Why is my calculation for the speed of a falling rod off by sqrt(2)?

  • Thread starter Thread starter faculaganymede
  • Start date Start date
  • Tags Tags
    Inertia Torque
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a thin uniform rod falling from a vertical position about a pivot point. Participants are exploring the calculation of the speed at which the free end of the rod strikes the ground, with a specific focus on the discrepancy in the original poster's answer compared to the expected result.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of gravitational potential energy into rotational kinetic energy, considering the height of the center of mass and the moment of inertia for the rod. There are questions regarding the setup and the calculations leading to the discrepancy of sqrt(2) in the original poster's answer.

Discussion Status

The discussion is ongoing, with some participants providing insights into the energy transformation involved in the problem. There is a lack of explicit consensus on the correct approach, and multiple interpretations of the problem setup are being explored.

Contextual Notes

One participant pointed out a potential issue with posting in the forum, indicating a need for adherence to community guidelines. The original poster acknowledges this as their first post, suggesting a possible lack of familiarity with the forum's expectations.

faculaganymede
Messages
3
Reaction score
0
A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!
 
Physics news on Phys.org
the gpe is being transformed into KE rotational. you would use the height as L/2 since the center of mass is changing that much distance.
the moment of inertia of a stick rotating about the end is 1/3mL^2

[tex]mg(0.5L) = 0.5I\omega^2[/tex]
 
faculaganymede said:
A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!

You should not have posted this here. Did you not read https://www.physicsforums.com/showthread.php?t=89899 thread ?


regards
marlon
 
sorry, that was my first post. didn't know better.
 
Thanks Andrew!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
7K