Question Number 217238 by Tawa11 last updated on 06/Mar/25
Answered by mr W last updated on 07/Mar/25
$$\alpha=\frac{{a}}{{R}} \\ $$$${I}=\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${Mg}−{T}={Ma}\:\:\:…\left({i}\right) \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\alpha={TR}\:\:\:…\left({ii}\right) \\ $$$${T}={M}\left({g}−{a}\right)=\frac{{Ma}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}{g}}{\mathrm{3}}\:\:\:\checkmark \\ $$$$\Rightarrow{T}=\frac{{Mg}}{\mathrm{3}}\:\:\checkmark \\ $$
Commented by Tawa11 last updated on 07/Mar/25
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Wuji last updated on 07/Mar/25
$$\Sigma\mathrm{F}=\mathrm{mg}−\mathrm{T}=\mathrm{ma} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{MR}^{\mathrm{2}} \\ $$$$\tau=\mathrm{T}×\mathrm{R}\:\Rightarrow\tau=\mathrm{I}\alpha\:\Rightarrow\mathrm{TR}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{MR}^{\mathrm{2}} \alpha \\ $$$$\alpha=\frac{\mathrm{2T}}{\mathrm{MR}} \\ $$$$\mathrm{a}=\alpha\mathrm{R}\:=\left(\frac{\mathrm{2T}}{\mathrm{MR}}\right)\mathrm{R}=\frac{\mathrm{2T}}{\mathrm{M}} \\ $$$$\mathrm{mg}−\mathrm{T}=\mathrm{ma} \\ $$$$\mathrm{mg}−\mathrm{T}=\mathrm{M}\left(\frac{\mathrm{2T}}{\mathrm{M}}\right)=\mathrm{2T}\:\Rightarrow\mathrm{mg}=\mathrm{3T} \\ $$$$\mathrm{T}=\frac{\mathrm{mg}}{\mathrm{3}} \\ $$$$\mathrm{a}=\frac{\mathrm{2T}}{\mathrm{M}}=\frac{\mathrm{2}}{\mathrm{M}}\left(\frac{\mathrm{Mg}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{g} \\ $$$$\therefore\:\mathrm{acceleration}\:\mathrm{a}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{g}\:\left(\downarrow\right) \\ $$$$\mathrm{Tension}\:\mathrm{T}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{mg} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 08/Mar/25
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$