# Problem using conservation of angular momentum

1. Oct 23, 2008

### john001

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

A 0.2kg ball and its supporting cord are revolving about the vertical axis on the fixed smooth conical surface with an angular velocity of 4 rad/s. The ball is held in the position b=0.3m by the tension T in the cord. If the distance b is reduced to the constant value of 0.2m by increasing the tension T in the cord, compute the new angular velocity w and the work done on the system by T.

2. Relevant equations

$$\sum$$M=dH/dt

H=r x mv

3. The attempt at a solution

I solved this problem using conservation of angular momentum and it came out correct, but I can't for the life of me understand why momentum is conserved. I understand that the force in the radial direction does not have a moment about the radial direction because the cross product is zero, but what about the vertical direction? Doesn't the ball have to accelerate in the vertical direction in order to change its b value from 0.3 to 0.2? And doesn't an acceleration in the vertical direction mean an unbalanced force in the vertical direction, which subsequently means having an unbalanced moment about the middle of the cone?

Please let me know if I'm not making sense.

#### Attached Files:

• ###### untitled.JPG
File size:
24.1 KB
Views:
473
2. Oct 24, 2008

### tiny-tim

Welcome to PF!

Hi john001! Welcome to PF!

If you take moments about the vertex of the cone, then the "vertical" force goes through the vertex, and so its moment is zero, and the angular momentum about the vertex will not change.

You are worrying abut "an unbalanced moment about the middle of the cone" … so you are taking moments about a point other than the vertex. Even so, this is to be expected … the moment (a vector) will be horizontal, so it should produce a rotation about the (instantaneously) parallel horizontal axis through the centre of the cone … which is exactly what does happen, since when it goes up the cone it is rotating about a horizontal axis!

3. Oct 25, 2008

### john001

Re: Welcome to PF!

Thank you very much tiny-tim.

It makes sense now.