Physics - Rotation Question with Solution | TestHub

Angle turned in three second and ${\theta }_{3}s=2\pi \times 10$ $=20\pi \mathrm{rad}$ ${\theta }_1={\omega }_{0}t+\frac{1}{2}\alpha t^2$ $\alpha =\frac{49\pi }{9}\mathrm{rad}/{\mathrm{s}}^2$ Now, angle turned in 6 s from the starting ${\theta }_2={\omega }_{0}t+\frac{1}{2}\alpha t^2$ $=0+\frac{1}{2}\times \left(\frac{40\pi }{9}\right)+(6{)}^2$ $=80\pi \mathrm{rad}$ Angle turned between $t=3\mathrm{s}$ to $t=6\mathrm{s}$ ${\theta }_{\text{lost }}={\theta }_2-{\theta }_3=80\pi -20\pi$ $=60\pi$ Number of revolution $=\frac{60\pi }{2\pi }=30$