Khimraj
Last Activity: 6 Years ago
Angular displacement in 2n rotations is4n\pi.Hence from equation of motion,\omega^2-\omega_0^2=2\theta \alpha\implies \alpha=\dfrac{50^2-100^2}{2(2n\pi)}=-\dfrac{7500}{8n\pi}Hence angular velocity after n rotations=\sqrt{\omega_0^2+2\alpha\theta_2}=\sqrt{100^2-2\dfrac{7500}{8n\pi}(2n\pi)}=79rad/sec