201902200814 Derivation 1.1

Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy:

$\displaystyle{\frac{\mathrm{d}T}{\mathrm{d}t}}=\mathbf{F}\cdot\mathbf{v}$ ,

while if the mass varies with time the corresponding equation is

$\displaystyle{\frac{\mathrm{d}(mT)}{\mathrm{d}t}}=\mathbf{F}\cdot\mathbf{p}$ .

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

Single particle with constant mass:

$\begin{aligned} \frac{\mathrm{d}T}{\mathrm{d}t} & =\frac{\mathrm{d}\big(\frac{1}{2}mv^2\big)}{\mathrm{d}t} \\ & =m\mathbf{v}\cdot \mathbf{\dot{v}} \\ & =m\mathbf{a}\cdot \mathbf{v} \\ & =\mathbf{F}\cdot \mathbf{v} \end{aligned}$

if mass varies with time:

$\begin{aligned} \frac{\mathrm{d}(mT)}{\mathrm{d}t} & =\frac{\mathrm{d}\big( m\times \frac{1}{2}mv^2\big)}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{(mv)^2}{2}\bigg) \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{p^2}{2}\bigg) \\ & =\mathbf{p}\cdot \mathbf{\dot{p}} \\ & =\mathbf{F}\cdot \mathbf{p} \end{aligned}$

Remark.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}v^2 & =\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot \mathbf{v}) \\ & =\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\cdot \mathbf{v}+\mathbf{v} \cdot \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} \\ & =2\mathbf{v}\cdot \mathbf{a} \end{aligned}$

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