The Force Law for Simple Harmonic Motion — Question 8

Section 15-2 · Physics pages 2266–2267

Question 8 A floating cylinder has a buoyant restoring stiffness A vertical cylinder of constant cross-sectional area S and mass m floats at rest in liquid of density ρ. Displace it downward by a small distance q while keeping it vertical and partially immersed. Ignore drag, added fluid inertia and surface tension. q cylinder Partly immersed cylinder; positive displacement is downward. (a) Derive its small vertical motion equation and period. (b) Find the period for S = .010 m2, m = 2.0 kg, ρ = 1000 kg/m3 and g = 10 m/s2. State the immersion constraint on the amplitude.
Show solutionHide solution
Question 8 - Solution A floating cylinder has a buoyant restoring stiffness q cylinder (a) Derivation and calculation Let d be the equilibrium draft. Equilibrium and the displaced-volume increment give mg = ρgSd, ∆V = Sq, ∆FB = ρgSq. m¨q = mg −ρgS(d + q) = −ρgSq. ω2 = ρgS m , T = 2π √m ρgS . (b) Result and interpretation keff= (1000)(10)(0.010) = 100 N/m, T = 2π √ 2/100 = 0.88858 s , d = 2 1000(0.010) = 0.200 m. 0 < d −A, d + A < H ⇒A < min(d, H −d) . Emergence or full submergence changes the buoyancy-displacement relation. Physical check The result excludes fluid added inertia and drag as specified; these would change the frequency or damping in an actual experiment.

Question 8

A floating cylinder has a buoyant restoring stiffness

A vertical cylinder of constant cross-sectional area SS and mass mm floats at rest in liquid of density ρ\rho. Displace it downward by a small distance qq while keeping it vertical and partially immersed. Ignore drag, added fluid inertia and surface tension.

Diagram shown in the original SVG worksheet.

  1. Derive its small vertical motion equation and period.

  2. Find the period for S=.010m2S=.010\,\mathrm{m^2}, m=2.0kgm=2.0\,\mathrm{kg}, ρ=1000kg/m3\rho=1000\,\mathrm{kg/m^3} and g=10m/s2g=10\,\mathrm{m/s^2}. State the immersion constraint on the amplitude.

Question 8 - Solution

A floating cylinder has a buoyant restoring stiffness

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Let dd be the equilibrium draft. Equilibrium and the displaced-volume increment givemg=ρgSd,ΔV=Sq,ΔFB=ρgSq.mg=\rho gSd,\qquad\Delta V=Sq,\qquad\Delta F_B=\rho gSq.mq̈=mg−ρgS(d+q)=−ρgSq.m\ddot q=mg-\rho gS(d+q)=-\rho gSq.ω2=ρgSm,T=2πmρgS.\boxed{\omega^2=\frac{\rho gS}{m}},\qquad\boxed{T=2\pi\sqrt{\frac{m}{\rho gS}}}.

(b) Result and interpretation

keff=(1000)(10)(0.010)=100N/m,k_{\mathrm{eff}}=(1000)(10)(0.010)=100\,\mathrm{N/m},T=2π2/100=0.88858s,d=21000(0.010)=0.200m.T=2\pi\sqrt{2/100}=\boxed{0.88858\,\mathrm s},\qquad d=\frac{2}{1000(0.010)}=0.200\,\mathrm m.0<d−A,d+A<H⇒A<min⁡(d,H−d).0<d-A,\quad d+A<H\Rightarrow\boxed{A<\min(d,H-d)}.Emergence or full submergence changes the buoyancy-displacement relation.

Physical check

The result excludes fluid added inertia and drag as specified; these would change the frequency or damping in an actual experiment.