Angular acceleration in moment of inertia

In summary: When I multiplied the angular acceleration by the radius I was actually trying to calculate the acceleration of the center of mass of the cylinder relative to the edge, which is not the same as the acceleration of the center of mass relative to the origin. Thanks for catching that!
  • #1
dumbdumNotSmart
41
3
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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  • #2
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.
 
  • #3
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?
 
  • #4
While explaining my reasoning in an upcoming reply I found what I was doing wrong.
 

1. What is angular acceleration?

Angular acceleration is a measure of how quickly the angular velocity of an object changes over time. It is defined as the rate of change of angular velocity, and is measured in radians per second squared (rad/s²).

2. How is angular acceleration related to moment of inertia?

Angular acceleration is directly proportional to the moment of inertia of an object. Moment of inertia is a measure of an object's resistance to rotational motion, and the larger the moment of inertia, the smaller the angular acceleration for a given torque applied.

3. How can angular acceleration be calculated?

Angular acceleration can be calculated by dividing the torque applied to an object by its moment of inertia. The formula for angular acceleration is α = τ / I, where α is the angular acceleration, τ is the torque, and I is the moment of inertia.

4. What are the units of angular acceleration?

The units of angular acceleration are radians per second squared (rad/s²). This unit is equivalent to the unit of acceleration in linear motion, meters per second squared (m/s²), but in angular motion, it measures the change in angular velocity instead of linear velocity.

5. How does angular acceleration affect the motion of an object?

Angular acceleration can affect the motion of an object in several ways. It can cause an object to speed up or slow down its rotational motion, change its direction of rotation, or cause it to remain in a constant rotational motion. The direction of the angular acceleration is always perpendicular to the plane of rotation and follows the right-hand rule.

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