Category: Continuum Mechanics *

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201902200820 Derivation 1.2

Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation

M^2R^2=M \displaystyle{\sum_i}m_ir_i^2-\displaystyle{\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2}.

Solution.

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

\displaystyle{M\mathbf{R}}=\sum m_i \mathbf{r}_i.

Taking squares on both sides,

\displaystyle{M^2\mathbf{R}^2}=\sum_{i,j}m_im_j\mathbf{r}_i\cdot \mathbf{r}_j.

Notice \mathbf{r}_{ij}=\mathbf{r}_i-\mathbf{r}_j, squaring it,

r_{ij}^2=r_i^2-2\mathbf{r}_i\cdot \mathbf{r_j}+r_j^2,

arranging,

\displaystyle{\mathbf{r}_i}\cdot \mathbf{r}_j=\frac{1}{2}(r_i^2+r_j^2-r_{ij}^2).

Plugging this,

\begin{aligned} \displaystyle{M^2R^2} & =\frac{1}{2}\sum_{i,j}m_im_jr_i^2+\frac{1}{2}\sum_{i,j}m_im_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{\frac{1}{2}M\sum_im_ir_i^2}+\frac{1}{2}M\sum_jm_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{M\sum_im_ir_i^2}-\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2 \\ \end{aligned}

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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}