Angular Velocity and Acceleration Along Curve

Suppose a particle moves along the curve  
\[y^2=x^3\]
. If the horizontal velocity is constant and equal to 2 units/s, find the angular velocity  
\[\omega\]
  and the angular acceleration  
\[\alpha\]
.
The particle is at a point  
\[\theta = tan^{-1} (y/x)=tan^{-1}(x^{3/2}/x)=tan^{-1}(\sqrt{x})\]
  anticlockwise from the  
\[x\]
  axis, and the angular velocity is  
\[\omega = \frac{d \theta }{dt} =\frac{d}{dt}(tan^{-1}(\sqrt{x}))= \frac{1}{1+x} \frac{1}{2 \sqrt{x}} \frac{dx}{dt}= \frac{1}{\sqrt{x}(1+x)}\]
.
The angular acceleration is  
\[\alpha = \frac{d \omega}{dt}= \frac{d}{dt} ( \frac{1}{\sqrt{x}(1+x)})=- \frac{1+3x}{2 \sqrt{x} (1+x)} \frac{dx}{dt}= - \frac{1+3x}{\sqrt{x} (1+x)}\]
.

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