Question and Answers Forum

All Questions      Topic List

Properties of Matter Questions

Previous in All Question      Next in All Question      

Previous in Properties of Matter      Next in Properties of Matter      

Question Number 57114 by Tinkutara last updated on 30/Mar/19

Commented by ajfour last updated on 30/Mar/19

Commented by mr W last updated on 31/Mar/19

nice solution sir! let′s say m>m_(min) . what is the periode of the swinging motion of the stick?

$${nice}\:{solution}\:{sir}! \\ $$$${let}'{s}\:{say}\:{m}>{m}_{{min}} .\:{what}\:{is}\:{the}\:{periode} \\ $$$${of}\:{the}\:{swinging}\:{motion}\:{of}\:{the}\:{stick}? \\ $$

Commented by ajfour last updated on 31/Mar/19

should viscosity be taken into consideration, Sir?

$$\mathrm{should}\:\mathrm{viscosity}\:\mathrm{be}\:\mathrm{taken}\:\mathrm{into} \\ $$$$\mathrm{consideration},\:\mathrm{Sir}? \\ $$

Commented by mr W last updated on 31/Mar/19

no. only weight of stick and buoyant force act. stability means the buoyant force brings the stick always to its original position.

$${no}.\:{only}\:{weight}\:{of}\:{stick}\:{and}\:{buoyant} \\ $$$${force}\:{act}.\:{stability}\:{means}\:{the}\:{buoyant} \\ $$$${force}\:{brings}\:{the}\:{stick}\:{always}\:{to}\:{its} \\ $$$${original}\:{position}. \\ $$

Answered by ajfour last updated on 30/Mar/19

for stability centre of buoyancy must be above or at same level as centre of gravity. ⇒ (x/2)≥((m×0+ρπR^2 L((L/2)))/(m+ρπR^2 L)) and m+ρπR^2 L=σπR^2 x ⇒ ((m+ρπR^2 L)/(2σπR^2 )) ≥ ((ρπR^2 L((L/2)))/(m+ρπR^2 L)) let πR^2 L = c , then (m+ρc)^2 ≥σρc^2 ⇒ m≥((√(σρ)) )c−ρc ⇒ m_(minimum) =((√(σρ))−ρ)πR^2 L .

$$\mathrm{for}\:\mathrm{stability}\: \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{buoyancy}\:\mathrm{must}\:\mathrm{be}\:\mathrm{above} \\ $$$$\mathrm{or}\:\mathrm{at}\:\mathrm{same}\:\mathrm{level}\:\mathrm{as}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{gravity}. \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{x}}{\mathrm{2}}\geqslant\frac{\mathrm{m}×\mathrm{0}+\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}\left(\frac{\mathrm{L}}{\mathrm{2}}\right)}{\mathrm{m}+\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}} \\ $$$$\mathrm{and}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{m}+\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}=\sigma\pi\mathrm{R}^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\:\:\frac{\mathrm{m}+\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}}{\mathrm{2}\sigma\pi\mathrm{R}^{\mathrm{2}} }\:\geqslant\:\frac{\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}\left(\frac{\mathrm{L}}{\mathrm{2}}\right)}{\mathrm{m}+\rho\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}} \\ $$$$\mathrm{let}\:\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}\:=\:\mathrm{c}\:,\:\mathrm{then} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{m}+\rho\mathrm{c}\right)^{\mathrm{2}} \:\geqslant\sigma\rho\mathrm{c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{m}\geqslant\left(\sqrt{\sigma\rho}\:\right)\mathrm{c}−\rho\mathrm{c} \\ $$$$\Rightarrow\:\:\:\:\mathrm{m}_{\mathrm{minimum}} =\left(\sqrt{\sigma\rho}−\rho\right)\pi\mathrm{R}^{\mathrm{2}} \mathrm{L}\:. \\ $$

Commented by Tinkutara last updated on 30/Mar/19

Thank you Sir! But what was wrong with previous answer?

Commented by ajfour last updated on 30/Mar/19

previous answer should go for maximum mass that be attached to stick to keep it afloat still, i think.

$$\mathrm{previous}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{go}\:\mathrm{for} \\ $$$$\mathrm{maximum}\:\mathrm{mass}\:\mathrm{that}\:\mathrm{be}\:\mathrm{attached} \\ $$$$\mathrm{to}\:\mathrm{stick}\:\mathrm{to}\:\mathrm{keep}\:\mathrm{it}\:\mathrm{afloat}\:\mathrm{still},\:\mathrm{i}\:\mathrm{think}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com