Question Number 55375 by Tawa1 last updated on 23/Feb/19
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
$${pls}\:{specify}\:{the}\:{point}\:{of}\:{applicstion}\:{of}\:{force} \\ $$$${on}\:{axle}... \\ $$
Answered by mr W last updated on 23/Feb/19
$${I}\alpha={Fr}_{{A}} \\ $$$$\Rightarrow\alpha=\frac{{Fr}_{{A}} }{{I}} \\ $$$$\omega=\alpha{t}=\frac{{Fr}_{{A}} {t}}{{I}} \\ $$$${KE}=\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\left({Fr}_{{A}} {t}\right)^{\mathrm{2}} }{\mathrm{2}{I}}=\frac{\left({Fr}_{{A}} {t}\right)^{\mathrm{2}} }{\mathrm{2}×\frac{{mr}_{{D}} ^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{{m}}\left(\frac{{Ftr}_{{A}} }{{r}_{{D}} }\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}\left(\frac{\mathrm{12}×\mathrm{1}.\mathrm{2}×\mathrm{1}.\mathrm{5}}{\mathrm{15}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{0}.\mathrm{104}\:{J} \\ $$
Commented by Tawa1 last updated on 23/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$
Commented by Tawa1 last updated on 23/Feb/19
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{see}\:\mathrm{question}\:\:\mathrm{55412},\:\:\:\mathrm{help}\:\mathrm{me}\:\mathrm{use}\:\:\mathrm{lambert}\:\mathrm{W}\:\mathrm{function}. \\ $$
Commented by mr W last updated on 23/Feb/19
Commented by mr W last updated on 23/Feb/19
$${it}\:{can}\:{not}\:{be}\:{solved}\:{analytically}. \\ $$