Moment of inertia/radius of gyration

  • Thread starter Nylex
  • Start date
  • #1
551
1
Can someone please tell me where I've gone wrong with this question?

"A thin rectangular sheet of sides 2a and 2b about an axis through its centre of mass and parallel to the side 2b. Show that the radius of gyration, k, of the sheet is given by [itex]k = \frac{a}{\sqrt{3}}[/itex]"

I drew a rectangle of sides 2a and 2b and then drew an axis through the centre of the rectangle and parallel to the 2b side and labelled it y.

Then

[tex]I_{y} = \int_{-b}^{b} y^2 dm[/tex]

I divided the sheet into strips perpendicular to the axis, each having a mass [itex]dm = \sigma dA = \sigma 2a dy[/itex] where [itex]\sigma[/itex] is the mass per unit area.

So, [tex]I_{y} = \sigma 2a \int_{-b}^{b} y^2 dy = \sigma 2a \left[ \frac{y^3} {3} \right]_{-b}^{b} \Rightarrow I_{y} = \sigma 2a \frac{2b^3} {3}[/tex]

Now [tex]M = \sigma A = \sigma 2a 2b \Rightarrow I_{y} = \frac{Mb^2} {3} = Mk^2 \Rightarrow k = \frac{b} {\sqrt{3}}[/tex]

I don't understand why my integration is wrong.

Thanks.
 

Answers and Replies

  • #2
FredGarvin
Science Advisor
5,067
9
You are mistaking your axes for the integration.

To find Iy you need to (using rho as mass per area):

[tex] I_{y} = \int_{-a}^{a} x^2 dm[/tex]

[tex]I_{y} = 2 \rho b \int_{-a}^{a} x^2 dx[/tex]

[tex]I_{y} = \frac{2 \rho b}{3} (a^3 - (-a^3)) [/tex]

[tex]I_{y} = \frac{4 \rho b}{3} a^3[/tex]

Now you can say that [tex]M = \rho 4 a b[/tex]

That leaves you with:
[tex]I_y = k^2 M[/tex]

[tex]\frac{4 \rho b}{3} a^3 = \rho 4 a b k^2[/tex]

[tex]k = \frac{a}{\sqrt{3}}[/tex]
 
  • #3
551
1
Oh damn it! Thank you :smile:.
 

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