\[y^2=x^3\]
. 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
\[\alpha = \frac{d \omega}{dt}= \frac{d}{dt} ( \frac{1}{\sqrt{x}(1+x)})=- \frac{1+3x}{2 \sqrt{x} (1+x)} \frac{dx}{dt}= - \frac{1+3x}{\sqrt{x} (1+x)}\]
.