Moment of inertia of a sign consisting of ring and rectangle

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

1. The problem statement, all variables and given/known data
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?

2. Relevant 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

3. 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.

 

http://img362.imageshack.us/img362/9395/qwks4.jpg [Broken]

here is a picture of the sign

 

I think the rectangle would be the the sum of the MOI, for a rectangle about its corner and for a disc since it is connected.

So for the rectangle

[tex]I=\frac{M}{3}(a^2+b^2) + \frac{3}{2}mr^2[/tex]

For disc AE it would be the moment of inertia of disc AE about E([itex]\frac{3}{2}mr^2[/itex]) + [itex]md^2[/itex] where d=1.2

right?

 

I think I understand, so that the only reason I have the correct equation is because the two discs are similar. Meaning that if one had a larger radius, the equation wouldn't work.
Also, I misunderstood the parallel axis theorem in that I thought that you could find the MOI of an object about any axis once you know its MOI about any other axis that is parallel to the required axis.

ok, according to my list of formula,the MOI about the centre is [itex]\frac{M}{12}(a^2+b^2)[/itex]

So I would need to find the distance of the centre of mass of the rectangle to point F. Which is [itex]\sqrt{(0.6)^2+(0.7)^2}[/itex] where the 0.6 is half of AD and the 0.7 is the half of DC+Diameter of DF.

and then say that the MOI of ABCD about F is [itex]I=\frac{M}{12}(a^2+b^2) +Md^2[/itex] where [itex]d^2=(0.6)^2+(0.7)^2[/itex]

 

But I don't get how the distance is .2^2+1.2^2, if the distance EF=1.2m

Wow,never thought of it like that. Makes more sense now!

Yep yep

 

So second part. Here is what I tried.

Taking the horizontal line through EF to have 0 potential energy.

When AD is horizontal i.e. when sign in equilibrium.
Total GPE,[itex]E_1=E_{AE}++E_{ABCD}+E_{FD}[/itex]=(0.25)(10)(0.2)+(M)(10)(0.7)+(0.25)(10)(0.2)=1+7M

When AD is vertical.

Total GPE,[itex]E_2=E_{AE}++E_{ABCD}+E_{FD}[/itex]=0+(0.6*10*M)+(1.2*10*0.25)=6M+3

and the change in GPE=6M+3-(1+7M)=2-M.

and that change in GPE is converted into rotational ke.

so that

[itex]\frac{1}{2}I \omega^2=2-M[/itex]

Making

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


How interesting I actually got that out. I thought my energy considerations were going off.

Is there any other way to get that for [itex]\omega^2[/itex] without considering energy?

and for the last part now.

If M>2, all I can really say is that, [itex]\omega^2=-ve[/itex] which is impossible since anything squared is positive and ang. speed can't be complex. So it just wouldn't rotate,or the sign might just break and fall down or some odd thing like that?