The motion of 2 objects in contact

[PhysicS FAN]

Homework Statement

An object of mass m=2kg rests in a horizontal level with a coeffcient of friction μ=0.5. A cylinder with radius R is placed near the first obect. The mass of cylinder is m=2kg and we know that Ι=1/2mR^2. At a some time instant we start pushing the system with a force of 23N, in such a way that the force passes through the center of the cylinder. What is the acceleration of the system?

Correct Answer: a=2m/s^2

Relevant Equations

ΣF=ma, Στ=Ia(angular), T=μmg

For the non-circular object of mass m: From Newtons second law we get that F-N-T=ma where N is the force that the cylinder acts on the object. Replacing numbers: 13-N=2a.
For the cylinder: From Στ=Τa(ang) we get that T'= 1/2ma or T'=a. Where T' is the friction that acts on the cylinder and we assume that its static which means a=a(ang)R.
From ΣF=ma we get N-T'=ma where N' is the reaction of N. N=3a.
Replacing: 13-3a=2a or a=13/5. and if we do the math for the firction we see that the friction for the cylinder is indeed static.
I don't know where I am wrong. Any Help?

 

[haruspex]

Homework Statement: An object of mass m=2kg rests in a horizontal level with a coeffcient of friction μ=0.5. A cylinder with radius R is placed near the first obect. The mass of cylinder is m=2kg and we know that Ι=1/2mR^2. At a some time instant we start pushing the system with a force of 23N, in such a way that the force passes through the center of the cylinder. What is the acceleration of the system?

Correct Answer: a=2m/s^2
Homework Equations: ΣF=ma, Στ=Ia(angular), T=μmg

For the non-circular object of mass m: From Newtons second law we get that F-N-T=ma where N is the force that the cylinder acts on the object. Replacing numbers: 13-N=2a.
For the cylinder: From Στ=Τa(ang) we get that T'= 1/2ma or T'=a. Where T' is the friction that acts on the cylinder and we assume that its static which means a=a(ang)R.
From ΣF=ma we get N-T'=ma where N' is the reaction of N. N=3a.
Replacing: 13-3a=2a or a=13/5. and if we do the math for the firction we see that the friction for the cylinder is indeed static.
I don't know where I am wrong. Any Help?

I get the given answer by assuming the same coefficient of friction applies between the two masses.

 

[PhysicS FAN]

I guess you are right but still if we assume that there is friction between the two masses then the friction will only affect the circular motion of the cylinder. So it would be written T'-μΝ= 1/2ma where N the force between the objects and its horizontal. So T'- 0.5(13-2a)=a and so T' - o.5*13=0 or T=o.5*13
and then because we know that 13-T'=4a we get 13-1/2*13 =4a or 4a=1/2*13 and so is not 2
(We know that 13-N=2a and N-T'=ma so 13-T'=4a)