Class Information

Newton’s laws of motion, one-dimensional motion, second order differential equations, harmonic oscillators (damped, forced), vector analysis, conservation laws, three-dimensional motion, central forces, motion in electromagnetic fields, collisions, center-of-mass transformations, two-body problem, numerical/computer solutions, coupled oscillators. Rigid body rotation, statics, elasticity, fluid equilibrium, gravitation.

Textbook: Classical Mechanics by John Taylor

2: Newton’s Laws of Motion

Newton’s Laws in 2D Polar Coordinates

See Wikipedia for an alternative derivation

We have $\hat{\rho }$ and $\hat{\varphi }$ for 2D polar coordinates
$$\overrightarrow{r}=\rho \hat{\rho }$$
Note that the position vector, $\overrightarrow{r}$, does not depend on $\varphi$ explicitly

$$\overrightarrow{v}=\frac{\mathrm{d}\overrightarrow{r}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\rho \hat{\rho }\right)=\rho \frac{\mathrm{d}\hat{\rho }}{\mathrm{d}t}+\dot{\rho }\hat{\rho }$$

Now we find $\frac{\mathrm{d}\hat{r}}{\mathrm{d}t}$
$$\left|\frac{\mathrm{d}\hat{r}}{\mathrm{d}t}\right|=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\hat{\rho }(t+\mathrm{\Delta }t)-\hat{\rho }(t)}{\mathrm{\Delta }t}=\underset{\mathrm{\Delta }\to 0}{lim}\frac{\mathrm{\Delta }\varphi \cdot |\hat{\rho }|}{\mathrm{\Delta }t}=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }\varphi \cdot 1}{\mathrm{\Delta }t}=\frac{d\varphi }{dt}$$

We can find that the direction is in the $\hat{\varphi }$ by looking at a diagram.

$$\frac{\mathrm{d}\hat{r}}{\mathrm{d}t}=\dot{\varphi }\hat{\varphi }$$

$$\overrightarrow{v}=\dot{\rho }\hat{\rho }+r\dot{\varphi }\hat{\varphi }$$

5: Oscillation

Damped Oscillations

Consider a mass attached to a spring which is attached to a wall. Consider horizontal motion only.

$$\overrightarrow{F}=m\overrightarrow{a}$$

• Assume negligible friction from ground
• We have drag from air $F_{fr}=-b\dot{x}$
  $$\overrightarrow{F}=-kx-b\dot{x}$$

$$\ddot{x}+\frac{b}{m}\dot{x}+\frac{k}{m}x=0$$

• Let $\frac{b}{m}=2\beta$, where $\beta$ is the damping constant
• ${\omega }_0=\sqrt{k/m}$ is the natural frequency
  $$\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=0$$
  Trial Solution
  $$x(t)=e^{rt}$$

$$r^2{e}^{rt}+2\beta r{e}^{rt}+{\omega }_0^2{e}^{rt}=0$$
$$r^2+2\beta r+{\omega }_0^2=0$$

Find roots of above

$$r_1=-\beta +\sqrt{{\beta }^2-{\omega }_0^2}$$
$$r_2=-\beta -\sqrt{{\beta }^2-{\omega }_0^2}$$

General solution
$$x(t)=c_1{e}^{r_{1}t}+c_2{e}^{r_{2}t}$$

$$x(t)=e^{-\beta t}[c_1{e}^{t\sqrt{{\beta }^2-{\omega }_0^2}}+c_2{e}^{-t\sqrt{{\beta }^2-{\omega }_0^2}}]$$

Undamped ($\beta =0$)

$$x(t)=e^{-\beta t}[c_1{e}^{t\sqrt{-{\omega }_0^2}}+c_2{e}^{-t\sqrt{-{\omega }_0^2}}]$$
$$x(t)=c_1{e}^{i{\omega }_{0}t}+c_2{e}^{-i{\omega }_{0}t}$$

