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rock.freak667

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**[SOLVED] Moment of inertia of a sign consisting of ring and rectangle**

## Homework Statement

A sign is formed from two uniform discs,each of mass 0.25kg and radius 0.2m,rigidly,fixed to a uniform rectangular lamina ABCD at A and D. This dis attached at A has diameter AE and BAE is a straight line. This disc at D has diameter DF and CDF is a straight line. This discs are in the same plane as ABCD. The mass of the rectangular lamina i M kg and AB=0.6m,BC-1.2m. The sign is hung from smooth supports at E and F with AD horizontal. Show that the moment of inertia of the sign about an axis through F perpendicular to ABCD is (0.39+M)kgm[itex]^2[/itex].

The Support at E breaks and the sign rotates freely under gravity, in a vertical plane, about the axis through F. Assuming that AD becomes vertical in the subsequent motion when the sign in rotating with angular speed [itex]\omega rad s^{-1}[/itex]. show that,

[tex]\omega^2=\frac{2(2-M)}{0.39+M}[/tex]

What happens if M>2?

## Homework Equations

Don't know how to prove these moments of intertia and would like to know how, since by using the def'n [itex]I=\int r^2 dm[/itex] I can't get the formulae.

[itex]I_{disc}=\frac{1}{2}mr^2[/itex]

[itex]I_{rectangle}=M \frac{a^2+b^2}{3}[/itex] through a corner

[itex]I=I_{centre}+md^2[/itex] i.e. parallel axis theorem(PAT)

[itex]I_z=I_x+I_y[/itex] i.e. perpendicular axis theorem

## The Attempt at a Solution

So I have the MOI for a disc about the centre,so I use the PAT to find the MOI about F

which is [itex]\frac{1}{2}mr^2+mr^2=\frac{3}{2}mr^2[/itex]

which is

[itex]I_{disc 1}=\frac{3}{2}*0.25*(0.2)^2=0.015[/itex]

[itex]I_{rectangle}=M(\frac{0.6^2+1.2^2}{3})=0.6M[/itex]

[itex]I_{disc 2}=\frac{3}{2}*0.25*(1.2^2)=0.54[/itex]

Not sure for disc 2, since I put the r=1.2 since the distance of E from F is 1.2m.

So need some help for the first part.