Calculating Oscillation Periods for a Floating Cylinder

[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.]

Homework Statement

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.

Homework 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

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 [?]


Are these answers correct? Thanks.

 

[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?

 

[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

 

[OlderDan]

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

(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 ?

 

[smithg86]

I misread my notes...after reading your reply and my textbook, I took another stab at it:
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]

 

[OlderDan]

I misread my notes...after reading your reply and my textbook, I took another stab at it:
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]

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.

 

[smithg86]

Thanks for your help!