# Universal Gravitation Ice Rink Problem!

1. Dec 4, 2007

### physicsbhelp

[SOLVED] Universal Gravitation Ice Rink Problem! :)

never mind

Last edited: Dec 4, 2007
2. Dec 4, 2007

### Staff: Mentor

maybe Ug= Gm1m2 / r2 would give a very small number, even though r is small.

The skater rotate in a mutual circle, so the centripetal force of one equals to force of the other.

What is the period of rotation, in which each skater turns 2$\pi$ rad?

Each skater represents a mass 60 kg at a distance of 0.8 m from the axis of rotation.

3. Dec 4, 2007

### t-money

Choose a skater, from their point of view the other skater is moving in a circle around them, the arms are applying a centripetal force (tension).

4. Dec 4, 2007

It seems as though you are are just picking random equations without any rhyme or reason. Why do you think this has something to do with universal gravitation?

The first thing you neee to ask yourself is "what is the question asking me?" In this case it is asking for a force.

Secondly ask "have I been given all of the information necessary to calculate this force?" In this case no.

Third "Do I have enough information to derive or generate the information needed to calculate the force?" In this case...definately.

Now what do you know about force in general?

Casey

5. Dec 4, 2007

Well...possibly. I myself have never used that equation. But no matter, you are missing my point. What do you know about what ALL forces equal?

Hint: Newton's 2nd

Casey

I think you are correct here.

6. Dec 4, 2007

Now we are onto something. You know mass, now what kind of acceleration is it?
Hint: It always points towards the center
Casey

Last edited: Dec 4, 2007
7. Dec 4, 2007

The mass is the combined mass. I am not sure why you insist that gravity has anything to do with this.

Picture two people holding hands a spinning in a circle. Now the circle being made is on the horizontal! Where does gravity come in?

Now what kind of acceleration always points to the center?

You already told me we are looking for CENTRIPETAL force right...

8. Dec 4, 2007

Yes!!! Now how do you find centripetal acceleration? What is the formula?

9. Dec 4, 2007

Well...if you want to go that route you DO know the period. What is the definition of the period T? Isn't it the amount of time in seconds it takes to complete one cycle?

Also $$a_{centripetal}=\frac{v^2}{r}$$

Have you learned about centripetal acceleration and tangential velocity, etc...?

10. Dec 4, 2007

Okay I see where that formula comes from... anyway, according to the definition of T that I gave you above, you should be able to see what T is in the 1st post.

Casey

11. Dec 4, 2007

Okay, so you have found T=3. Now using your teachers formula: if
f=ma then
f=m*a_centripetal
=m*4pi^2R / T^2

12. Dec 4, 2007

The formula you had is correct m*4pi^2R / T^2

If you want to know why, it is because:

$$f_c=m*a_c$$

$$\Rightarrow f_c=m*\frac{v^2}{r}$$ where v= the amount of time T it takes to complete a circle where the distance around a circle is 2pi*r so substituting that into the formula gives

$$f_c=m*(\frac{2\pi*r}{T})^2*\frac{1}{r}$$

Casey

13. Dec 4, 2007

I don't have a calculator handy, but it looks much more reasonable now!

14. Dec 4, 2007