #1
kimikims
36
0
This is a hard problem!! :cry: 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

  • #2
Doc Al
Mentor
45,433
1,885
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!
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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.
 
  • #4
kimikims
36
0
kimikims said:
This is a hard problem!! :cry: 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?
 
  • #5
thermodynamicaldude
60
1
How did you get a mass of 11.71 kg? The passenger's normal weight is 367 newtons....and

weight = mass * g.
 
  • #6
kimikims
36
0
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 ?
 

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