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 \[\alpha =\frac{d\omega }{dt}\] |
| i.e., \[d\omega =\alpha \,dt\] |
| 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 |
| \[\frac{d\theta }{dt}={{\omega }_{0}}+\alpha \,t\] |
| i.e., \[d\theta =({{\omega }_{0}}+\alpha \,t)\,dt\] |
| 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, \[\alpha =3.0\,rd/{{s}^{2}},\,{{\omega }_{0}}=2.0\,\,rad/s,\,\,t=2\,s\] |
| 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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