Amusement park ride gravitation problem

  • Thread starter Thread starter holmeskaei
  • Start date Start date
  • Tags Tags
    Gravitation
Click For Summary
SUMMARY

The discussion focuses on the physics of an amusement park ride involving a 5.4-meter diameter hollow steel cylinder that rotates, creating centripetal force to keep passengers against the wall after the floor drops. The static coefficient of friction ranges from 0.64 to 1.0, and the kinetic coefficient ranges from 0.40 to 0.70. The user attempts to calculate the necessary velocity for passengers to remain adhered to the wall but incorrectly assumes the normal force equals the weight of the passengers. Correct application of the equations of motion and friction is essential for accurate results.

PREREQUISITES
  • Understanding of centripetal force and its relation to circular motion
  • Knowledge of static and kinetic friction coefficients
  • Familiarity with Newton's laws of motion
  • Ability to apply basic algebra to physics equations
NEXT STEPS
  • Review the concept of centripetal acceleration and its formula: a = v²/r
  • Study the relationship between normal force and gravitational force in circular motion
  • Learn how to apply friction equations in dynamic scenarios
  • Explore the implications of varying coefficients of friction in practical applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts of friction and circular motion in real-world applications.

holmeskaei
Messages
8
Reaction score
0

Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.4-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.64 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."


Homework Equations


f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r


The Attempt at a Solution


f=(0.64)(9.8*30)=188.16N
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.
 
Physics news on Phys.org


holmeskaei said:

Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.4-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.64 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."


Homework Equations


f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r


The Attempt at a Solution


f=(0.64)(9.8*30)=188.16N use your 3rd relevant equation, f(fric)= mg. You are incorrectly assuming that the Normal force is the person's weight. [/color]
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.
Please see above in red[/color]
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 6 ·
Replies
6
Views
6K
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
10K
  • · Replies 3 ·
Replies
3
Views
3K