What is the sweet spot for rotational kinematics?

Click For Summary

Homework Help Overview

The discussion revolves around a problem in rotational kinematics, specifically focusing on the concept of a 'sweet spot' related to a rotating object. Participants are exploring the dynamics of a rotating bar and its properties at different positions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to define the 'sweet spot' and its implications for the rotational motion. Some have calculated angular velocity and tangential velocity, while others are exploring the relationship between various parameters like distance and length in their expressions.

Discussion Status

There is ongoing exploration of the problem with some participants providing expressions and relationships between variables. Guidance has been offered regarding the use of specific ratios in the equations, but no consensus has been reached on the overall approach or solution.

Contextual Notes

Participants are working under the constraints of a homework problem, with specific variables and relationships that need to be addressed in their solutions. The definition of the 'sweet spot' remains a point of contention and inquiry.

Pogorz
Messages
15
Reaction score
0

Homework Statement



i've altered this image to be easier to read. ignore problem 1 except for the information given. the picture of the rotating object is for problem 2, which is the problem i need help with
http://img32.imageshack.us/img32/3690/screenshot20091121at931x.th.png

2. The attempt at a solution

i've solved for 'w' when the rotating bar is vertical, as well as velocity tangential. my issue is the whole 'sweet spot' definition.
 
Last edited by a moderator:
Physics news on Phys.org
You must have d, L and x=L/2 in your answer for v.
Hopefully the expression can be arranged so these appear together as d/L.
You are just asked to replace each d/L with Z.
 
I've attempted to solve the problem, found the centre of mass to be Xcm = (L+2d)/3. Not sure where to go from there, though.
 
Use your expression for the linear speed of the tip.
Replace every d/L with z.
 

Similar threads

  • · Replies 39 ·
2
Replies
39
Views
7K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
7
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K