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Question-46299

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Question Number 46299 by rahul 19 last updated on 23/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
area remaining portion=πR^2 −π((R/3))^2 =πR^2 (1−(1/9))=((8πR^2 )/9) mass per unit area(ρ)=(M/((8πR^2 )/9))=((9M)/(8πR^2 )) so mass for cut disc=π(R^2 /9)×((9M)/(8πR^2 ))=(M/8) mass of uncut complete disc=M+(M/8)=((9M)/8) moment of inertia of uncut complete disc about the axis ⊥ to the plane of disc tbrough centre =((9M)/8)×(R^2 /2)=((9MR^2 )/(16)) ((9MR^2 )/(16))←I_o momeng of inertia cut small disc about mentioned axix =(M/8)×(R−(R/3))^2 +(M/8)×((R/3))^2 ×(1/2) =(M/8)×((4R^2 )/9)+(M/8)×(R^2 /(18))=((MR^2 )/8)((4/9)+(1/(18)))=((MR^2 )/8)×(1/2) =((MR^2 )/(16)) sl required ans is =((9MR^2 )/(16))−((MR^2 )/(16))=((MR^2 )/2) pls check...
$${area}\:{remaining}\:{portion}=\pi{R}^{\mathrm{2}} −\pi\left(\frac{{R}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$=\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)=\frac{\mathrm{8}\pi{R}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${mass}\:{per}\:{unit}\:{area}\left(\rho\right)=\frac{{M}}{\frac{\mathrm{8}\pi{R}^{\mathrm{2}} }{\mathrm{9}}}=\frac{\mathrm{9}{M}}{\mathrm{8}\pi{R}^{\mathrm{2}} } \\ $$$${so}\:{mass}\:{for}\:{cut}\:{disc}=\pi\frac{{R}^{\mathrm{2}} }{\mathrm{9}}×\frac{\mathrm{9}{M}}{\mathrm{8}\pi{R}^{\mathrm{2}} }=\frac{{M}}{\mathrm{8}} \\ $$$${mass}\:{of}\:{uncut}\:{complete}\:{disc}={M}+\frac{{M}}{\mathrm{8}}=\frac{\mathrm{9}{M}}{\mathrm{8}} \\ $$$${moment}\:{of}\:{inertia}\:{of}\:{uncut}\:{complete}\:{disc}\: \\ $$$${about}\:{the}\:{axis}\:\bot\:{to}\:{the}\:{plane}\:{of}\:{disc}\:{tbrough} \\ $$$${centre} \\ $$$$=\frac{\mathrm{9}{M}}{\mathrm{8}}×\frac{{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{9}{MR}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\frac{\mathrm{9}{MR}^{\mathrm{2}} }{\mathrm{16}}\leftarrow{I}_{{o}} \\ $$$${momeng}\:{of}\:{inertia}\:{cut}\:{small}\:{disc}\:{about} \\ $$$${mentioned}\:{axix}\:=\frac{{M}}{\mathrm{8}}×\left({R}−\frac{{R}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{{M}}{\mathrm{8}}×\left(\frac{{R}}{\mathrm{3}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{{M}}{\mathrm{8}}×\frac{\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{9}}+\frac{{M}}{\mathrm{8}}×\frac{{R}^{\mathrm{2}} }{\mathrm{18}}=\frac{{MR}^{\mathrm{2}} }{\mathrm{8}}\left(\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{18}}\right)=\frac{{MR}^{\mathrm{2}} }{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{{MR}^{\mathrm{2}} }{\mathrm{16}} \\ $$$${sl}\:{required}\:{ans}\:{is}\:=\frac{\mathrm{9}{MR}^{\mathrm{2}} }{\mathrm{16}}−\frac{{MR}^{\mathrm{2}} }{\mathrm{16}}=\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$
Commented by rahul 19 last updated on 24/Oct/18
Sir, i′m getting same answer. But i wonder moment of inertia of disc of mass m = ((mR^2 )/2). (normal disc) if we take another disc which has a hole, of mass m,M.I is still same!
$${Sir},\:{i}'{m}\:{getting}\:{same}\:{answer}. \\ $$$${But}\:{i}\:{wonder}\:{moment} \\ $$$${of}\:{inertia}\:{of}\:\:{disc}\:{of}\:{mass}\:{m}\:=\:\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}. \\ $$$$\left({normal}\:{disc}\right) \\ $$$${if}\:{we}\:{take}\:{another}\:{disc}\:{which}\:{has}\:{a}\:{hole}, \\ $$$${of}\:{mass}\:{m},{M}.{I}\:{is}\:{still}\:{same}! \\ $$
Commented by MrW3 last updated on 24/Oct/18
your comparation is not really correct. the disc without hole and the disc with hole do have in fact not the same MoI w.r.t. their CoM. I_(disc without hole) =((MR^2 )/2) I_(disc with hole) <((MR^2 )/2) the MoI of a object w.r.t. its CoM is the smallest. its MoI w.r.t. to any other point is bigger. for the disc with hole if we calculate its MoI not w.r.t. its CoM, but w.r.t. to an other point, we will get a bigger value. in fact we can get any value if we choose this point correspondingly. for the disc with hole if we the point O which is not its CoM, we′ll get the same value as the MoI of a disc without hole w.r.t. its CoM.
