# Please can someone check this

1. Apr 2, 2009

### a.mlw.walker

THIS IS NOT HOMEWORK, IT IS FOR MY OWN UNDERSTANDING

I have written a few notes to myself on circular motion, if it is right, i think it will help others trying to understand it. Please can someone check the attached pdf.

Imagine having a marble, in a tin can, and you are shaking the can in circles, causing the marble to stay on the side. As you slow the tin, the marble will slow, and at some point fall to the bottom.

The pdf is trying to show the maths of this, if you know the first two times for revolution.

It assumes constant deceleration.

I'm not sure about my use of the constant acceleration equation to estimate how far round its last revolution (as it decelerates), the ball will get.

thanks

#### Attached Files:

• ###### Equations to find the position of a ball at any point.pdf
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Last edited: Apr 2, 2009
2. Apr 2, 2009

### Cantab Morgan

I think this is a really hard problem, because the centripetal force of the can on the marble is acting inwards (it's a central force), and it doesn't act directly against gravity. What you're really doing is tilting the can as you're moving it, so that there's a slight upward vertical component of the normal force of the wall of the can on the marble.

It also just occurs to me that since the can is moving, it's not obvious that the marble's trajectory is a circle. Another question, is the marble sliding without friction, or is it rotating along?

3. Apr 3, 2009

### a.mlw.walker

Damn, you caught me out, I was trying to think of a simple illustration, but as i wrote it i also noticed it was flawed, and hoped nothing of it. Better example is one of those theme park rides where you all stand against the edge of the cylindrical cage, the cage starts spinning and then the floor drops away, leaving you stuck to the wall and not falling.

Another example is a roulette ball spinning in its track before it falls.

Do they help?

4. Apr 3, 2009

### rcgldr

Assuming no losses in energy you have two cases.

If the interaction between ball and can is frictionless, the ball falls as if it were in free fall, it's forward and angular velocity have no affect on the vertical acceleration due to gravity.

If the interaction between ball and can involves no slippages, then the ball is rolling, and the ball will accelerate downwards at bit slower, because angular energy is increased as well as linear energy as the ball descends.

5. Apr 3, 2009

### a.mlw.walker

Lets forget the can now, and look at the other two examples i mentioned above, the can was a rubbish idea.

Last edited: Apr 3, 2009