Roller coaster and spiderman swing circular motion problems

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
rain_ex
Messages
6
Reaction score
0
Circular Motion and Universal Law of Gravitation Homework help!

Homework Statement



The passengers in a roller coaster car feel twice as heavy as their true weight as the car goes through a dip with a 30 m radius of curvature. What is the car's speed at the bottom of this dip?

Given: r = 30m, mass = twice the force due to gravity(?)
Find: Velocity

Homework Equations



[tex]\sum[/tex] F = m [tex]\frac{v^{2}}{r}[/tex]

The Attempt at a Solution



I'm not sure on where to start or set this one up. Also, I don't know where to find the original "true" weight of the car, if the feeling on the roller coaster is twice this.

Homework Statement



Spiderman plans to cross a gap between two buildings by swinging in an arc from his web. If his arms are capable of exerting a force of 1900N on the webbing, what is the maximum speed he can tolerate at the lowest point of his swing? Spiderman's mass is 80 kg and the webbing is 4.8 m long.

Given: F = 1900N, m = 80kg, r = 2.4m[?] (the webbing is 4.8m long total so that would be the diameter, so half that would be the radius? or is 4.8m the actual radius?)

Homework Equations



[tex]\sum[/tex] F = m [tex]\frac{v^{2}}{r}[/tex]

The Attempt at a Solution



With the radius as 2.4m, I got 7.55 m/s as the velocity.
With the radius as 4.8m, I got 10.68 m/s as the velocity.

Which one am I supposed to use and are these even the correct solutions? Help would be greatly appreciated!
 
Physics news on Phys.org


.
Welcome to Physics Forums.

rain_ex said:

Homework Statement



The passengers in a roller coaster car feel twice as heavy as their true weight as the car goes through a dip with a 30 m radius of curvature. What is the car's speed at the bottom of this dip?

Given: r = 30m, mass = twice the force due to gravity(?)
Find: Velocity

Homework Equations



[tex]\sum[/tex] F = m [tex]\frac{v^{2}}{r}[/tex]
Okay, and what are the different forces that would act on one of the passengers? These forces would go in that summation on the left side of your equation.

The Attempt at a Solution



I'm not sure on where to start or set this one up. Also, I don't know where to find the original "true" weight of the car, if the feeling on the roller coaster is twice this.
Apparent weight is the reading you'd see if the car were placed on a scale. Tough to do while it is moving, but you could imagine a passenger sitting on a scale inside the car, and the scale reading would be the apparent weight of that passenger. In other words, apparent weight is the normal force acting on the passenger or car.

Just to check, have you drawn a force diagram for the car? That's usually a good way to start one of these problems.