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Problem involving tension, circular motion, and equilibrium

  • Thread starter crazyog
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  • #1
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The Problem:
An air puck of mass (m1) is tied to a string and allowed to revolve in a circle of radius (R) on a frictionless horizontal table. The other end of the string passes througth a hole in the center of the table, and a counterwieght of mass (m2) is tied to it. The suspended object remains in equilibrium while the puck on teh tabletop revolves.
What are (a) the tension in the string, (b) the radial force acting on the puck,
(c) the speed of the puck?

I believe relevant equations include
Tension=(mass)(gravity)
Centripal force= (m1)(v^2)/(r)
T= ((2*m1*m2)/(m1+m2))(g)

Ok for part a)
I understand tension would be T= m2*g for the counterweight. Is there a tension in the other part of the string? (I'm not sure if I fully understand the concept of "tension")
or would I use an equation I found it my book when I was fidning the tension of similar to the Atwood Machine problem - T= ((2*m1*m2)/(m1+m2))(g)

b) I believe this has to do with centripal force. (Is that what they mean by "radial force"?)
Is this the opposite of tension?

c)
I would set (m1)(v^2)/r=m2*g
and I get v=sqrt((m2*g*r)/(m1))
However I was thinking and I'm not sure if the bottom part would be (m1+m2)...

I attached the image in a document for a visual representation.

Any help at all would be appreciated! Thank you!!!
 

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Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
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Hi crazyog,

Your attachment isn't yet ready for viewing, but I think I understand the problem.

The Problem:
An air puck of mass (m1) is tied to a string and allowed to revolve in a circle of radius (R) on a frictionless horizontal table. The other end of the string passes througth a hole in the center of the table, and a counterwieght of mass (m2) is tied to it. The suspended object remains in equilibrium while the puck on teh tabletop revolves.
What are (a) the tension in the string, (b) the radial force acting on the puck,
(c) the speed of the puck?

I believe relevant equations include
Tension=(mass)(gravity)
Centripal force= (m1)(v^2)/(r)
T= ((2*m1*m2)/(m1+m2))(g)

Ok for part a)
I understand tension would be T= m2*g for the counterweight. Is there a tension in the other part of the string? (I'm not sure if I fully understand the concept of "tension")
or would I use an equation I found it my book when I was fidning the tension of similar to the Atwood Machine problem - T= ((2*m1*m2)/(m1+m2))(g)
This case will be different from the Atwood machine, because both masses are accelerating in the Atwood machine. In this problem, there is no acceleration of the hanging mass, so it's total force must cancel, and your initial answer looks correct.


b) I believe this has to do with centripal force. (Is that what they mean by "radial force"?)
Is this the opposite of tension?
The radial force is the total force acting on the moving mass toward the center of its circular path.


c)
I would set (m1)(v^2)/r=m2*g
and I get v=sqrt((m2*g*r)/(m1))
This work looks correct to me.
 

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