Discussion
A particle of mass m is constrained to move on a horizontal xy-plane which is rotating about the vertical z-axis with angular velcoity \(\omega\). If T is the total kinetic energy of the particle, then \(\dot{x}\frac{\partial T}{\partial \dot{x}}\) + \(\dot{y}\frac{\partial T}{\partial \dot{y}}\) is equal to :
where \(T = T_{2} + T_{1} + T_{0}\),
\[ \begin{aligned} T_{2} &= \tfrac{1}{2}m(\dot{x}^{2} + \dot{y}^{2}), \\ T_{1} &= m\omega(x\dot{y} - y\dot{x}), \\ T_{0} &= \tfrac{1}{2}m\omega^{2}(x^{2} + y^{2}) \end{aligned} \]
Answer: B
The sum equals \(2T_{2} + T_{1}\).
T splits into parts that are homogeneous in the velocities: \(T_{2}\) of degree 2, \(T_{1}\) of degree 1 and \(T_{0}\) of degree 0. By Euler's theorem on homogeneous functions, \(\sum \dot{q}\,\frac{\partial T_{k}}{\partial \dot{q}} = k\,T_{k}\). Hence
\[ \begin{aligned} &\dot{x}\frac{\partial T}{\partial \dot{x}} + \dot{y}\frac{\partial T}{\partial \dot{y}} \\ &= 2T_{2} + T_{1} + 0 \cdot T_{0} \end{aligned} \]
Check on \(T_{1} = m\omega(x\dot{y} - y\dot{x})\): \(\dot{x}(-m\omega y) + \dot{y}(m\omega x) = T_{1}\).
2T would be correct only if T were purely quadratic in the velocities, which fails in a rotating frame. Option C wrongly counts \(T_{0}\), which has no velocity dependence.
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