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Question Number 3337 by Yozzi last updated on 11/Dec/15

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ⇠⇠ ⇠⇠ h_1 h_2 h_1 Figure 1 Let us model a yoyo by three combined, uniform cylindrical bodies which yield the above profile in Fig. 1. The mass of the entire body is M. h_1 is the side length of the two outter cylindrical bodies and r_1 is the side radius of these cylinders. h_(2 ) is the side length of the inner cylindrical body whose side radius is r_2 . Here, r_1 >r_2 and h_1 >h_2 . The axis of each cylindrical body, passing through their respective centre of mass and perpendicular to their circular faces, all coincide with each other. Determine the 3D moment of inertia of the yoyo about its axis normal to its outter circular faces and passing through its centre of mass.

$$\:\: \\ $$$$\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast \\ $$$$\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast \\ $$$$\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast \\ $$$$\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\ast \\ $$$$\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast \\ $$$$\:\:\:\:\:\dashleftarrow\dashleftarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\dashleftarrow\dashleftarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:{h}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:{h}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Figure}}\:\mathrm{1} \\ $$$${Let}\:{us}\:{model}\:{a}\:{yoyo}\:{by}\:{three}\:{combined},\:{uniform} \\ $$$${cylindrical}\:{bodies}\:{which}\:{yield}\:{the}\:{above}\:{profile} \\ $$$${in}\:{Fig}.\:\mathrm{1}.\:{The}\:{mass}\:{of}\:{the}\:{entire}\:{body} \\ $$$${is}\:{M}. \\ $$$${h}_{\mathrm{1}} \:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{two}\:{outter} \\ $$$${cylindrical}\:{bodies}\:{and}\:{r}_{\mathrm{1}} \:{is}\:{the}\:{side}\:{radius}\:{of}\: \\ $$$${these}\:{cylinders}.\:{h}_{\mathrm{2}\:} {is}\:{the}\:{side}\:{length} \\ $$$${of}\:{the}\:{inner}\:{cylindrical}\:{body}\:{whose}\:{side} \\ $$$${radius}\:{is}\:{r}_{\mathrm{2}} .\:{Here},\:{r}_{\mathrm{1}} >{r}_{\mathrm{2}} \:{and}\:{h}_{\mathrm{1}} >{h}_{\mathrm{2}} . \\ $$$${The}\:{axis}\:{of}\:{each}\:{cylindrical}\:{body}, \\ $$$${passing}\:{through}\:{their}\:{respective} \\ $$$${centre}\:{of}\:{mass}\:{and}\:{perpendicular}\:{to} \\ $$$${their}\:{circular}\:{faces},\:{all}\:{coincide}\:{with} \\ $$$${each}\:{other}.\:{Determine}\:{the}\:\mathrm{3}{D}\:{moment} \\ $$$${of}\:{inertia}\:{of}\:{the}\:{yoyo}\:{about}\:{its}\:{axis} \\ $$$${normal}\:{to}\:{its}\:{outter}\:{circular}\:{faces}\:{and}\:{passing} \\ $$$${through}\:{its}\:{centre}\:{of}\:{mass}. \\ $$$$ \\ $$$$ \\ $$

