Question

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

• A. $k\omega _{0} \, e^{- k t}$
• B. $\frac{\omega _{0}}{k} \, e^{- k t}$
• C. $\frac{\omega_0}{k}\left(1-e^{-k t}\right)$
• D. $\frac{k}{\omega_0}\left(e^{-k t}-1\right)$

Answer 1

Answer: (C)

$\because \omega=\omega_{0}-k \phi$ or $\frac{d \phi}{d t}=\omega_{0}-k \phi$
Or $\int_{0}^{\phi} \frac{d \phi}{\omega_{0}-k \phi}=\int_{0}^{t} d t$
Or $-\frac{1}{k}\left[\ln \left(\omega_{0}-k \phi\right)\right]_{0}^{\phi}=t$
Or $\ln \left(\frac{\omega_{0}-k \phi}{\omega_{0}}\right)=-k t$
$\therefore \phi=\frac{\omega_{0}}{k}\left(1-e^{-k t}\right)$