$${your}\:{comparation}\:{is}\:{not}\:{really}\:{correct}. \\ $$$${the}\:{disc}\:{without}\:{hole}\:{and}\:{the}\:{disc}\:{with} \\ $$$${hole}\:{do}\:{have}\:{in}\:{fact}\:{not}\:{the}\:{same}\:{MoI} \\ $$$${w}.{r}.{t}.\:{their}\:{CoM}. \\ $$$${I}_{{disc}\:{without}\:{hole}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{{disc}\:{with}\:{hole}} <\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${the}\:{MoI}\:{of}\:{a}\:{object}\:{w}.{r}.{t}.\:{its}\:{CoM}\:{is} \\ $$$${the}\:{smallest}.\:{its}\:{MoI}\:{w}.{r}.{t}.\:{to}\:{any} \\ $$$${other}\:{point}\:{is}\:{bigger}.\:{for}\:{the}\:{disc}\:{with} \\ $$$${hole}\:{if}\:{we}\:{calculate}\:{its}\:{MoI}\:{not}\:{w}.{r}.{t}.\:{its} \\ $$$${CoM},\:{but}\:{w}.{r}.{t}.\:{to}\:{an}\:{other}\:{point},\:{we} \\ $$$${will}\:{get}\:{a}\:{bigger}\:{value}.\:{in}\:{fact}\:{we}\:{can} \\ $$$${get}\:{any}\:{value}\:{if}\:{we}\:{choose}\:{this}\:{point} \\ $$$${correspondingly}.\:{for}\:{the}\:{disc}\:{with}\:{hole} \\ $$$${if}\:{we}\:{the}\:{point}\:{O}\:{which}\:{is}\:{not}\:{its}\:{CoM}, \\ $$$${we}'{ll}\:{get}\:{the}\:{same}\:{value}\:{as}\:{the}\:{MoI} \\ $$$${of}\:{a}\:{disc}\:{without}\:{hole}\:{w}.{r}.{t}.\:{its}\:{CoM}. \\ $$
Answered by ajfour last updated on 24/Oct/18
I_(reqd) = (1/2)MR^( 2) .
$${I}_{{reqd}} =\:\frac{\mathrm{1}}{\mathrm{2}}{MR}^{\:\mathrm{2}} \:. \\ $$
Commented by rahul 19 last updated on 24/Oct/18
Sir, is this method correct ? M.I= (1/2)×M×R^2 where M=mass of remaining disc......or we need to calculate the way Tanmay sir has done.
$${Sir},\:{is}\:{this}\:{method}\:{correct}\:? \\ $$$${M}.{I}=\:\frac{\mathrm{1}}{\mathrm{2}}×{M}×{R}^{\mathrm{2}} \:{where}\:{M}={mass}\:{of} \\ $$$${remaining}\:{disc}……{or}\:{we}\:{need}\:{to} \\ $$$${calculate}\:{the}\:{way}\:{Tanmay}\:{sir}\:{has}\:{done}. \\ $$
Commented by ajfour last updated on 24/Oct/18
yes i had myself solved the same way; only dint post; but it could be reduced a little; ((M+m)/m) = (R^2 /((R/3)^2 )) = 9 ⇒ M=8m I=(((M+m)R^2 )/2)−[(m/2)((R/3))^2 +m(((2R)/3))^2 ] = ((9mR^2 )/2)−((mR^2 )/(18))−((4mR^2 )/9) = ((mR^2 )/(18))(81−1−8) = 4mR^2 = 4((M/8))R^2 = ((MR^2 )/2) .
$${yes}\:{i}\:{had}\:{myself}\:{solved}\:{the}\:{same} \\ $$$${way};\:{only}\:{dint}\:{post};\:{but}\:{it}\:{could} \\ $$$${be}\:{reduced}\:{a}\:{little}; \\ $$$$\frac{{M}+{m}}{{m}}\:=\:\frac{{R}^{\mathrm{2}} }{\left({R}/\mathrm{3}\right)^{\mathrm{2}} }\:=\:\mathrm{9}\:\:\Rightarrow\:{M}=\mathrm{8}{m} \\ $$$${I}=\frac{\left({M}+{m}\right){R}^{\mathrm{2}} }{\mathrm{2}}−\left[\frac{{m}}{\mathrm{2}}\left(\frac{{R}}{\mathrm{3}}\right)^{\mathrm{2}} +{m}\left(\frac{\mathrm{2}{R}}{\mathrm{3}}\right)^{\mathrm{2}} \right] \\ $$$$\:\:=\:\frac{\mathrm{9}{mR}^{\mathrm{2}} }{\mathrm{2}}−\frac{{mR}^{\mathrm{2}} }{\mathrm{18}}−\frac{\mathrm{4}{mR}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\:\:=\:\frac{{mR}^{\mathrm{2}} }{\mathrm{18}}\left(\mathrm{81}−\mathrm{1}−\mathrm{8}\right)\:=\:\mathrm{4}{mR}^{\mathrm{2}} \\ $$$$\:=\:\mathrm{4}\left(\frac{{M}}{\mathrm{8}}\right){R}^{\mathrm{2}} \:=\:\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\:. \\ $$

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