Question

A wheel has angular acceleration of $ 3.0\,rad/s^{2} $ and an initial angular speed of $2.00 \,rad/s$. In a time of $2 \,s$ it has rotated through an angle (in radian) of

• A. 6
• B. 10
• C. 12
• D. 4

Answer 1

The Correct Option is B

Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement. By definition $ \alpha=\frac{d \omega}{d t}$ ie, $ d \omega=\alpha d t$ So, if in time $t$ the angular speed of a body changes from $\omega_{0}$ to $\omega$ $\int\limits_{\omega_{0}}^{\omega} d \omega=\int\limits_{0}^{t} \alpha d t$ If $\alpha$ is constant $\omega-\omega_{0} =\alpha t $ or $ \omega =\omega_{0}+\alpha t$ ... (i) Now, as by definition $\omega=\frac{d \theta}{d t}$ E (i) becomes $\frac{d \theta}{d t}=\omega_{0}+\alpha t $ $d \theta=\left(\omega_{0}+\alpha t\right) d t$ So, if in the time $t$ angular displacement is $\theta$. $\int\limits_{0}^{\theta} d \theta=\int\limits_{0}^{t}\left(\omega_{0}+\alpha t\right) d t$ or $ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$ ... (ii) Given, $\alpha = 3.0\, rad / s ^{2}$, $\omega_{0} = 2.0 \,rad / s , t = 2 \,s$ Hence, $ \theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2}$ or $ \theta=4+6=10 \,rad$ Note Eqs. (i) and (ii) are similar to first and second equations of linear motion.