Angular acceleration in moment of inertia

  • #1
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.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,267
6,320
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.
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.
 
  • #3
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?
 
  • #4
While explaining my reasoning in an upcoming reply I found what I was doing wrong.
 

Related Threads on Angular acceleration in moment of inertia

Replies
3
Views
2K
Replies
2
Views
3K
Replies
7
Views
5K
Replies
2
Views
906
Replies
8
Views
6K
Replies
21
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
8
Views
4K
Top