Undergrad Time dependence of kinetic energy in Lagrangian formulation

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Kinetic energy can depend explicitly on time within the Lagrangian formulation if the inertial Cartesian coordinates are functions of generalized coordinates that also vary with time. This scenario typically arises in non-inertial reference frames. The relationship is established through the time derivative of position, which incorporates both the generalized coordinates and their time derivatives. Consequently, the expression for kinetic energy becomes explicitly time-dependent. This highlights the complexity of kinetic energy in non-inertial frames within Lagrangian mechanics.
Ahmed1029
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Could kinetic energy possibly depend explicitly on time in the lagrangian for some arbitrary set of generalized coordinates?
 
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Yes, if the inertial Cartesian coordinates as functions of the generalized coordinates depend explicitly on time (describing the motion in a non-inertial frame of reference) you get from
$$\vec{x}=\vec{x}(q^k,t), \quad k \in \{1,\ldots,f \}$$
the time derivative (Einstein summation convention applies)
$$\dot{\vec{x}}=\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x}$$
and thus
$$T=\frac{m}{2} \dot{\vec{x}}^2 = \frac{m}{2} \left (\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x} \right)^2,$$
which is in general explicitly time dependent.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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