Central forces: spinning puck tied to hanging mass

[unscientific]

Homework Statement

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

Homework Equations

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

 

[TSny]

Did you include the kinetic energy of the dangling mass?

 

[unscientific]

Did you include the kinetic energy of the dangling mass?

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

 

[TSny]

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

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

 

[unscientific]

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

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

 

[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.

 

[unscientific]

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.

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

 

[TSny]

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

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.

 

[unscientific]

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.

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

 

[unscientific]

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.

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..

 

[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.

 

[unscientific]

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.

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

 

[TSny]

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

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.

 

[unscientific]

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.

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

 

[unscientific]

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.

 

[unscientific]

bumppp anyone??

 

[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 $r$.

 

[unscientific]

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 $r$.

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

 

[TSny]

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

 

[unscientific]

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

ah, that is right!

 

[unscientific]

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

Yup, and this is what I end up with:

Nevermind, solved it!