February 20, 2019 – Physicspupil

February 20, 2019January 19, 2023

201902200847 Exercise 11 Chapter 2

When two billiard balls collide, the instantaneous forces between them are very large but act only in an infinitesimal time $\Delta t$ , in such a manner that the quantity

$\displaystyle{\int_{\Delta t}}F\,\mathrm{d}t$

remains finite. Such forces are described as impulsive forces, and the integral over $\Delta t$ is known as the impulse of the force. Show that if impulsive forces are present Lagrange’s equations may be transformed into

$\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{q}_j}} \bigg)_f-\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{q}_j}} \bigg)_i=S_j$ ,

where the subscripts $i$ and $f$ refer to the state of the system before and after the impulse, $S_j$ is the impulse of the generalized impulsive force corresponding to $q_j$ , and $\mathcal{L}$ is the Lagrangian including all the non-impulsive forces.

Solution.

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

For billiard-balls collision, the Euler-Lagrange (E-L) equation is

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)-\bigg( \frac{\partial L}{\partial q_j}\bigg)=Q_j$ ,

where $Q_j$ is the generalised impulsive force corresponding to $q_j$ and not derivable from the potential.

Taking integral over $\Delta t$ on both sides,

LHS becomes

$\displaystyle{\int_{\Delta t}\frac{\displaystyle{\mathrm{d}\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)}}{\mathrm{d}t}\,\mathrm{d}t}-\int_{\Delta t}\bigg( \frac{\partial L}{\partial q_j}\bigg) \,\mathrm{d}t$ .

The second term upon integration is zero,

because $\displaystyle{\bigg( \frac{\partial L}{\partial q_j}}\bigg) \Delta t=0$ for infinitesimal time $\Delta t\to 0$ .

The first term is

$\displaystyle{\int_{\Delta t}\mathrm{d}\bigg( \frac{\partial L}{\partial \dot{q_j}} \bigg)}=\bigg[ \frac{\partial L}{\partial \dot{q_j}}\bigg]^{t+\Delta t}_{t}$ .

Rename $t+\Delta t$ the (final) state of system $f$ after the impulse and $t$ the (initial) state of system $i$ before the impulse.

LHS reads

$\displaystyle{\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_f}-\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_i$

whereas RHS reads

$\displaystyle{\int_{\Delta t}Q_j \,\mathrm{d}t=S_j}$ ,

i.e., the impulse of generalised impulsive force.

The transformed E-L equation in the presence of impulsive forces is

$\displaystyle{\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_f}-\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_i=S_j$ ,

as desired.

February 20, 2019January 19, 2023

201902200839 Exercise 5 Chapter 2

A particle is subjected to the potential $V(x)=-Fx$ , where $F$ is a constant. The particle travels from $x=0$ to $x=a$ in a time interval $t_0$ . Assume the motion of the particle can be expressed in the form $x(t)=A+Bt+Ct^2$ . Find the values of $A$ , $B$ , and $C$ such that the action is a minimum.

Solution.

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

1D-case:

$\mathcal{L}=T-V=\displaystyle{\frac{1}{2}}m\dot{x}^2+Fx$ .

Euler-Lagrange (E-L) equation:

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{x}}}\bigg) =\frac{\partial \mathcal{L}}{\partial x}$

gives the path over which the action is stationary.

That is,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}( m\dot{x}) & = F \\ \Rightarrow \ddot{x} & =\displaystyle{\frac{F}{m}}\\ \end{aligned}$ .

On $x(t)=A+Bt+Ct^2$ taking derivative twice,

$\ddot{x}(t)=2C$ .

Equate them,

$C=\displaystyle{\frac{F}{2m}}$ .

The event $(x=0,t=0)$ gives

$A=0$ .

And the event $(x=a,t=t_0)$ gives

$a=x(t_0)=Bt_0+\displaystyle{\frac{F}{2m}}t_0^2$ .

Express it as

$B=\displaystyle{\frac{a-\frac{F}{2m}t_0^2}{t_0}}$ .

Thus,

$x(t)=(0)+\Bigg( \displaystyle{\frac{a-\frac{F}{2m}t_0^2}{t_0}}\Bigg) t+\bigg( \displaystyle{\frac{F}{2m}}\bigg) t^2$

is recovered.

February 20, 2019January 17, 2023

201902200829 Derivation 2.3

Prove that the shortest distance between two points in space is a straight line.

Solution.

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

Assume the path (of any curve $C$ ) connecting two points $(a,y(a))$ and $(b,y(b))$ is given by a function $C(x)=(x,y(x))$ , with $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}C(x)=(1,y')$ being the first derivative of the curve.

To minimise the path distance

$\displaystyle{\mathcal{L}=\int \| C'\| \,\mathrm{d}x=\int_a^b\sqrt{1+y'^2}\,\mathrm{d}x}$ ,

define now

$f(x,y,y')=\sqrt{1+y'^2}$ ,

having $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y}}=0$ and $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y'}}=\frac{y'}{\sqrt{1+y'^2}}$ .

From Euler-Lagrange (E-L) equation it follows that

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}\bigg( \displaystyle{\frac{y'}{\sqrt{1+y'^2}}}\bigg) =0$ ,

i.e., $y'=\mathrm{Const.}$

In conclusion, the shortest distance between two points in space is a straight line.

Lemma. (Fundamental lemma of the calculus of variations)

If $\displaystyle{\int_{x_1}^{x_2}M(x)\eta (x)\,\mathrm{d}x=0}$ for any $\eta (x)$ continuous through second derivative, then $M(x)$ must identically vanish in the interval $x_1,x_2$ .

_{Text on pg.38, Goldstein}

February 20, 2019January 17, 2023

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,

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

February 20, 2019January 17, 2023

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.

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

M	T	W	T	F	S	S
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