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Moment of inertia of a sign consisting of ring and rectangle

  • #1
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.
 

Answers and Replies

  • #2
alphysicist
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Hi rock.freak667,

I don't understand the setup; could you post a picture somewhere with all of these points labelled?

However, I believe you did make a mistake in applying the parallel axis theorem. The theorem applied to the case of a solid disk is:

[tex]
I = \frac{1}{2}mr^2 + m d^2
[/tex]

where r is the radius of the disk and d is the distance between the center of mass axis the new axis. So this would not be equal to 3/2 mr^2 unless r and d are equal, which is when the axis is on the edge of the disk.
 
  • #3
rock.freak667
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Last edited by a moderator:
  • #4
alphysicist
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So the moment of inertia of disk DF would be 3/2 mr^2, since the axis is on the edge of the disk (point F). What are the moments of inertia of the rectangle and the other disk about an axis at point F?
 
  • #5
rock.freak667
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So the moment of inertia of disk DF would be 3/2 mr^2, since the axis is on the edge of the disk (point F). What are the moments of inertia of the rectangle and the other disk about an axis at point F?
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?
 
  • #6
alphysicist
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The one we know is disk FD. It's MOI is

I = (1/2) m r^2 + m d^2

What is d? It is the distance from the center of the disk to the axis; in this case it is equal to the radius, so we get I=(3/2) mr^2. So that one is right.

Now for the other disk. The thing is, you got the right answer, but based on your equation I think you were probably thinking about it incorrectly. The parallel axis theorem is used to to move from an axis at the center of mass to some other parallel axis. But your equation makes it look like you're moving the axis from point E to point F. The reason it works in this particular case is because the direction from the center to point E happens to be perpendicular to the direction from point E to point F. But you wouldn't be able to do this, for example to go from point E to point B.

Sorry about the wordy paragraph; does that make sense?

Now do the same thing for the rectangle; find it's moment of inertia at it's center, and then find the distance from the center to point F to use the parallel axis theorem. Your current formula for the rectangle calculates the MOI for the corner, but that is not where the axis is.
 
  • #7
rock.freak667
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Now for the other disk. The thing is, you got the right answer, but based on your equation I think you were probably thinking about it incorrectly. The parallel axis theorem is used to to move from an axis at the center of mass to some other parallel axis. But your equation makes it look like you're moving the axis from point E to point F. The reason it works in this particular case is because the direction from the center to point E happens to be perpendicular to the direction from point E to point F. But you wouldn't be able to do this, for example to go from point E to point B.

Sorry about the wordy paragraph; does that make sense?
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.

Now do the same thing for the rectangle; find it's moment of inertia at it's center, and then find the distance from the center to point F to use the parallel axis theorem. Your current formula for the rectangle calculates the MOI for the corner, but that is not where the axis is.
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]
 
Last edited:
  • #8
alphysicist
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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.
Well, that's related to the reason; here is what happened. Applying the parallel axis theorem:

[tex]
I = \frac{1}{2}mr^2 + md^2
[/tex]

and [itex] d^2 = .2^2 + 1.2^2[/itex]. Plugging those in:

[tex]
I = \frac{1}{2}mr^2 + m(.2^2 + 1.2^2)
[/tex]

and what you effectively did is, since now [itex]m(.2^2 + 1.2^2)[/itex] can easily be separated into two terms, combine the first one with the center of mass term. So there were a lot of coincidences that made it work out right.



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.
I remember believing that when I was learning this, but it's not true. Here are two ways you can see its definitely not true: You know the MOI about point E is 3/2 mr^2; could you use the PAT to move the axis back to the center of mass? If you could, then you could "prove" that the center of mass MOI is 5/2 mr^2.

Or, you could just keep going back and forth between two points (say, E and A) and make the MOI as large as you want.


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]
Sounds good; do you get the right answer?
 
  • #9
rock.freak667
Homework Helper
6,230
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Well, that's related to the reason; here is what happened. Applying the parallel axis theorem:

[tex]
I = \frac{1}{2}mr^2 + md^2
[/tex]

and [itex] d^2 = .2^2 + 1.2^2[/itex]. Plugging those in:

[tex]
I = \frac{1}{2}mr^2 + m(.2^2 + 1.2^2)
[/tex]

and what you effectively did is, since now [itex]m(.2^2 + 1.2^2)[/itex] can easily be separated into two terms, combine the first one with the center of mass term. So there were a lot of coincidences that made it work out right.
But I don't get how the distance is .2^2+1.2^2, if the distance EF=1.2m




I remember believing that when I was learning this, but it's not true. Here are two ways you can see its definitely not true: You know the MOI about point E is 3/2 mr^2; could you use the PAT to move the axis back to the center of mass? If you could, then you could "prove" that the center of mass MOI is 5/2 mr^2.

Or, you could just keep going back and forth between two points (say, E and A) and make the MOI as large as you want.

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


Sounds good; do you get the right answer?
Yep yep :smile:
 
  • #10
rock.freak667
Homework Helper
6,230
31
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?
 
  • #11
alphysicist
Homework Helper
2,238
1
But I don't get how the distance is .2^2+1.2^2, if the distance EF=1.2m
That's the distance because we want to use the parallel axis theorem to go from the center of the disk to point F. So it's 0.2 up and 1.2 over to the right.




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?
It means one of our assumptions was wrong. In the problem, they asked what was the angular speed after the sign had rotated enough that AD was vertical. But what if AD is never vertical?

If M < 2, the sign swings past the point where AD is vertical, and so at that point it still has some speed. When M = 2, it just barely makes it to that point. And when the mass is even larger, it just does not swing that far.


It's the same kind of idea as in this problem: If you throw a ball up at speed [itex]v_0[/itex], what is it's speed at a height of 5m? Well, we can use energy (for example) and we find that:

[tex]
v=\sqrt{v_0^2 - 2 g h}
[/tex]

[tex]
v=\sqrt{v_0^2 - 98}
[/tex]

But if [itex]v_0 [/itex] is less than about 9.899, the part under the square root is negative, and that occurs because the assumption that the ball reaches 5m is wrong.
 

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