# Central forces: spinning puck tied to hanging mass

• unscientific
In summary, the conversation discusses the concepts of angular momentum and total energy in a system involving a hanging mass and a mass on a surface connected by a string. The experts clarify that the equations for angular momentum and energy include both the radial and angular components of velocity. They also explain that the tension in the string and the weight of the hanging mass are not equal, but they can be related through the application of the second law. Finally, they discuss how to approach part (b) of the problem, which involves finding the equation of motion in polar coordinates for the mass on the surface.
unscientific

## Homework Statement

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

## 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

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Did you include the kinetic energy of the dangling mass?

TSny said:
Did you include the kinetic energy of the dangling mass?

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

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

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

TSny said:
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?

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.

TSny said:
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?

unscientific said:
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.

TSny said:
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!

TSny said:
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..

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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.

TSny said:
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?

unscientific said:
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.

TSny said:
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!

TSny said:
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.

bumppp anyone??

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##.

TSny said:
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.

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

TSny said:
Multiply your second boxed equation by r and compare to the derivative of ##r^2\dot\theta##

ah, that is right!

TSny said:
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!

## 1. What is a central force?

A central force is a type of force that acts towards or away from a fixed point, known as the center of force. This type of force is always directed along the line connecting the center of force to the object experiencing the force.

## 2. How does a spinning puck tied to a hanging mass exhibit a central force?

The spinning puck tied to a hanging mass exhibits a central force because the string connecting the puck and the hanging mass always pulls the puck towards the center of the circle it is moving in. This force acts along the radius of the circle and is directed towards the center, making it a central force.

## 3. What is the relationship between centripetal force and central force in this scenario?

In this scenario, the centripetal force is the force that keeps the puck moving in a circular motion and is provided by the central force. The central force is responsible for changing the direction of the puck's velocity, keeping it moving in a circular path.

## 4. How does the mass of the hanging mass affect the motion of the spinning puck?

The mass of the hanging mass affects the motion of the spinning puck by changing the magnitude of the central force. A larger mass will result in a larger central force, which will cause the puck to move in a smaller circle at a faster speed. On the other hand, a smaller mass will result in a smaller central force and a larger circle with a slower speed.

## 5. How does the speed of the spinning puck change as the radius of the circle it is moving in changes?

According to the relationship between centripetal force and central force, the speed of the spinning puck will increase as the radius of the circle decreases. This is because the centripetal force, provided by the central force, needs to increase in order to keep the puck moving in a smaller circle. Similarly, the speed will decrease as the radius increases.

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