Pulley problem with moment of inetia

[Iansno1]

Problem

As shown below, Block A has mass 'M' and rests on a surface with a coefficient of kinetic friction 'U_k' .The cord attached to A passes over a pulley at C and is attached to a block B of mass '2M'. If B is released calculate the acceleration of A.

Assume the cord does not slip over the pulley. The Pulley can be approximated as a thin disk of radius 'r' and mass 'M/4' . Neglect the mass of the cord
Relevant equations

moment of inertia of the disk: I=0.5*M*r²
Not sure where to start

 

[ehild]

with free body diagrams?

ehild

 

[Iansno1]

How does the Moment of inertia come into the equation? I can do these problems when the pulley is assumed to have no mass and no friction.

 

[gneill]

How does the Moment of inertia come into the equation? I can do these problems when the pulley is assumed to have no mass and no friction.

Moment of inertia is the analog of mass for rotational motion. The moment of inertia resists angular acceleration in the same fashion as mass resists linear acceleration.

In this problem the pulley adds resistance to the acceleration of the overall system because the cord does not slip, and so must cause angular acceleration of the pulley. The angular acceleration is related to the linear acceleration (how?).

 

[Nickie]

with this problem, but I know that the key to solving it will be to use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration (F=ma). In this case, we have two objects, A and B, connected by a cord and a pulley. We also have the coefficient of kinetic friction, which will affect the motion of Block A.

To start, we can draw a free body diagram for Block A, showing all the forces acting on it. We have the force of gravity pulling it down (mg), the normal force from the surface (N), and the force of friction (Ff) opposing its motion. We can also include the tension in the cord (T) pulling Block A to the right.

Next, we can consider the forces acting on Block B. We have the force of gravity pulling it down (2mg), and the tension in the cord (T) pulling it to the left.

Using Newton's second law, we can set up the following equations for each block:

Block A: ΣF=ma
mg-Ff-T=ma

Block B: ΣF=ma
2mg-T=2ma

We also know that the force of friction is equal to the coefficient of kinetic friction times the normal force: Ff=Uk*N. We can use this to substitute into our equations and solve for the acceleration of Block A:

mg-Uk*N-T=ma
2mg-T=2ma

mg-Uk*(mg-N)-T=ma
2mg-T=2ma

Next, we can use the fact that the cord does not slip over the pulley to say that the acceleration of Block A is equal to the acceleration of Block B divided by the radius of the pulley (a=rα). We can also use the moment of inertia equation for the pulley, I=0.5*M*r^2, to substitute into our equations:

mg-Uk*(mg-N)-T=mrα
2mg-T=0.5*M*r^2*α

Solving these equations simultaneously, we can find the acceleration of Block A, which is also the acceleration of Block B:

a=(mg-Uk*(mg-N)-T)/m
a=(2mg-T)/(0.5*M*r^2)

We can then substitute the known values and solve for the acceleration:

a=(Mg-0.