Calculating Rotational Inertia and Speed of a Square Plate

[postfan]

From 1 to 2 I found that the center of mass move downs by .4.
From 2 to 3 I found that the center of mass move down by .4+ sqrt(2)/2*.8.

 

[TSny]

From 1 to 2 I found that the center of mass move downs by .4.
From 2 to 3 I found that the center of mass move down by .4+ sqrt(2)/2*.8.

I agree that h₁ = .4 for the first part. I don't agree that h₂ = .4+ sqrt(2)/2*.8 for the additional distance from question 2 to question 4.

 

[postfan]

Does h2 = .8+ sqrt(2)/2*.8

 

[TSny]

Does h2 = .8+ sqrt(2)/2*.8

No. See the attached diagram. As corner B moves from B' to B'', the center of mass moves downward a distance h₂. You can see that h₂ is half the diagonal of the square.

 

[postfan]

Ok so for part 4 h= sqrt(2)/2*.8, right?

 

[TSny]

Ok so for part 4 h= sqrt(2)/2*.8, right?

That represents the correct value of h₂. What is the total h in going from where the plate is released from rest to where B is at the bottom?

 

[postfan]

The total h is equal to .4+sqrt(2)/2*.8.

 

[TSny]

Yes, that's right.

 

[postfan]

Ok then we use the equation mgh=.5mv^2+.5Iw*2, substitute values convert angular velocity to linear velocity and then solve for v, right?

 

[TSny]

Yes. [EDIT: NO]

 

[TSny]

Sorry, it's late here and I'm half asleep! The total kinetic energy is just (1/2)Iw² (as you had in post #14). So,
mgh = (1/2)Iw².

 

[postfan]

Why is that so? The question is asking for the linear speed of B.

 

[TSny]

The plate is in pure rotation about D. So, as long as you use the moment of inertia about D, the kinetic energy can be expressed as just rotational KE about D.

Equivalently, you could express the kinetic energy as kinetic energy due to translation of the center of mass plus kinetic energy of rotation about the center of mass:

KE = (1/2)mV_cm² + (1/2)I_cmω². Here you would use the moment of inertia about the center of mass rather than about point D.

Either way will yield the same answer.

Once you find ω you can then find the linear speed of B.

 

[postfan]

I got the answer, but I am still confused about when/how I can just use (1/2)Iw2 instead of the full (1/2)mV^2 + (1/2)Iω^2. I know it is rather late, so would you rather want to talk about it tomorrow?

 

[TSny]

As I said before, you can use either KE = (1/2)I_Dω² or KE = (1/2)mV_cm² + (1/2)I_cmω². Both are correct.

In this problem, the first question was to find I_D. So, it will be easier to use KE = (1/2)I_Dω² when setting up mgh = KE. You can then solve for ω. Then use ω to find V_B.

It is late here (about 1 am) and I do need my beauty rest. So I will quit for now and check back tomorrow. :zzz:

 

[postfan]

Alright, thanks for all your help! You're awesome!

 

[postfan]

TSny,How did you draw the diagrams?

 

[TSny]

I used Microsoft Paint.

 

[postfan]

How did you get the square to rotate?

 

[TSny]

In some cases I just redrew it in a rotated position. Sometimes I copy a drawing that I drew in Paint into another program that will allow me to easily rotate the figure and then copy the rotated figure back into Paint.

For example, Powerpoint allows you to easily rotate a figure any amount you want.

 

[postfan]

Oh, I see.