Question

if angular velocity of a disc depends an angle rotated THITA as OMEGA =thita sq. + 2thita , then its angular acceleration alpha at thita = 1 is

Answer 1

$$\begin{gathered} Given,angularvelocity,\omega =\frac{d\theta }{dt}={\theta }^2+2\theta where,\theta isangulardisplacement \\ Since,angularacceleration,\alpha =\frac{d\omega }{dt}=\left(\frac{d\omega }{d\theta }\right)\left(\frac{d\theta }{dt}\right) \\ \alpha =\left(\frac{d\omega }{d\theta }\right)\omega \\ \alpha =\left[\frac{d}{d\theta }\left({\theta }^2+2\theta \right)\right]{\theta }^2+2\theta \\ \alpha =\left(2\theta +2\right)\left({\theta }^2+2\theta \right) \\ \alpha =2{\theta }^3+6{\theta }^2+4\theta \\ Putting\theta =1inaboveequation,weget \\ \alpha =12rad{s}^{-2} \end{gathered}$$