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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

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

Commented by Tawa11 last updated on 06/Mar/25

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

Answered by Wuji last updated on 06/Mar/25

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

Commented by Tawa11 last updated on 06/Mar/25

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

Answered by MrGaster last updated on 07/Mar/25

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

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