# Oscillation/Buoyancy Question

1. Dec 17, 2006

### smithg86

[This was a 4 part question. The first 2 parts were correctly done (so I didn't show much work for them). I'm not sure about the last 2 parts. I only need help with the last 2 parts.]

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

A closed hollow cylinder of length L = 0.5 m, cross sectional area A= 0.0004 m^2 and a negligible mass has a lead weight of mass m=0.1 kg inside at the bottom so that it floats vertically when placed in water.

[1st part]

Determine the distance, d, from the bottom of the cylinder to the surface of the water.

I calculated d = 0.25m

[2nd part]

The cylinder is now pushed down a distance x from the equilibrium position, d, determined above. What is the additional force on the cylinder trying to restore it to its equilibrium position?

I calculated F = 3.924x

[3rd part]

What is the period of the vertical oscillations of the cylinder?

[4th part]

Estimate the period of rotational oscillations, where the axis of the cylinder oscillates back and forth in a vertical plane.

2. Relevant equations

[For part 1]

(.1kg mass) = (mass of displaced water) = (density of water)(volume of submersed part of cylinder)

[For part 2]

(.1 kg + F) = (new displaced volume of water)(density of water)

[For part 3]

w = (k/m)^(1/2)
T = (2pi)/w

[For part 4]

(torque) = -k(theta)
w = (k/I)^(1/2)
T = 2pi/w

3. The attempt at a solution

[Part 3]

T = 2pi/w = 2pi(m/k)^(1/2) = 2pi(.1/3.924)^(1/2) = 1.00303 seconds [?]

[Part 4]

T = 2pi/w = 2pi/ 25.0567 = 0.250758 seconds [?]

2. Dec 17, 2006

### OlderDan

Part 3) looks OK assuming the earlier parts are correct and that the units of k that you have not included in your answer to part 2) are N/m

I have absolutely no idea what you did in part 4) Throwing in a dimensionless number of 25.0567 without any explanation where it comes from does not exactly help anyone here help you. A lot of people here can do the problem. We should not have to do your work to see if you got the right answer. Where does that number come from?

3. Dec 17, 2006

### smithg86

Sorry about that. Part 2's answer is in [N/m]. Heres what I did for part 4:
moment of inertia for cylinder, I = md^2 = (.1 kg)(.25 m)^2= 0.00625 [kg m^2]
angular frequency, w = (k/I)^(1/2) = (3.924/0.00625)^(1/2) = 25.0567 [1/seconds]
period, T = 2pi/w = 2pi/25.0567 = 0.250758 seconds

4. Dec 17, 2006

### OlderDan

(k/I)^½ has the dimensions of [(N/m)/(kg*m²)]^½ = [(kg*m/s²)/(kg*m³)]^½ = 1/(m*s)

These are not the units of frequency. Where did you get ω = (k/I)^½ ???

Last edited: Dec 17, 2006
5. Dec 17, 2006

### smithg86

I treated the cylinder as a pendulum, rotating around its center of buoyancy, which I took to be d=0.25 [m].
w=(g/d)^(1/2) = (9.81/0.25)^(1/2) = 6.26418 [1/s]
T = 2pi/w = 2pi/6.26 = 1.00303 [seconds]

6. Dec 17, 2006

### OlderDan

That looks much better. It does of course neglect some very real effects related to how the buoyant force is distributed over the surface of the cylinder, and the resistance to motion through the water. However, for very small oscillations it should be about right, and I doubt that you are expected to worry about other effects.

7. Dec 17, 2006