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Question Number 69878 by jagannath19 last updated on 28/Sep/19

Commented by mr W last updated on 29/Sep/19

(3) I

$$\left(\mathrm{3}\right)\:{I} \\ $$

Commented by jagannath19 last updated on 29/Sep/19

sir can you please tell how

$${sir}\:{can}\:{you}\:{please}\:{tell}\:{how} \\ $$

Commented by mr W last updated on 29/Sep/19

circular ring of mass M and radius R  has the MoI=I_(circular ring)   I_(circular ring) =∫r^2 dm=R^2 ∫dm=R^2 M=I    MoI of semi−circular ring=I_(semi)   I_(semi) =∫r^2 dm=R^2 ∫dm=R^2 M=I    both rings have the same MoI.  this is easy to see: they have the  same mass M, their mass has the  same distance to the reference axis,  MoI is the product of mass and square  of distance, therefore their MoI is  the same.

$${circular}\:{ring}\:{of}\:{mass}\:{M}\:{and}\:{radius}\:{R} \\ $$$${has}\:{the}\:{MoI}={I}_{{circular}\:{ring}} \\ $$$${I}_{{circular}\:{ring}} =\int{r}^{\mathrm{2}} {dm}={R}^{\mathrm{2}} \int{dm}={R}^{\mathrm{2}} {M}={I} \\ $$$$ \\ $$$${MoI}\:{of}\:{semi}−{circular}\:{ring}={I}_{{semi}} \\ $$$${I}_{{semi}} =\int{r}^{\mathrm{2}} {dm}={R}^{\mathrm{2}} \int{dm}={R}^{\mathrm{2}} {M}={I} \\ $$$$ \\ $$$${both}\:{rings}\:{have}\:{the}\:{same}\:{MoI}. \\ $$$${this}\:{is}\:{easy}\:{to}\:{see}:\:{they}\:{have}\:{the} \\ $$$${same}\:{mass}\:{M},\:{their}\:{mass}\:{has}\:{the} \\ $$$${same}\:{distance}\:{to}\:{the}\:{reference}\:{axis}, \\ $$$${MoI}\:{is}\:{the}\:{product}\:{of}\:{mass}\:{and}\:{square} \\ $$$${of}\:{distance},\:{therefore}\:{their}\:{MoI}\:{is} \\ $$$${the}\:{same}. \\ $$

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