Linear Acceleration of a Rigid Object with Attached Disks and Tension on Cord

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The discussion focuses on calculating the linear acceleration of a rigid object with two attached disks and a tension force. The initial attempt incorrectly assumes that the moment of inertia is the same relative to the edge of the disk as it is to its center. Participants clarify that the instantaneous point of rotation is at the contact point with the ground, which affects the moment of inertia calculation. The correct formula for linear acceleration includes a factor of 3 in the denominator, indicating a misunderstanding in the initial approach. The conversation highlights the importance of accurately determining the point of rotation in rigid body dynamics.
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Homework Statement



The object of the figure below has 2 disks attached on both sides through and axel of negligible mass. Both disks of mass M. I need to find the linear acceleration of this object knowing there exists a tension T on the cord.
Sin título.jpg


Homework Equations

:[/B]

T=αI

The Attempt at a Solution

:[/B]

First we know that :
τ=αI ⇒
T(R-r)=I(a/R)⇒
T(R-r)=(2(MR^2)/2)(a/R)⇒
T(R-r)=MRa⇒
a=(T(R-r))/(MR)

But the solution of the exercise is : a=(T(R-r))/(3MR)

What am I doing wrong?
 

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Nicolas Gallardo said:
I(a/R)⇒
T(R-r)=(2(MR^2)/2)(a/R)
You are assuming that the moment of inertia relative to a point on the edge of a disk is the same as that relative to its centre. This is not true.
 
Orodruin said:
You are assuming that the moment of inertia relative to a point on the edge of a disk is the same as that relative to its centre. This is not true.
But how would the moment of inertia change? Both of the disks are rotating with respect to the center axis...
 
Nicolas Gallardo said:
Both of the disks are rotating with respect to the center axis...
No they are not, the instantaneous point of rotation is the contact point with the ground ... and you certainly are not computing the torsion relative to the central axis.
 
Orodruin said:
No they are not, the instantaneous point of rotation is the contact point with the ground ... and you certainly are not computing the torsion relative to the central axis.
Yes! You are right! How could I be so dumb. I am sorry I am new with rigids body dinamics. Thank you for your help.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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