Question

A spinning top has an angular retardation of $\alpha =k{\omega }^2$ (in ${\text{rad/s}}^2$), where $\omega$ is angular velocity of the top and $k=1{\text{rad}}^{-1}$. At ${\theta }_0=0\text{rad}$, if ${\omega }_0=120\text{rad/s}$, then the relation between angular displacement ($\theta$) and angular velocity is

• A. $\omega =120e^{-\theta }$
• B. $\omega =120e^{\theta }$
• C. $\omega =60e^{-\theta }$
• D. $\omega =60e^{\theta }$

Answer 1

The correct option is A $\omega =120e^{-\theta }$

Initial angular velocity $\left({\omega }_0\right)=120\text{rad/s}$
Angular acceleration, $\alpha =k{\omega }^2={\omega }^2$
($k=1$)
To find relation between $\omega$ and $\theta$, we use $\alpha =-\omega \frac{d\omega }{d\theta }$ (negative sign due to retardation)
$\Rightarrow -\omega \frac{d\omega }{d\theta }={\omega }^2$
$\Rightarrow -\frac{d\omega }{\omega }=d\theta$
Integrating on both sides
${\int }_{{\omega }_0}^{\omega }\frac{d\omega }{\omega }=-{\int }_{{\theta }_0}^{\theta }d\theta$
$\Rightarrow {\begin{array}{c}\ln \omega \end{array}|}_{{\omega }_0}^{\omega }=-(\theta -{\theta }_0)$
$\Rightarrow \ln \left(\frac{\omega }{{\omega }_0}\right)=-(\theta -{\theta }_0)$
$\Rightarrow \omega ={\omega }_0{e}^{-(\theta -{\theta }_0)}$
$\Rightarrow \omega ={\omega }_0{e}^{-\theta }=120e^{-\theta }$
(because ${\theta }_0=0$)