Moment of inertia/radius of gyration

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In summary, the conversation discusses the calculation of the radius of gyration, k, for a thin rectangular sheet with sides 2a and 2b when rotated about an axis through its center of mass and parallel to the 2b side. The correct integration is shown, using the mass per unit area, and the final equation is presented as k = a/√3.
  • #1
Nylex
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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.
 
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  • #2
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
Oh damn it! Thank you :smile:.
 

Related to Moment of inertia/radius of gyration

1. What is moment of inertia and how is it calculated?

Moment of inertia, also known as rotational inertia, is a measure of an object's resistance to change in its rotational motion. It is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

2. What is the relationship between moment of inertia and radius of gyration?

Radius of gyration is the distance from the axis of rotation to a point where the entire mass of an object can be considered to be concentrated. The moment of inertia is directly proportional to the square of the radius of gyration, meaning that as the radius of gyration increases, the moment of inertia also increases.

3. How does moment of inertia affect an object's rotational motion?

The moment of inertia determines how easily an object can be rotated. Objects with a larger moment of inertia require more force to rotate, while objects with a smaller moment of inertia are easier to rotate.

4. What factors affect the moment of inertia of an object?

The moment of inertia depends on the mass and shape of an object. Objects with a larger mass and a greater distance from the axis of rotation will have a larger moment of inertia. The distribution of mass within an object also affects its moment of inertia.

5. How is the moment of inertia used in real-world applications?

Moment of inertia is used in various engineering and physics applications, such as designing rotating machinery, analyzing the stability of structures, and calculating the motion of objects in space. It is also important in sports, for example, in determining the inertia of a baseball bat or a gymnast's body during a routine.

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