Question Number 35174 by ajfour last updated on 16/May/18
Commented by ajfour last updated on 16/May/18
$${Q}.\mathrm{35136}\:\:\:\left({solution}\right) \\ $$
Answered by ajfour last updated on 16/May/18
$$\omega=\omega_{\mathrm{0}} −\frac{{fRt}}{\frac{\mathrm{2}}{\mathrm{5}}{mR}^{\mathrm{2}} } \\ $$$${v}=\frac{{ft}}{{m}}\:,\:\:\:\:{V}=−\frac{{ft}}{{m}} \\ $$$${when}\:{pure}\:{rolling}\:{on}\:{plank}\:{begins} \\ $$$$\:\:\:\:\:\:\:−\omega{R}+{v}={V}\:=\:\frac{−{ft}}{{m}} \\ $$$$\Rightarrow\:\:\:\:−\omega_{\mathrm{0}} {R}+\frac{\mathrm{5}{ft}}{\mathrm{2}{m}}\:+\frac{{ft}}{{m}}\:=\:−\frac{{ft}}{{m}} \\ $$$$\:\:\:\:\:\frac{\mathrm{9}{ft}}{\mathrm{2}{m}}\:=\:\omega_{\mathrm{0}} {R}\:\:\:{with}\:\:{f}=\mu{mg} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{t}}\:=\:\frac{\mathrm{2}\boldsymbol{\omega}_{\mathrm{0}} \boldsymbol{{R}}}{\mathrm{9}\boldsymbol{\mu{g}}}\:. \\ $$
Commented by Tinkutara last updated on 16/May/18
Thank you very much Sir! I got the answer. ��������