Import Question JSON

\text{Given: Kinetic energy } K = \beta s\text{, where } s \text{ is the distance travelled} \text{We know that } K = \frac{1}{2}Mv^2\text{, so } \frac{1}{2}Mv^2 = \beta s \text{Therefore, } v^2 = \frac{2\beta s}{M} \text{For circular motion, the relationship between linear distance } s \text{ and angular displacement } \theta \text{ is } s = r\theta \text{The angular velocity } \omega = \frac{v}{r}\text{, so } v = r\omega \text{Substituting: } (r\omega)^2 = \frac{2\beta s}{M} \text{Since } s = r\theta\text{, we get } r^2\omega^2 = \frac{2\beta r\theta}{M} \text{Therefore, } \omega^2 = \frac{2\beta \theta}{Mr} \text{To find angular acceleration } \alpha = \frac{d\omega}{dt}\text{, we differentiate both sides with respect to time:} 2\omega \frac{d\omega}{dt} = \frac{2\beta}{Mr} \frac{d\theta}{dt} \text{Since } \frac{d\theta}{dt} = \omega\text{, we have } 2\omega \alpha = \frac{2\beta \omega}{Mr} \text{Therefore, } \alpha = \frac{\beta}{Mr} \text{The correct answer is option 4: } \frac{\beta}{Mr}