What is the required speed to successfully complete the spinning cylinder game?

  • Thread starter Thread starter stevo13
  • Start date Start date
Click For Summary
SUMMARY

The spinning cylinder game requires precise calculations to determine the necessary rotational speed for successful gameplay. With a cylinder diameter of 0.5 meters, the calculated angular velocity needed is 2.45 rad/s, while a reference book suggests 1.6 rad/s. The discrepancy arises from the varying acceleration of the object as it falls through the hole, which is influenced by the cylinder's orientation. Utilizing the equations of motion, specifically S = 0.5at² and a = gsin(θ), is essential for accurate calculations.

PREREQUISITES
  • Understanding of angular velocity and its calculation
  • Knowledge of basic physics concepts, including gravitational acceleration
  • Familiarity with kinematic equations of motion
  • Ability to analyze rotational dynamics
NEXT STEPS
  • Study the principles of angular motion and its applications
  • Learn how to apply kinematic equations in rotational systems
  • Explore the effects of varying acceleration on falling objects
  • Investigate the relationship between linear and angular velocity
USEFUL FOR

Physics students, educators, game designers, and anyone interested in the mechanics of rotational motion and its practical applications.

stevo13
Messages
7
Reaction score
0
the hollow cylinder shown in fig 2.2 (see attachment) is free to rotate on a horizontal axis. one hole is cut in the side of the cylinder. the object of the game is to spin the cylinder so fast that the object dropped through the hole when it is in the uppermost position will fall through the same hole when it has rotated to the bottom position. if the diameter of the cylinder is .5 meters how fast must the hole be moving (hint: first calculate the time for the object to fall the appropriate distance, then use that result to determine the number of rad/s.) I get 2..45 rad/s but the book gets 1.6? Try it and see what you get. Thanks

Stevo
 

Attachments

  • phys1.JPG
    phys1.JPG
    8.5 KB · Views: 400
Physics news on Phys.org
Although I haven't see your attachments yet, you have to remember that the acceleration of the ball isn't constant (it temporarily becomes 0 when the cylinder is horizintal and is "g" when vertical).
Try to relate the distance traveled with the acceleration at some instant of time and the corresponding acceleration at that instant.
Hint: Use S = 0.5at^2, a = gsin(theta) where theta = ...

Can you go from here ?

Arun
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 8 ·
Replies
8
Views
12K
  • · Replies 45 ·
2
Replies
45
Views
7K
  • · Replies 54 ·
2
Replies
54
Views
7K
Replies
20
Views
9K
  • · Replies 52 ·
2
Replies
52
Views
8K
Replies
3
Views
5K
  • · Replies 6 ·
Replies
6
Views
9K
Replies
2
Views
13K
  • · Replies 5 ·
Replies
5
Views
5K