Rotational Inertia Four point masses problem-diagonal

Click For Summary
SUMMARY

The discussion focuses on calculating the rotational inertia of four point masses, each weighing 3.0 kg, arranged in a square configuration with a side length of 0.50 m. The rotational inertia is calculated for an axis passing through points A and C. The formula used is I = mr², where the radius for points B and D is derived as 0.5 m * √2 / 2, based on the sine of 45 degrees. This approach clarifies the geometric relationship between the masses and the axis of rotation.

PREREQUISITES
  • Understanding of rotational inertia and its formula (I = mr²)
  • Basic knowledge of trigonometry, specifically sine functions
  • Familiarity with geometric arrangements of point masses
  • Concept of axes of rotation in physics
NEXT STEPS
  • Study the derivation of rotational inertia for different geometric configurations
  • Learn about the application of trigonometric functions in physics problems
  • Explore the concept of moment of inertia in rigid body dynamics
  • Investigate the effects of changing the axis of rotation on rotational inertia
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking to explain concepts of rotational inertia and geometric configurations of mass.

en bloc
Messages
19
Reaction score
0
Rotational Inertia "Four point masses problem--diagonal"

Homework Statement



Four point masses of 3.0 kg each are arranged in a square on mass-less rod. The length of a side of the square is .50 m. What is the rotational Inertia for rotation about an axis passing through masses A and C?

A B


D C




Homework Equations



I=mr^2


The Attempt at a Solution



I have the solution but just don't understand it! I can't on my own find the radius if it's in a diagonal problem like this one. For B and D, the radius is .5m * [itex]\sqrt{}2[/itex] / 2.

Where does the [itex]\sqrt{}2[/itex] / 2 come from?
 
Physics news on Phys.org


Points B and D are each 0.5sin45m away from the axis.OK?
 


and sin45 = [itex]\frac{\sqrt{2}}{2}[/itex]
 

Similar threads

Replies
17
Views
1K
  • · Replies 19 ·
Replies
19
Views
5K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
5
Views
2K
Replies
52
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K