# Central forces: spinning puck tied to hanging mass

1. Jan 30, 2013

### unscientific

1. The problem statement, all variables and given/known data

http://i50.tinypic.com/mmws94.png

2. Relevant equations

3. The attempt at a solution

1. Angular momentum is constant; J = constant

2. Total energy is constant; E = constant

Not sure why I am ending up with a 2gr instead..as U = ∫ f dr = ∫ mg dr

Last edited by a moderator: Feb 3, 2013
2. Jan 30, 2013

### TSny

Did you include the kinetic energy of the dangling mass?

3. Jan 30, 2013

### unscientific

That's the problem, I know there is none.

4. Jan 30, 2013

### TSny

Why do you say there is no KE of the hanging mass?

5. Jan 30, 2013

### unscientific

If it's just hanging there and not moving, how can there be KE?

6. Jan 30, 2013

### TSny

The hanging mass is not necessarily at rest. If it were, then $r$ would be constant and $\dot{r}$ would be zero. The first part of the problem is asking for the general equation for $\dot{r}$ when the mass on the surface is not necessarily moving in a circle.

7. Jan 30, 2013

### unscientific

but how do you know both are moving at the same speed such that you can combine their KE into one?

8. Jan 30, 2013

### TSny

They don't necessarily have the same speed. The hanging mass moves along a straight line and its speed is due just to the rate of change of $r$. The mass on the surface moves in two dimensions and its velocity can be broken up into a radial component due to the rate of change of $r$ and a component perpendicular to the string due to the rate of change of the angular polar coordinate $\theta$. The $\dot{r}$ term of your energy expression takes account of the radial component of velocity and your angular momentum term takes care of the other component.

It should be clear that the speed of the hanging mass is just $|\dot{r}|$. So, you can easily add in the kinetic energy of the hanging mass into your energy expression.

9. Jan 30, 2013

### unscientific

That is very true. If both masses are connected by the same string, then by definition $\dot{r}$ is the same for both masses!

10. Jan 30, 2013

### unscientific

Part (b) i'm not sure what they want. I've written the EOM in polar coordinates but don't see how that will lead me to the answer..

Last edited: Jan 30, 2013
11. Jan 30, 2013

### TSny

The EOM that you wrote in polar coordinates applies to the particle on the surface. You are correct that there is no force in the $\theta$ direction for this particle. So, your equation $2\dot r\dot \theta + r \ddot\theta = 0$ is correct.

However, the next equation for the $r$ component is incorrect on the right side. How did you conclude that the right hand side is $g$? Think about what force acts on the particle in the $r$ direction.

12. Jan 31, 2013

### unscientific

the force would be the tension in the string, i.e. the weight of the other bob?

13. Jan 31, 2013

### TSny

The force would be the tension in the string, but the tension in the string does not equal the weight of the other particle. You can relate the tension to the weight of the other particle by applying the 2nd law to the other particle.

14. Jan 31, 2013

### unscientific

That is true, I forgot the other particle is accelerating as well! Thanks!

15. Jan 31, 2013

### unscientific

16. Feb 2, 2013

### unscientific

bumppp anyone??

17. Feb 2, 2013

### TSny

I think T should be T/m in your equations, but it won't make a difference in your final equations in the boxes. They look correct.

You can integrate the last equation by noting what you get if you take the time derivative of $r^2\dot{\theta}$.

The result will allow you to write $\dot{\theta}$ in terms of $r$ and a constant. You can then use this to eliminate $\dot{\theta}$ in your first boxed equation to get a differential equation for $r$.

18. Feb 5, 2013

### unscientific

By differentiating $r^2\dot{\theta}$ that gives you the correct first term, but the second term has r2.

19. Feb 5, 2013

### TSny

Multiply your second boxed equation by r and compare to the derivative of $r^2\dot\theta$

20. Feb 5, 2013

### unscientific

ah, that is right!