Question

A solid body rotates about a stationary axis so that its angular velocity depends on the rotational angle $\varphi$ as $\omega ={\omega }_0-k\varphi$ where ${\omega }_0$ and $k$ are positive constants. At the moment $t=0,\varphi =0,$ find the time dependence of rotation angle

• A. $k{\omega }_0{e}^{-kt}$
• B. $\frac{{\omega }_0}{k}{e}^{-kt}$
• C. $\frac{{\omega }_0}{k}\left(1-e^{-kt}\right)$
• D. $\frac{k}{{\omega }_0}\left(e^{-kt}-1\right)$

Answer 1

Answer: (C)

$\because \omega ={\omega }_0-k\varphi$ or $\frac{d\varphi }{dt}={\omega }_0-k\varphi$
Or ${\int }_0^{\varphi }\frac{d\varphi }{{\omega }_0-k\varphi }={\int }_0^{t}dt$
Or $-\frac{1}{k}{\left[\ln \left({\omega }_0-k\varphi \right)\right]}_0^{\varphi }=t$
Or $\ln \left(\frac{{\omega }_0-k\varphi }{{\omega }_0}\right)=-kt$
$\therefore \varphi =\frac{{\omega }_0}{k}\left(1-e^{-kt}\right)$