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## Homework Statement

We have a closed system above. M

_{1}is a cylinder of 2 Kilograms, moment of inertia of a cylinder ( MR

^{2}/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 M

_{1}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 = MR

^{2}/2 . α

α.R=a

F=ma

μ

_{E}=.1

## The Attempt at a Solution

let a be acceleration of the string (mass m) and a

_{cm}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=Ma

_{cm}(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=(MR

^{2}/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.