Calculating Work on a Frictionless Horizontal Table

In summary, the work done by the person pulling the string was W = \int F . dr. The force required to do this work was F(r) = \frac{mv^2 } {r} = \frac{m \omega_0^2 r^2}{r}.
  • #1
roeb
107
1

Homework Statement


A particle of mass m is moving on a frictionless horizontal table, attached to a massless string, the other end of which passes through a hole in the table. It was rotating with angular velocity [tex]\omega_0[/tex], at a distance [tex]r_0[/tex] from the hole. Assuming that I pull the string so slowly that we can always approximate the path of the particle at any time by a circle of slowly shrinking radius, calculate the work I did while pulling the string. Show that the work-energy theorem is satisfied in this case.


Homework Equations


[tex]W = \int F . dr[/tex]
[/tex]\delta W = KE_f - KE_i[/tex]


The Attempt at a Solution


Here is how I initially attempted the problem:
[tex]KE_i = \frac{1}{2}*m*(r_0 \omega_0)^2 [/tex]
[tex]KE_f = \frac{1}{2}*m*(r \omega_f)^2[/tex]
And by taking the difference I was hoping that I would get the work done. Unfortunately it turns out that I need to do something a bit more complicated.

I know [tex]W = \int F . dr[/tex] but I'm not really sure what to do in order to apply it to this situation.

Anyone have any hints?
 
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  • #2
Find the force required as a function of radius. Hint: What's conserved? Apply Newton's 2nd law.
 
  • #3
For this situation, I believe energy conservation is important to consider (or perhaps conservation of angular momentum?)

I'm not sure if this is what you were referring to, but centripetal force comes to mind when trying to find a force as a function of radius.
[tex]F(r) = \frac{ mv^2 } {r} = \frac{m \omega_0^2 r^2}{r} [/tex]
[tex]\int_{r_0}^{r} F(r) . dr = 1/2 m \omega_0^2 (r^2 - r_0^2)[/tex]
Unfortunately this isn't correct, hmm. In this case I assumed that [tex]\omega_0[/tex] was a constant for both radii. Physically, this seems like a bad assumption, so I will have to try to think of something else.

Am I correct that I should be using centripetal force?
 
Last edited:
  • #4
Yes, you need to use centripetal force. (That tells you how hard you must pull on the string.)

No, ω is definitely not constant as the radius changes. But another quantity is constant, which will allow you to figure out the force as a function of radius. (And it's not energy!)
 
  • #5
Thank you Doc Al, I was able to figure it out.
 

Related to Calculating Work on a Frictionless Horizontal Table

1. What is the formula for calculating work on a frictionless horizontal table?

The formula for calculating work on a frictionless horizontal table is W = F x d, where W is work, F is the applied force, and d is the displacement.

2. How do you determine the direction of work on a frictionless horizontal table?

The direction of work on a frictionless horizontal table is determined by the direction of the applied force. Work is positive when the force and displacement are in the same direction, and negative when they are in opposite directions.

3. How is kinetic friction taken into account when calculating work on a frictionless horizontal table?

Kinetic friction is not taken into account when calculating work on a frictionless horizontal table. In a frictionless system, there is no force acting against the motion, so there is no work done to overcome friction.

4. Can work be negative on a frictionless horizontal table?

Yes, work can be negative on a frictionless horizontal table if the applied force and displacement are in opposite directions. This indicates that the object is losing energy, rather than gaining it.

5. How does the angle of the applied force affect the calculation of work on a frictionless horizontal table?

The angle of the applied force does not affect the calculation of work on a frictionless horizontal table. Only the magnitude and direction of the force and displacement are taken into account in the formula for work.

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