What is the Speed of The Rocket at the Carnival?

  • Thread starter Thread starter splac6996
  • Start date Start date
  • Tags Tags
    Motion Uniform
Click For Summary

Homework Help Overview

The problem involves calculating the speed of rocket-shaped cars on a carnival ride, where the cars swing out at an angle as the ride spins. The setup includes a disk with a diameter of 5 m and chains that are 6 m long, with the cars swinging out to 20 degrees from vertical.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the cars and suggest drawing free-body diagrams. There are attempts to derive equations for tangential speed and questions about the correct radius to use in calculations.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the radius and the components of the chain's length. Some guidance has been offered regarding the need to reevaluate the radius based on the setup.

Contextual Notes

There are mentions of confusion regarding the radius and the time factor mentioned in the problem statement. Participants are also addressing potential miscalculations in their attempts to find the speed.

splac6996
Messages
37
Reaction score
0

Homework Statement


You've taken your neighbor's young child to the carnival to ride the rides. She wants to ride The Rocket. Eight rocket-shaped cars hang by chains from the outside edge of a large steel disk. A vertical axle through the center of the ride turns the disk, causing the cars to revolve in a circle. You've just finished taking physics, so you decide to figure out the speed of the cars while you wait. You estimate that the disk is 5 m in diameter and the chains are 6 m long. The ride takes 10 s to reach full speed, then the cars swing out until the chains are 20 from vertical.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
You've been told a few times already...

c'mon!
 
I am not sure what is going on
 
Forget the neighbor's kid. Picture a ball on a string going around an upright pole.

Draw that picture.

Make a free-body-diagram of the forces on the ball.
 
Fx=Nsin(theta)=ma=m*v^2/r
Fy=Ncos(theta)-mg=o
 
OK, now you need to use the information to find the tangential speed. Rotational kinematics.
 
Using that information I get a speed of 8.93m/s which is not correct.

v=sqrt(rgtan(theta)

am I missing some idea
 
Last edited:
splac6996 said:
Using that information I get a speed of 8.93m/s which is not correct.

v=sqrt(rgtan(theta)

am I missing some idea

it's 4.033?
 
splac6996 said:
Using that information I get a speed of 8.93m/s which is not correct.

v=sqrt(rgtan(theta)

am I missing some idea

I misread the problem. The radius has to be re-evaluated. Rather than a string attached to a pole, is is a string attached to a 2.5 m radius disk. The radius of the circle taken by the chair will be 2.5 m + the horizontal component of the chain's length.
 
  • #10
thanks
 
  • #11
please help

I have done

v=\sqrt{8.5*9.80*tan(20)}
= 13.65

computer told me that it is not correct.

where did I went wrong?

Do I have to consider the "time factor" of 10 seconds as described in this problem?

please help.
 
  • #12
the radius isn't 8.5, it's 2.5 plus the "horizontal component of the chain's length", which is 6sin(20).
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 9 ·
Replies
9
Views
12K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 20 ·
Replies
20
Views
6K
  • · Replies 11 ·
Replies
11
Views
7K