# Small amplitude oscillations of a ball with water

1. Feb 2, 2010

### glueball8

1. The problem statement, all variables and given/known data
The interior of a thin spherical shell of mass M and radius R is completely filled with water and hangs from a ceiling on a light thread. The distance from the sphere's center to the hanging point is L, and the mass of water is m. Determine the change in the frequency of small amplitude oscillations of this system when the water freezes. (Neglect the viscosity of water, and the change of volume upon freezing.)

2. Relevant equations
$$f = c \sqrt{g/L}$$

3. The attempt at a solution
Well its even hard to see there's a difference. But for the water part there's no viscosity, it some how changes the frequency. No idea how, or how to do this.

The ice ball follows $$f = c \sqrt{g/L}$$.

Ideas?

2. Feb 3, 2010

### ehild

This is a physical pendulum. If you use conservation of energy, you have to take the rotational energy of the ball into account.

In case of ice, the spherical shell with the ice inside constitute a solid body. The ice rotates together the sphere. If water is inside, and viscosity is neglected, no torque acts on the water. It orientation remains the same, it does not rotate with respect to the ground.

ehild

3. Feb 3, 2010

### glueball8

What do you mean no torque acts on the water? If no torque then how would the water rotate?

What do you mean orientation remains the same. It won't go in a projectile motion but more linear.

4. Feb 3, 2010

### ehild

I mean that the water will not rotate around its centre of mass, but the shell will.

The whole thing is a pendulum, the sphere oscillates along a piece of circle.

There is no friction between water and shell, the energy is conserved.

Write down the expression of energy for both cases: water and ice.

ehild

Last edited: Feb 3, 2010