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Question Number 26181 by NECx last updated on 21/Dec/17
Find the moment of inertia and  the radius of gyration of a   circular plate about an axis through  its centre,perpendicular to the  plane of the plate.
$${Find}\:{the}\:{moment}\:{of}\:{inertia}\:{and} \\ $$$${the}\:{radius}\:{of}\:{gyration}\:{of}\:{a}\: \\ $$$${circular}\:{plate}\:{about}\:{an}\:{axis}\:{through} \\ $$$${its}\:{centre},{perpendicular}\:{to}\:{the} \\ $$$${plane}\:{of}\:{the}\:{plate}. \\ $$
Answered by mrW1 last updated on 22/Dec/17
ρ=(M/(πR^2 ))  I=∫_0 ^R 2πrρr^2 dr=2πρ(R^4 /4)=((πR^2 ρR^2 )/2)=((MR^2 )/2)  r_g =(√(I/M))=(R/( (√2)))
$$\rho=\frac{{M}}{\pi{R}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{{R}} \mathrm{2}\pi{r}\rho{r}^{\mathrm{2}} {dr}=\mathrm{2}\pi\rho\frac{{R}^{\mathrm{4}} }{\mathrm{4}}=\frac{\pi{R}^{\mathrm{2}} \rho{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${r}_{{g}} =\sqrt{\frac{{I}}{{M}}}=\frac{{R}}{\:\sqrt{\mathrm{2}}} \\ $$
Answered by ajfour last updated on 21/Dec/17
I=∫r^2 dm =∫_0 ^(  R) r^2 ((m/(πR^2 )))(2πrdr)   =((2m)/R^2 )((R^( 4) /4)) = ((mR^( 2) )/2)   r_g =(√(I/m)) =(R/( (√2))) .
$${I}=\int{r}^{\mathrm{2}} {dm}\:=\int_{\mathrm{0}} ^{\:\:{R}} {r}^{\mathrm{2}} \left(\frac{{m}}{\pi{R}^{\mathrm{2}} }\right)\left(\mathrm{2}\pi{rdr}\right) \\ $$$$\:=\frac{\mathrm{2}{m}}{{R}^{\mathrm{2}} }\left(\frac{{R}^{\:\mathrm{4}} }{\mathrm{4}}\right)\:=\:\frac{{mR}^{\:\mathrm{2}} }{\mathrm{2}}\: \\ $$$${r}_{{g}} =\sqrt{\frac{{I}}{{m}}}\:=\frac{{R}}{\:\sqrt{\mathrm{2}}}\:. \\ $$

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