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Rotational Dynamics-Moment of inertia

  • #1

Homework Statement


A meter stick of mass 0.44kg rotates in the horizontal plane about a vertical axis passing through the 30cm mark. What is the moment of intertia of the stick? (Treat it as a long uniform rod).


Homework Equations


Moment of inertia for a long uniform rod (axis through center): (1/12)ML2
Moment of inertia for a long uniform rod (axis through end): (1/3)ML2
I=mr2

The Attempt at a Solution



LA=1.0m;
LB=0.70m (I'm not sure how to determine the length/axis for LB???)
I=1/3LA2 + 1/3LB2
I=1/3(1.0)2 + 1/3(0.70)2
I= 0.50 m x N
 

Answers and Replies

  • #2
34
0
[tex]L_{B} = 0.2m [/tex] (distance between center of mass and axis if you use moment of inertia on center)

[tex]I = \frac{1}{12}ML^2 + M(0.2)^2 [/tex]
 
  • #3
Ok thanks.

However, wouldn't the formula for both terms on the right side of the equation be 1/12ML2 (I.e., LA and LB, respectively)

Also, if the L is determined by the distance from the centre of gravity, does that mean LA=0.5? or does is it still equal to 1.0?

Sorry for so many questions, I just want to make sure I understand this.
 
  • #4
rock.freak667
Homework Helper
6,230
31
Ok thanks.

However, wouldn't the formula for both terms on the right side of the equation be 1/12ML2 (I.e., LA and LB, respectively)

Also, if the L is determined by the distance from the centre of gravity, does that mean LA=0.5? or does is it still equal to 1.0?

Sorry for so many questions, I just want to make sure I understand this.
The parallel axis theorem: I=Ic+mr2

r= distance from the centre of rotation

Ic= moment of inertia about the centre.
 
  • #5
The parallel axis theorem: I=Ic+mr2

r= distance from the centre of rotation

Ic= moment of inertia about the centre.
I'm not sure what the parallel axis theorem is, as we haven't covered it in any of the lessons for this course...Is there a way to solve the problem without that theorem?
Thanks :)
 
  • #6
34
0
of course, you can always use its definition:

[tex] I = \int_{M} r^2dm[/tex]
 
  • #7
of course, you can always use its definition:

[tex] I = \int_{M} r^2dm[/tex]
The course I am in is a very basic course, and as such, calculus is outside the scope of this course. The only formulas we have been given are for moment of inertia (i.e. I=1/12ML2, etc.) Can the problem be solved using those formulas?

Thanks very much for your help though :)
 
  • #8
rock.freak667
Homework Helper
6,230
31
The course I am in is a very basic course, and as such, calculus is outside the scope of this course. The only formulas we have been given are for moment of inertia (i.e. I=1/12ML2, etc.) Can the problem be solved using those formulas?

Thanks very much for your help though :)
It can. You know the moment of inertia about the centre as being 1/12 ML2.


You want to move the axis to 0.3m mark. At the centre, the axis is at 0.5m. So the distance from the centre to the 0.3m mark is 0.2m.

So the moment of interia about the 0.3m mark = 1/12 ML2+ M(0.2)2

Which is how kyiydnlm got the answer.
 
  • #9
Oh okay - thanks I understand what you're saying now.

For the eq'n: 1/12 ML2+ M(0.2)2

Will the L be equal to the full distance of 1.0m, or will it equal 0.30 (the distance from the end to the axis)?
 
  • #10
rock.freak667
Homework Helper
6,230
31
Oh okay - thanks I understand what you're saying now.

For the eq'n: 1/12 ML2+ M(0.2)2

Will the L be equal to the full distance of 1.0m, or will it equal 0.30 (the distance from the end to the axis)?
L is the total length of the rod...or metre stick as it may be
 
  • #11
Okay, thanks for all your help!!
 

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