Angular acceleration in moment of inertia

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The discussion revolves around understanding the relationship between moment of inertia and angular acceleration in a system involving a cylinder and a hanging mass. The problem involves determining the maximum mass that allows the cylinder to roll without sliding, factoring in static friction and unknown gravity. A key point of confusion is the application of the moment of inertia equation and the distinction between the acceleration of the center of mass and the edge of the cylinder. Participants clarify that the teacher's approach effectively reverses the typical application of acceleration definitions. Ultimately, the original poster realizes their misunderstanding while explaining their reasoning.
dumbdumNotSmart
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I'm having a hard time undertanding a concept of moment of inertia and Angular acceleration.

Homework Statement


Capture.JPG

We have a closed system above. M1 is a cylinder of 2 Kilograms, moment of inertia of a cylinder ( MR2 /2 ) with a string tightly rolled around it. This string connects to a free hanging mass (m kilograms) from a ideal massless pulley like in the picture. Gravity is present but has uknown value. The surface under M1 has a static friction coefficient of .1 . When the System is released from Rest the mass descends. The Radius of the cylinder (R) is unknown.

Determine the maximum value of m so that the cylinder rolls without sliding

Homework Equations


Sum of Moment of inertias on Cylinder = MR2 /2 . α
α.R=a
F=ma
μE =.1

The Attempt at a Solution


let a be acceleration of the string (mass m) and acm be the acceleration of M's center of mass. Ψ will be the force of friction.
mg-T=ma
m= T/(g-a)
T+Ψ=Macm (Ψ and T vectors have same direction.

My problem with this case stems from the moment of inertia equation.
We have the following:
TR -ΨR=(MR2 /2)α

Simplifying we get 2T -2Ψ=MRα
In the solution given to us by the teacher, multiplying the angular acceleration by the radius would give us the acceleration of the center of mass of the cylinder. I feel something is wrong since usually the equation gives the acceleration for the edge of the cylinder which is not the same as the center.
 
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It's not a closed system. The frictional force is external, as are the forces on the string from the pulley, but they don't affect your attempted solution.
dumbdumNotSmart said:
usually the equation gives the acceleration for the edge of the cylinder which is not the same as the center.
That would be the acceleration of a perimeter point relative to the centre. Here your teacher is simply reversing that, using it for the acceleration of the centre relative to the edge.
 
haruspex said:
That would be the acceleration of a perimeter point relative to the centre. Here your teacher is simply reversing that, using it for the acceleration of the centre relative to the edge.

Excuse me?
 
While explaining my reasoning in an upcoming reply I found what I was doing wrong.
 
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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