Commented by prakash jain last updated on 11/Dec/15

A ↑ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ −− − ∗ ∗ ∗ −−→B ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∣ The yoyo in my diagram is 90 degree rotated. You intend to calculated moment of intertia about axis A or B? Axis A (corrected based on comment below) m_1 mass of outer cylinder m_2 mass of inner cylinder moment of inertia of inner cylider=((m_2 r_2 ^2 )/2) moment of inertia of outer cylider=((m_1 r_1 ^2 )/2) Total moment of inertia=m_1 r_1 ^2 +(1/2)m_2 r_1 ^2 Assuming all cylinders of the same density d M=d(2πr_1 ^2 h_1 +πr_2 ^2 h_2 )⇒d=(M/(π(2r_1 ^2 h_1 +r_2 ^2 h_2 ))) m_1 =((Mr_1 ^2 h_1 )/((2r_1 ^2 h_1 +r_2 ^2 h_2 ))) m_2 =((Mr_2 ^2 h_2 )/((2r_1 ^2 h_1 +r_2 ^2 h_2 ))) I=((Mr_1 ^2 h_1 r_1 ^2 )/((2r_1 ^2 h_1 +r_2 ^2 h_2 )))+((Mr_2 ^2 h_2 r_2 ^2 )/(2(2r_1 ^2 h_1 +r_2 ^2 h_2 ))) =((M(2r_1 ^4 h_1 +r_2 ^4 h_2 ))/(2(2r_1 ^2 h_1 +r_2 ^2 h_2 )))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\uparrow \\ $$$$\:\:\:\:\:\:\:\ast\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\ast \\ $$$$\:\:\:\:\:\:\:\ast\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\ast \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\: \\ $$$$−−\:−\:\:\ast\:\:\:\ast\:\:\:\ast\:−−\rightarrow\mathrm{B}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\ast\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\ast \\ $$$$\:\:\:\:\:\:\:\ast\:\:\:\:\ast\:\:\:\ast\:\:\:\ast\:\:\:\:\ast \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid \\ $$$$\mathrm{The}\:\mathrm{yoyo}\:\mathrm{in}\:\mathrm{my}\:\mathrm{diagram}\:\mathrm{is}\:\mathrm{90}\:\mathrm{degree}\:\mathrm{rotated}. \\ $$$$\mathrm{You}\:\mathrm{intend}\:\mathrm{to}\:\mathrm{calculated}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{intertia} \\ $$$$\mathrm{about}\:\mathrm{axis}\:\mathrm{A}\:\mathrm{or}\:\mathrm{B}? \\ $$$$\mathrm{Axis}\:\mathrm{A}\:\left(\mathrm{corrected}\:\mathrm{based}\:\mathrm{on}\:\mathrm{comment}\:\mathrm{below}\right) \\ $$$${m}_{\mathrm{1}} \:\mathrm{mass}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{cylinder} \\ $$$${m}_{\mathrm{2}} \:\mathrm{mass}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{cylinder} \\ $$$$\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{cylider}=\frac{{m}_{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{cylider}=\frac{{m}_{\mathrm{1}} {r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{Total}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}={m}_{\mathrm{1}} {r}_{\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{Assuming}\:\mathrm{all}\:\mathrm{cylinders}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{density}\:{d} \\ $$$${M}={d}\left(\mathrm{2}\pi{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +\pi{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)\Rightarrow{d}=\frac{{M}}{\pi\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)} \\ $$$${m}_{\mathrm{1}} =\frac{{Mr}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} }{\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)} \\ $$$${m}_{\mathrm{2}} =\frac{{Mr}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} }{\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)} \\ $$$${I}=\frac{{Mr}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} {r}_{\mathrm{1}} ^{\mathrm{2}} }{\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)}+\frac{{Mr}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)} \\ $$$$=\frac{{M}\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{4}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{4}} {h}_{\mathrm{2}} \right)}{\mathrm{2}\left(\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} +{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} \right)} \\ $$

Commented by Yozzi last updated on 11/Dec/15

I′d like the M.I about axis A.

$${I}'{d}\:{like}\:{the}\:{M}.{I}\:{about}\:{axis}\:{A}. \\ $$

Commented by Yozzi last updated on 11/Dec/15

Thanks for the calculation. I think you calculated M.I about axis A. I′d check if the calculus approach in 3D yields a different answer.

$${Thanks}\:{for}\:{the}\:{calculation}.\:{I}\:{think} \\ $$$${you}\:{calculated}\:{M}.{I}\:{about}\:{axis}\:{A}. \\ $$$${I}'{d}\:{check}\:{if}\:{the}\:{calculus}\:{approach} \\ $$$${in}\:\mathrm{3}{D}\:{yields}\:{a}\:{different}\:{answer}. \\ $$$$ \\ $$

Commented by prakash jain last updated on 11/Dec/15

Is the assumption about uniform density correct. I intended to calculate about axis B.

$$\mathrm{Is}\:\mathrm{the}\:\mathrm{assumption}\:\mathrm{about}\:\mathrm{uniform}\:\mathrm{density}\:\mathrm{correct}. \\ $$$$\mathrm{I}\:\mathrm{intended}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{about}\:\mathrm{axis}\:\mathrm{B}. \\ $$

