Homework Help: Problem involving tension, circular motion and equilibrium

1. May 29, 2006

walker

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 through a hole in the center of the table, and a counterweight of mass m2 is tied to it (see the figure below). The suspended object remains in equilibrium while the puck on the tabletop revolves.

Find the tension T on the string.
Find the radial force Fc acting on the puck
Find speed v of the puck

What i've come up with is

T = m2g - m1(v^2/R)
Fc = m1(v^2/R)
v = sqrt[(Fc*R)/m1]

For some reason i'm finding the question to be too ambigious and i am second guessing what i am doing. Also I have nothing to really go by in the way of lecture notes or example textbook problems that can aid me with this question. if someone could shed some light that would be awesome.

ive included a diagram and drawn on it what i think the forces are.

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Last edited: May 29, 2006
2. May 30, 2006

Pyrrhus

Remember the tension along a masless string is the same, and that the radial force or central force is just a role given to forces components. In this case the tension is the radial force according to description. I haven't seen the picture yet.

3. May 30, 2006

arunbg

It is always useful to consider the diff. forces acting on each mass independently.
The only force acting on the rotating mass is T, the tension in the string (normal and gravitational forces cancel), which provides the centripietal force required.
Both gravitational force and tension act on the hanging mass, keeping it in equilibrium.
Can you form the necessary equations now ?

4. May 30, 2006

walker

so then T is just m2g? that makes sense i guess. im having trouble actually understanding the concept of tension in general, if anyone has any suggested websites i can read that would help me out that would be great too. thanks for the help.

5. May 30, 2006

Hootenanny

Staff Emeritus
Yes, the tension in the string is simply m2g. Usually, the tension in a light inextensible string is constant throughout its length.

Consider the forces acting on the hanging mass. What would happen if the tension was greater than m2g? Does that make sense now?

~H

6. May 30, 2006

walker

yeah it makes sense, im just having a hard time visualizing a freebody diagram with tension...need more practice i guess. my prof this year seems to rush her answers and doesnt provide much help 8/ so thanks for the help.

7. May 30, 2006

Hootenanny

Staff Emeritus
Well, I would recommend more practise, soon you'll be able to solve them without thinking. You could always ask you prof for a more complete solution, but we'll always be happy to help you out here.

Are you ok for solving the rest of the problem now?

~H

8. Jul 12, 2006

thiotimoline

The correct ans is, the same tension is felt throughout the string, so the weight of m2 is the same as the tension maintaining m1's revolution around the table. So, tension = weight of m2 = radial force. To find v, simply equate radial force with weight of m2.