# Angular acceleration in moment of inertia

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1. Dec 15, 2015

### dumbdumNotSmart

I'm having a hard time undertanding a concept of moment of inertia and Angular acceleration.
1. The problem statement, all variables and given/known data

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

2. Relevant equations
Sum of Moment of inertias on Cylinder = MR2 /2 . α
α.R=a
F=ma
μE =.1
3. 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.

2. Dec 15, 2015

### haruspex

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.
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.

3. Dec 15, 2015

### dumbdumNotSmart

Excuse me?

4. Dec 15, 2015

### dumbdumNotSmart

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