Question Number 50808 by Tinkutara last updated on 20/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18
![at temparature t M.I is I=(1/2)(πR^2 hρ)R^2 +Md^2 at temparature t d=distance between axis of cylinder and axis of rotation... I=((πρ)/2)(R^4 h) (dI/dt)=((πρ)/2)(R^4 (dh/dt)+4hR^3 (dR/dt))+((dM/dt))d^2 dM=0 no change of mass with temparature rise α=(dl/(ldt))=((change of length)/(original length×change of temparsture)) so (dI/dt)=((πρ)/2)(R^4 h×(dh/(hdt))+4hR^4 ×(dR/(Rdt))) dI=((πρR^4 h)/2)αdt+4×((πρhR^4 )/2)×(−α)dt[dR=−ve] dI=Iαdt−4×Iαdt=−3Iαdt since value of h increases so value of R decreases hence M.I get decreases... pls check is it right...](https://www.tinkutara.com/question/Q50884.png)
$${at}\:{temparature}\:{t}\:{M}.{I}\:{is} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi{R}^{\mathrm{2}} {h}\rho\right){R}^{\mathrm{2}} \:\:\:+{Md}^{\mathrm{2}} \:{at}\:{temparature}\:{t} \\ $$$${d}={distance}\:{between}\:{axis}\:{of}\:{cylinder}\:{and}\:{axis} \\ $$$${of}\:{rotation}… \\ $$$${I}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} {h}\right) \\ $$$$\frac{{dI}}{{dt}}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} \frac{{dh}}{{dt}}+\mathrm{4}{hR}^{\mathrm{3}} \frac{{dR}}{{dt}}\right)+\left(\frac{{dM}}{{dt}}\right){d}^{\mathrm{2}} \\ $$$${dM}=\mathrm{0}\:\:\:{no}\:{change}\:{of}\:{mass}\:{with}\:{temparature} \\ $$$${rise} \\ $$$$\alpha=\frac{{dl}}{{ldt}}=\frac{{change}\:{of}\:{length}}{{original}\:{length}×{change}\:{of}\:{temparsture}} \\ $$$${so}\: \\ $$$$\frac{{dI}}{{dt}}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} {h}×\frac{{dh}}{{hdt}}+\mathrm{4}{hR}^{\mathrm{4}} ×\frac{{dR}}{{Rdt}}\right) \\ $$$${dI}=\frac{\pi\rho{R}^{\mathrm{4}} {h}}{\mathrm{2}}\alpha{dt}+\mathrm{4}×\frac{\pi\rho{hR}^{\mathrm{4}} }{\mathrm{2}}×\left(−\alpha\right){dt}\left[{dR}=−{ve}\right] \\ $$$${dI}={I}\alpha{dt}−\mathrm{4}×{I}\alpha{dt}=−\mathrm{3}{I}\alpha{dt} \\ $$$${since}\:{value}\:{of}\:{h}\:{increases}\:{so}\:\:{value}\:{of}\:{R}\:{decreases} \\ $$$${hence}\:{M}.{I}\:{get}\:{decreases}… \\ $$$${pls}\:{check}\:{is}\:{it}\:{right}… \\ $$
Commented by Tinkutara last updated on 22/Dec/18
No sir your answer is wrong.
Answered by ajfour last updated on 22/Dec/18
![I = ((mr^2 )/2) △I = (m/2)(2αr^2 )△t ⇒ △I = 2αI△t [option (c)].](https://www.tinkutara.com/question/Q50922.png)
$${I}\:=\:\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\: \\ $$$$\bigtriangleup{I}\:=\:\frac{{m}}{\mathrm{2}}\left(\mathrm{2}\alpha{r}^{\mathrm{2}} \right)\bigtriangleup{t} \\ $$$$\Rightarrow\:\:\:\bigtriangleup{I}\:=\:\:\mathrm{2}\alpha{I}\bigtriangleup{t}\:\:\:\:\:\left[{option}\:\left({c}\right)\right]. \\ $$
Commented by Tinkutara last updated on 23/Dec/18
Thanks Sir!