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Question Number 217228 by Tawa11 last updated on 06/Mar/25

Answered by mr W last updated on 06/Mar/25

I=((mr^2 )/2) KE=((Iω^2 )/2)+((m(rω)^2 )/2)=((Iω^2 )/2)+Iω^2 =((3Iω^2 )/2)

$${I}=\frac{{mr}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${KE}=\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}}+\frac{{m}\left({r}\omega\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}}+{I}\omega^{\mathrm{2}} =\frac{\mathrm{3}{I}\omega^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 06/Mar/25

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by Wuji last updated on 06/Mar/25

v=ωR K_(trans) =(1/2)Mv^2 =(1/2)M(ωR)^2 K_(rot) =(1/2)Iω^2 K_(total) =K_(trans) +K_(rot) =(1/2)M(ωR)^2 +(1/2)Iω^2 K_(total) =(1/2)(Mω^2 R^2 +Iω^2 )=(1/2)ω^2 (MR^2 +I) moment of inertial about its central axis I=(1/2)MR^2 ⇒MR^2 =2I K_(total) =(1/2)ω^2 (2I+I)=(1/2)ω^2 (3I) K_(total) =(3/2)Iω^2 ✓

$$\mathrm{v}=\omega\mathrm{R} \\ $$$$\mathrm{K}_{\mathrm{trans}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}\left(\omega\mathrm{R}\right)^{\mathrm{2}} \\ $$$$\mathrm{K}_{\mathrm{rot}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\omega^{\mathrm{2}} \\ $$$$\mathrm{K}_{\mathrm{total}} =\mathrm{K}_{\mathrm{trans}} +\mathrm{K}_{\mathrm{rot}} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}\left(\omega\mathrm{R}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\omega^{\mathrm{2}} \\ $$$$\mathrm{K}_{\mathrm{total}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{M}\omega^{\mathrm{2}} \mathrm{R}^{\mathrm{2}} +\mathrm{I}\omega^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} \left(\mathrm{MR}^{\mathrm{2}} +\mathrm{I}\right) \\ $$$$\mathrm{moment}\:\mathrm{of}\:\mathrm{inertial}\:\mathrm{about}\:\mathrm{its}\:\mathrm{central}\:\mathrm{axis} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{MR}^{\mathrm{2}} \:\:\Rightarrow\mathrm{MR}^{\mathrm{2}} =\mathrm{2I} \\ $$$$\mathrm{K}_{\mathrm{total}} =\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} \left(\mathrm{2I}+\mathrm{I}\right)=\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} \left(\mathrm{3I}\right) \\ $$$$\mathrm{K}_{\mathrm{total}} =\frac{\mathrm{3}}{\mathrm{2}}\mathrm{I}\omega^{\mathrm{2}} \checkmark \\ $$

Commented by Tawa11 last updated on 06/Mar/25

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by MrGaster last updated on 07/Mar/25

K.E.=(1/2)Iω^2 +(1/2)mv^2 v=ωR K.E.=(1/2)Iω^2 +(1/2)m(ωR)^2 K.E.=(1/2)Iω^2 +(1/2)mR^2 ω^2 ∵I=(1/2)mR^2 for a solid cylimder K.E.=(1/2)ω^2 ((1/2)mR^2 +mR^2 ) K.E.=(1/2)ω^2 ((3/2)mR^2 ) K.E.=(3/4) mR^2 ω^2 K.E.=(3/2)((1/2)mR^2 ω^2 ) K.E.= determinant ((((3/2)Iω^2 )))

$${K}.{E}.=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$${v}=\omega{R} \\ $$$${K}.{E}.=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}\left(\omega{R}\right)^{\mathrm{2}} \\ $$$${K}.{E}.=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mR}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$\because{I}=\frac{\mathrm{1}}{\mathrm{2}}{mR}^{\mathrm{2}} \mathrm{for}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{cylimder} \\ $$$${K}.{E}.=\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{mR}^{\mathrm{2}} +{mR}^{\mathrm{2}} \right) \\ $$$${K}.{E}.=\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}{mR}^{\mathrm{2}} \right) \\ $$$${K}.{E}.=\frac{\mathrm{3}}{\mathrm{4}}\:{mR}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$${K}.{E}.=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}{mR}^{\mathrm{2}} \omega^{\mathrm{2}} \right) \\ $$$${K}.{E}.=\begin{array}{|c|}{\frac{\mathrm{3}}{\mathrm{2}}{I}\omega^{\mathrm{2}} }\\\hline\end{array} \\ $$

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