Commented by Yozzi last updated on 11/Dec/15

If we assume uniformity of a regular body, its mass density is the same at any observed point of the volume the body takes. So, yes. This would imply constant density, given that influencial environmental factors are disregarded or invariant in the given scenario. Each part could be assigned different densities also, but for a yoyo it is required that the outter materials are identical in composition to ensure its usual functioning as a toy. If you wished to calculate M.I for the body about axis B, which I assume is parallel to the planes of the outter circular faces, both the parallel and perpendicular axis theorems are (possibly) needed. Obtain the M.I about the axis of one of the outter cylinders parallel to its side length (h_1 ) as I_x . By the perpendicular axes theorem, if I_y is the M.I of the cylinder about an axis parallel to axis B and I_(z ) is the M.I about an axis such that axis X,axis Y and axis Z are mutually perpendicular, then I_x =I_y +I_z . Since the body is regular and uniform, I_y =I_z . ∴I_x =2I_y ⇒I_y =(1/2)I_x . Let m_1 be the mass of each outter cylinder. Then, by the parallel axis theorem, the M.I of each outter cylinder about axis B is given by I_b ^((1)) =I_y +(((h_1 +h_2 )/2))^2 m_1 . I_b ^((1)) =(1/2)I_x +((m_1 (h_1 +h_2 )^2 )/4) ((h_1 +h_2 )/2) is the ⊥ distance between axis Y and axis B. The inner cylinder has I_b ^((2)) =(1/2)I_x ^((1)) . because of the perpendicular axis theorem. The M.I of the full figure about B is then I_B =2I_b ^((1)) +I_b ^((2)) I_B =I_x +((m_1 (h_1 +h_2 )^2 )/2)+I_x ^((1)) . I_x =(1/2)m_1 r_1 ^(2 ) and I_x ^((1)) =(1/2)m_2 r_2 ^2 .

$${If}\:{we}\:{assume}\:{uniformity}\:{of}\:{a}\:{regular} \\ $$$${body},\:{its}\:{mass}\:{density}\:{is}\:{the}\:{same}\:{at} \\ $$$${any}\:{observed}\:{point}\:{of}\:{the}\:{volume}\:{the} \\ $$$${body}\:{takes}.\:{So},\:{yes}.\:{This}\:{would}\:{imply} \\ $$$${constant}\:{density},\:{given}\:{that}\:{influencial} \\ $$$${environmental}\:{factors}\:{are}\: \\ $$$${disregarded}\:{or}\:{invariant}\:{in}\:{the}\: \\ $$$${given}\:{scenario}.\: \\ $$$${Each}\:{part}\:{could}\:{be}\:{assigned}\:{different} \\ $$$${densities}\:{also},\:{but}\:{for}\:{a}\:{yoyo}\:{it}\:{is} \\ $$$${required}\:{that}\:{the}\:{outter}\:{materials} \\ $$$${are}\:{identical}\:{in}\:{composition}\:{to} \\ $$$${ensure}\:{its}\:{usual}\:{functioning}\:{as}\:{a}\:{toy}. \\ $$$${If}\:{you}\:{wished}\:{to}\:{calculate}\:{M}.{I}\:{for}\:{the} \\ $$$${body}\:{about}\:{axis}\:{B},\:{which}\:{I}\:{assume} \\ $$$${is}\:{parallel}\:{to}\:{the}\:{planes}\:{of}\:{the}\:{outter}\:{circular}\:{faces}, \\ $$$${both}\:{the}\:{parallel}\:{and}\:{perpendicular} \\ $$$${axis}\:{theorems}\:{are}\:\left({possibly}\right)\:{needed}. \\ $$$${Obtain}\:{the}\:{M}.{I}\:{about}\:{the}\:{axis}\:{of}\:{one}\:{of}\:{the} \\ $$$${outter}\:{cylinders}\:\:{parallel}\:{to}\:{its}\:{side} \\ $$$${length}\:\left({h}_{\mathrm{1}} \right)\:{as}\:{I}_{{x}} . \\ $$$${By}\:{the}\:{perpendicular}\:{axes}\:{theorem}, \\ $$$${if}\:{I}_{{y}} \:{is}\:{the}\:{M}.{I}\:{of}\:{the}\:{cylinder}\:{about}\:{an}\:{axis}\:{parallel} \\ $$$${to}\:{axis}\:{B}\:{and}\:{I}_{{z}\:} \:{is}\:{the}\:{M}.{I}\:{about} \\ $$$${an}\:{axis}\:{such}\:{that}\:{axis}\:{X},{axis}\:{Y} \\ $$$${and}\:{axis}\:{Z}\:{are}\:{mutually}\:{perpendicular}, \\ $$$${then}\:{I}_{{x}} ={I}_{{y}} +{I}_{{z}} .\:{Since}\:{the}\:{body}\:{is}\: \\ $$$${regular}\:{and}\:{uniform},\:{I}_{{y}} ={I}_{{z}} . \\ $$$$\therefore{I}_{{x}} =\mathrm{2}{I}_{{y}} \Rightarrow{I}_{{y}} =\frac{\mathrm{1}}{\mathrm{2}}{I}_{{x}} . \\ $$$${Let}\:{m}_{\mathrm{1}} \:{be}\:{the}\:{mass}\:{of}\:{each}\:{outter} \\ $$$${cylinder}.\:{Then},\:{by}\:{the}\:{parallel}\:{axis} \\ $$$${theorem},\:{the}\:{M}.{I}\:{of}\:{each}\:{outter}\: \\ $$$${cylinder}\:{about}\:{axis}\:{B}\:{is}\:{given}\:{by} \\ $$$${I}_{{b}} ^{\left(\mathrm{1}\right)} ={I}_{{y}} +\left(\frac{{h}_{\mathrm{1}} +{h}_{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} {m}_{\mathrm{1}} . \\ $$$${I}_{{b}} ^{\left(\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{2}}{I}_{{x}} +\frac{{m}_{\mathrm{1}} \left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\frac{{h}_{\mathrm{1}} +{h}_{\mathrm{2}} }{\mathrm{2}}\:{is}\:{the}\:\bot\:{distance}\:{between} \\ $$$${axis}\:{Y}\:{and}\:{axis}\:{B}. \\ $$$${The}\:{inner}\:{cylinder}\:{has}\:{I}_{{b}} ^{\left(\mathrm{2}\right)} =\frac{\mathrm{1}}{\mathrm{2}}{I}_{{x}} ^{\left(\mathrm{1}\right)} . \\ $$$${because}\:{of}\:{the}\:{perpendicular}\:{axis} \\ $$$${theorem}. \\ $$$${The}\:{M}.{I}\:{of}\:{the}\:{full}\:{figure}\:{about}\:{B}\:{is} \\ $$$${then} \\ $$$${I}_{{B}} =\mathrm{2}{I}_{{b}} ^{\left(\mathrm{1}\right)} +{I}_{{b}} ^{\left(\mathrm{2}\right)} \\ $$$${I}_{{B}} ={I}_{{x}} +\frac{{m}_{\mathrm{1}} \left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}}+{I}_{{x}} ^{\left(\mathrm{1}\right)} . \\ $$$${I}_{{x}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {r}_{\mathrm{1}} ^{\mathrm{2}\:\:} {and}\:{I}_{{x}} ^{\left(\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {r}_{\mathrm{2}} ^{\mathrm{2}} . \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by prakash jain last updated on 11/Dec/15

