A) 6
B) 10
C) 12
D) 4
Correct Answer: B
Solution :
Key Idea Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement. By definition \[\alpha =\frac{d\omega }{dt}\] \[ie,\] \[d\omega =\alpha \,\,dt\] So, if in time \[t\] the angular speed of a body changes from \[{{\omega }_{0}}\] to\[\omega \] \[\int_{{{\omega }_{0}}}^{\omega }{d\omega =}\int_{0}^{t}{\alpha \,\,dt}\] If \[\alpha \] is constant \[\omega -{{\omega }_{0}}=\alpha \,\,t\] or \[\omega ={{\omega }_{0}}+\alpha \,t\] ... (i) Now, as by definition \[\omega =\frac{d\theta }{dt}\] Eq. (i) becomes \[\frac{d\theta }{dt}={{\omega }_{0}}+\alpha \,t\] \[ie,\] \[d\theta =({{\omega }_{0}}+\alpha \,t)dt\] So, if in time \[t\] angular displacement is\[\theta \]. \[\int_{0}^{\theta }{d\theta =}\int_{0}^{t}{({{\omega }_{0}}+\alpha t)dt}\] 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.
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