Calculating Rotational Inertia and Speed of a Square Plate

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Homework Help Overview

The discussion revolves around calculating the rotational inertia and speed of a uniform square plate pivoted at one vertex. The plate has a mass of 0.8 kg and a side length of 0.8 m, and it swings downward under the influence of gravity, with resistive forces neglected. Participants are tasked with finding the rotational inertia relative to the pivot point, the angular speed when a specific side is horizontal, and the linear speeds at different positions during the swing.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of the parallel axis theorem to calculate the moment of inertia and question the distances involved in the calculations. There are attempts to clarify the height changes of the center of mass during the swing and how these affect the energy conservation equations.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem setup and calculations. Some guidance has been provided regarding the use of energy conservation principles, and there is an ongoing examination of the values used for height and inertia. Participants are encouraged to clarify their reasoning and assumptions as they progress through the problem.

Contextual Notes

There is some confusion regarding the dimensions of the square plate, specifically whether the side length is 0.8 m or 1.2 m, which impacts the calculations. Participants are also discussing the correct interpretation of the height changes as the plate swings, which is critical for applying energy conservation correctly.

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Homework Statement



A uniform square plate ABCD has mass 0.8 kg and side length .8 m. The square is pivoted at vertex D and initially held at rest so that sides AB and CD are horizontal (see the diagram). After it is released, the plate swings downward, rotating about the pivot point. Resistive force can be neglected. The acceleration due to gravity is g = 9.8 m/s2. The rotational inertia of a square plate of side d relative to the axis perpendicular to the plate and passing through the center of mass is md^2/6.

Find the rotational inertia of the plate relative to the axis of rotation.
Find the angular speed of the plate at the moment when BD is horizontal.
Find the linear speed of B at the moment when BD is horizontal.
Find the linear speed of B at the moment when B is at the bottom position.

Homework Equations



The Attempt at a Solution


I used the parallel axis theorem to find the moment of inertia around the axis of rotation. I got
1/6(.8)(1.2)^2+.8(sqrt(2)/2)^2=.592 but that isn't correct.
 

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postfan said:
I used the parallel axis theorem to find the moment of inertia around the axis of rotation. I got
1/6(.8)(1.2)^2+.8(sqrt(2)/2)^2=.592 but that isn't correct.

What does the quantity √(2)/2 represent?
 
never mind
 
Last edited:
√(2)/2 represents the distance from point D to the center of mass.
 
postfan said:
√(2)/2 represents the distance from point D to the center of mass.

Shouldn't the 1.2 m come into play here?
 
Oh ok, that makes sense. I got .768, is that right?
 
I believe that's correct. (Add units of course.)
 
Ok, then how do I start solving part B?
 
Is anything conserved while the plate swings down?
 
  • #10
Energy,right?
 
  • #11
Right. Give it a shot.
 
  • #12
Ok I got 4.937. Is that right?
 
  • #13
I don't know, I haven't worked it out. Please show your work and we can see if you are setting it up correctly.
 
  • #14
m*g*h=.5*I*w^2

m*g*h=.5*.768*m*r^2*(v/r)^2

g*h=.384*v^2

9.8*.6=.384*v^2

v=3.91 (not 4.937 due to arithmetic error)
 
  • #15
Your set up looks good!

But note that in the second question, you are asked to find the angular speed rather than the linear speed.

Also, I don't agree with the value that you used for h in the calculation. How far does the center of mass of the square move downward for this question?
 
  • #16
I still think that its .6. Can you tell me why I'm wrong?
 
  • #17
I'm a little confused about the value of the side length of the square plate. I thought it was 1.2 m, but now when I look at your original post, it appears to be 0.8 m.

But either way, h would not be 0.6 m for the second question. Did you draw a picture showing the orientation of the square at the moment of release and then another picture for when BD is horizontal? Be sure to mark the center of mass in each case.

Since a diagram wasn't included, I am interpreting the geometry as best I can from the wording of the problem.
 
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  • #18
OK, now I see the picture. I had misinterpreted it. I thought that initially point B was directly above D so that BD would be a side of the square. But, BD is a diagonal of the square. So, if the length of a side is 1.2 m, then I agree that h would be 0.6 m for the second question. Sorry about that.
 
  • #19
For parts 3 and 4 is h= to .8 and .8(2+sqrt(2)/2) respectively?
 
  • #20
h for part 3 should be the same as for part 2.

Can we clarify if the length of the side of the square is 1.2 m or is it 0.8 m?
 
  • #21
Length is .8.
 
  • #22
postfan said:
Length is .8.

Oh Oh. That's going to change the value of I that you calculated using 1.2 m:

postfan said:
I used the parallel axis theorem to find the moment of inertia around the axis of rotation. I got
1/6(.8)(1.2)^2+.8(sqrt(2)/2)^2=.592 but that isn't correct.
 
  • #23
I know, I already accounted for it.
 
  • #24
So does h= .8(2+sqrt(2)/2) in part 4?
 
  • #25
OK. I agree with your value of h for part 3, but not for part 4. Can you explain how you got your expression for part 4?
 
  • #26
postfan said:
So does h= .8(2+sqrt(2)/2) in part 4?

I don't agree with the 2 that is shown in red above.
 
  • #27
Is it .8(1+sqrt(2)/2)?
 
  • #28
OK, I must be getting tired. The value of h for parts 2 and 3 is not the length of the side of the square. So, h is not .8 for parts 2 and 3. It's a certain fraction of the length of the side.

For part 4, you still don't quite have the right answer.
 
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  • #29
I relooked at it and still got .8(1+sqrt(2)/2). Can you tell me why I am wrong?
 
  • #30
postfan said:
I relooked at it and still got .8(1+sqrt(2)/2). Can you tell me why I am wrong?

I don't know why you are wrong because I don't know the thought process that you used to get your answer. That's why I wanted you to explain how you got your answer. Your answer is pretty close to the correct answer. It might help if you answer the following:

(A) How far does the center of mass move downward in going from question 1 to question 2?

(B) How much additional distance does it go down in going from question 2 to question 4?

It looks to me that you probably have the correct answer to (B) but not the correct answer to (A).
 

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