Sorry about the previous comments. The calculation that I did is applicable for axis A. Which is normal rotation of yoyo. You were correct!

$$\mathrm{Sorry}\:\mathrm{about}\:\mathrm{the}\:\mathrm{previous}\:\mathrm{comments}.\:\mathrm{The} \\ $$$$\mathrm{calculation}\:\mathrm{that}\:\mathrm{I}\:\mathrm{did}\:\mathrm{is}\:\mathrm{applicable}\:\mathrm{for}\:\mathrm{axis}\:\mathrm{A}. \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{normal}\:\mathrm{rotation}\:\mathrm{of}\:\mathrm{yoyo}. \\ $$$$\mathrm{You}\:\mathrm{were}\:\mathrm{correct}! \\ $$

Commented by Yozzi last updated on 11/Dec/15

No problem. I′m hoping to write an analysis of the mechanics of a yoyo when subjected to certain scenarios. Probability also arises since rotational motion is affected by the occasional friction between the ordinary string and the inner ′walls′ of the yoyo as the wounded string is spun out. That probability section might be a bit challenging to formalise.

$${No}\:{problem}.\:{I}'{m}\:{hoping}\:{to}\:{write}\:{an} \\ $$$${analysis}\:{of}\:{the}\:{mechanics}\:{of}\:{a}\:{yoyo} \\ $$$${when}\:{subjected}\:{to}\:{certain}\:{scenarios}. \\ $$$${Probability}\:{also}\:{arises}\:{since}\:{rotational} \\ $$$${motion}\:{is}\:{affected}\:{by}\:{the}\:{occasional} \\ $$$${friction}\:{between}\:{the}\:{ordinary}\:{string} \\ $$$${and}\:{the}\:{inner}\:'{walls}'\:{of}\:{the}\:{yoyo}\:{as} \\ $$$${the}\:{wounded}\:{string}\:{is}\:{spun}\:{out}.\: \\ $$$${That}\:{probability}\:{section}\:{might}\:{be} \\ $$$${a}\:{bit}\:{challenging}\:{to}\:{formalise}. \\ $$

Commented by prakash jain last updated on 12/Dec/15

Nwot related to question directly but when you are working on a topic are you taking to a logical conclusion?

