A machine part rotates at an angular speed of 0.057 rad/s; its speed is then increased to 2.0 rad/s at an angular acceleration of 0.60 rad/s2. (a) Find the angle through which the part rotates before reaching this final speed. rad (b) If both the initial and final angular speeds are doubled and the angular acceleration remains the same, by what factor is the angular displacement changed? by a factor of 2 by a factor of 4 by a factor of 16 The angular displacement doesn't change. Why?

If [tex]\omega_{i}[/tex] , [tex]\omega_{f}[/tex], [tex]\alpha[/tex] and [tex]\delta \theta[/tex] represents the initial angular velocity, the final angular velocity , the angular acceleration and the change of angular displacement s, then from angular kinematic equation, we can write

[tex]\omega_{f}^{2} = \omega_{i}^{2} + 2~\alpha~\delta \theta[/tex]

Part(a):

Given, [tex]\omega_{i} = 0.057~rad~s^{-1}[/tex], [tex]\omega_{f} = 2.0~rad~s^{-1}[/tex] and [tex]\alpha = 0.60~rad~s^{-2}[/tex]. Substituting ll the values in the above equation, we have

[tex]&& 2^{2} = 0.057^{2} + 2 \times 0.6 \times \delta \theta\\&or,& \delta \theta = \dfrac{4 - 0.057^{2}}{1.2} = 3.33~rad[/tex]

So the angle through which the parts rotates before reaching the final speed is [tex]3.33~rad[/tex].

Part(b):

Given, [tex]\omega_{i}^{new} = 2~\omega_{i}[/tex], [tex]\omega_{f}^{new} = 2~\omega_{f}[/tex] and [tex]\alpha^{new} = \alpha[/tex]. So,

[tex]&& (\omega_{f}^{new})^{2} = (\omega_{i}^{new})^{2} + 2~\alpha^{new} \delta \theta^{new}\\&or,& \delta \theta^{new} = \dfrac{( (\omega_{f}^{new})^{2} - (\omega_{i}^{new})^{2})}{2 \alpha}\\&or,& \delta \theta^{new} = \dfrac{4(\omega_{f}^{2} - \omega_{i}^{2})}{2~\alpha}\\&or,& \delta \theta^{new} = 2 \delta \theta[/tex]

Therefore, the angular displacement will be changes by a factor of 2.