• Support PF! Buy your school textbooks, materials and every day products via PF Here!

The motion of 2 objects in contact

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 dont know where im wrong. Any Help?
 

Attachments

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
31,387
4,596
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 dont know where im wrong. Any Help?
I get the given answer by assuming the same coefficient of friction applies between the two masses.
 
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)
 
Last edited:

Want to reply to this thread?

"The motion of 2 objects in contact" You must log in or register to reply here.

Related Threads for: The motion of 2 objects in contact

Replies
2
Views
1K
Replies
1
Views
597
  • Posted
Replies
6
Views
2K
Replies
1
Views
2K
Replies
5
Views
767

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top