Question

Consider motion of a free particle having mass m
in a plane. Express its kinetic energy in terms of
plane polar coordinates and their time derivatives?​

Answer 1

The kinetic energy in terms of planar coordinates and their time derivative is

$\boxed{K=\frac{1}{2}m[(\frac{dx}{dt})^2+(\frac{dy}{dt})^2]}$

Step-by-step explanation:

If the displacement of the particle in the XY plane is given by (x, y)

Then

The displacement vector can be written as

$\vec d=x\hat i+y\hat j$

The velocity vector will be the derivative of displacement vector

$\vec v=\frac{dx}{dt}\hat i+\frac{dy}{dt}\hat j$

The magnitude of velocity is

$v=|\vec v|=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$

Therefore, the kinetic energy of the particle having mass m

$K=\frac{1}{2}m[\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}]^2$

$\implies K=\frac{1}{2}m[(\frac{dx}{dt})^2+(\frac{dy}{dt})^2]$

Hope this answer is helpful.

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