$$\mathrm{Nwot}\:\mathrm{related}\:\mathrm{to}\:\mathrm{question}\:\mathrm{directly}\:\mathrm{but}\:\mathrm{when} \\ $$$$\mathrm{you}\:\mathrm{are}\:\mathrm{working}\:\mathrm{on}\:\mathrm{a}\:\mathrm{topic}\:\mathrm{are}\:\mathrm{you}\:\mathrm{taking}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{logical}\:\mathrm{conclusion}?\: \\ $$

Commented by Yozzi last updated on 12/Dec/15

Well, nowadays I am trying at math slightly higher than I was taught with the approach of deductive reasoning in mind. So yes. I′m searching for the building blocks of any solution I see. And many times that includes knowing and understanding definitions and my mathematical knowledge bank is very young. I hope I could approach number theory with deductive reasoning when I formally begin to learn it.

$${Well},\:{nowadays}\:{I}\:{am}\:{trying}\:{at}\:{math} \\ $$$${slightly}\:{higher}\:{than}\:{I}\:{was}\:{taught}\:{with} \\ $$$${the}\:{approach}\:{of}\:{deductive}\:{reasoning} \\ $$$${in}\:{mind}.\:{So}\:{yes}.\:{I}'{m}\:{searching}\:{for} \\ $$$${the}\:{building}\:{blocks}\:{of}\:{any}\:{solution} \\ $$$${I}\:{see}.\:{And}\:{many}\:{times}\:{that}\: \\ $$$${includes}\:{knowing}\:{and}\:{understanding} \\ $$$${definitions}\:{and}\:{my}\:{mathematical} \\ $$$${knowledge}\:{bank}\:{is}\:{very}\:{young}.\:{I} \\ $$$${hope}\:{I}\:{could}\:{approach}\:{number}\:{theory} \\ $$$${with}\:{deductive}\:{reasoning}\:{when}\:{I}\: \\ $$$${formally}\:{begin}\:{to}\:{learn}\:{it}.\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by prakash jain last updated on 11/Dec/15

For axis B inertia of inner cylinder=(m_2 /(12))(3r_2 ^2 +h_2 ^2 ) inertia of outer cylinder =(m_1 /(12))(3r_1 ^2 +h_1 ^2 ) for axis ∥ to faces of cylinder. Also distance of axis B from center of gravity=((h_1 +h_2 )/2) (this is same for both cylinders) inertia of outer cylinders (axis B)= m_1 (((3r_1 ^2 +h_1 ^2 )/(12))+(((h_1 +h_2 )^2 )/4)) Total moment of inertia =2m_1 (((3r_1 ^2 +h_1 ^2 )/(12))+(((h_1 +h_2 )^2 )/4))+(m_2 /(12))(3r_2 ^2 +h_2 ^2 )

$$\mathrm{For}\:\mathrm{axis}\:\mathrm{B} \\ $$$$\mathrm{inertia}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{cylinder}=\frac{{m}_{\mathrm{2}} }{\mathrm{12}}\left(\mathrm{3}{r}_{\mathrm{2}} ^{\mathrm{2}} +{h}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$$\mathrm{inertia}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{cylinder}\:=\frac{{m}_{\mathrm{1}} }{\mathrm{12}}\left(\mathrm{3}{r}_{\mathrm{1}} ^{\mathrm{2}} +{h}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\mathrm{for}\:\mathrm{axis}\:\parallel\:\mathrm{to}\:\mathrm{faces}\:\mathrm{of}\:\mathrm{cylinder}. \\ $$$$\mathrm{Also}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{axis}\:\mathrm{B}\:\mathrm{from}\:\mathrm{center} \\ $$$$\mathrm{of}\:\mathrm{gravity}=\frac{{h}_{\mathrm{1}} +{h}_{\mathrm{2}} }{\mathrm{2}}\:\left(\mathrm{this}\:\mathrm{is}\:\mathrm{same}\:\mathrm{for}\:\mathrm{both}\:\mathrm{cylinders}\right) \\ $$$$\mathrm{inertia}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{cylinders}\:\left(\mathrm{axis}\:\mathrm{B}\right)= \\ $$$$\:\:\:\:\:{m}_{\mathrm{1}} \left(\frac{\mathrm{3}{r}_{\mathrm{1}} ^{\mathrm{2}} +{h}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{12}}+\frac{\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\mathrm{Total}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia} \\ $$$$\:\:\:\:=\mathrm{2}{m}_{\mathrm{1}} \left(\frac{\mathrm{3}{r}_{\mathrm{1}} ^{\mathrm{2}} +{h}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{12}}+\frac{\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{{m}_{\mathrm{2}} }{\mathrm{12}}\left(\mathrm{3}{r}_{\mathrm{2}} ^{\mathrm{2}} +{h}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$

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