This is the equation is what we got earlier for a SHO without any force damping

Weak Damping ($\beta <{\omega }_0$)

$$\beta <{\omega }_0⟹\sqrt{{\beta }^2-{\omega }_0^2}=\text{Imaginary}⟹\text{Oscillations}$$

• Define ${\omega }_1=\sqrt{{\omega }_0^2-{\beta }^2}$

$$x(t)=e^{-\beta t}[c_1{e}^{i{\omega }_{1}t}+c_2{e}^{-i{\omega }_{1}t}]$$

• Notice that the solution is very similar to the undamped case except for the $e^{-\beta t}$ and a different frequency, ${\omega }_1<{\omega }_0$
• The $e^{-\beta t}$ term controls the damping and the smaller frequency means longer periods compared to the undamped case

Phase shifted Solution:
$$x(t)=A{e}^{-\beta t}\cos ({\omega }_{1}t-\delta )$$

• Larger $\beta$ means more damping and a quicker decrease in amplitude

Strong Damping ($\beta >{\omega }_0$)

$$\beta >{\omega }_0⟹\sqrt{{\beta }^2-{\omega }_0^2}\in \mathbb{R}⟹\text{No oscillations}$$

$$x(t)=e^{-\beta t}[c_1{e}^{-t\sqrt{{\beta }^2-{\omega }_0^2}}+c_2{e}^{t\sqrt{{\beta }^2-{\omega }_0^2}}]$$
$$x(t)=c_1{e}^{-(\beta -\sqrt{{\beta }^2-{\omega }_0^2})t}+c_2{e}^{-(\beta +\sqrt{{\beta }^2-{\omega }_0^2})t}$$

• After a long period of time the first term dominates since the second term goes to 0 faster. Thus damping is mostly controlled by the first term
• Damping Parameter: $\beta -\sqrt{{\beta }^2-{\omega }_0^2}$

Critical Damping ($\beta ={\omega }_0$)

$$r_1=-\beta +\sqrt{{\beta }^2-{\omega }_0^2}=-\beta$$
$$r_2=-\beta -\sqrt{{\beta }^2-{\omega }_0^2}=-\beta$$
$r_1=r_2$ is call the degeneracy

• First solution: $e^{-\beta t}$
• Second Solution $t{e}^{-\beta t}$

$$x(t)=c_1{e}^{-\beta t}+c_{2}t{e}^{-\beta t}$$

• Dampin Parameter: $\beta ={\omega }_0$

Driven Damped Oscillation

• In the real world there is friction, which eventually dampens oscillations until there are no oscillations as we saw above. To combat this, we can drive the oscillations by using a motor to provide an external force, $F(t)$

$${\overrightarrow{F}}_{\text{net}}=\text{spring force}+\text{Damping force}+\text{External force}$$
$$m\ddot{x}=-kx-b\dot{x}+F(t)$$

$$m\ddot{x}+kx+b\dot{x}=F(t)$$
$$\ddot{x}+\frac{k}{m}x+\frac{b}{m}\dot{x}=\frac{F(t)}{m}$$

$$\frac{k}{m}={\omega }_0^2,\frac{b}{m}=2\beta ,f(t)=\frac{F(t)}{m}$$

$$⟹\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=f(t)$$

In order to solve the above non-homogenous linear differential equation, we need the homogenous, $x_h$ and particular solution, $x_p$

• We have the linear operator, $D$
  $$Dx=f(t)$$

• Since $D$ is a linear operator
  $$D(x_h+x_p)=D{x}_h+D{x}_p=0+f(t)=f(t)$$
  This means that $x(t)=x_h+x_p$ is a solution

• Homogenous (weakly damped case)
  $$\ddot{x_h}+2\beta \dot{x_h}+{\omega }_0^2{x}_h=0$$
  $$x_h=A{e}^{-\beta t}\cos ({\omega }_{1}t-\delta )$$

• Particular
  Suppose $f(t)=f_0\cos \left(\omega t\right)$, where the frequency of the motor, $\omega$ can be different than the frequency the mass oscillates at

$$f(t)=Re(f_0{e}^{i\omega t})$$

Consider the complex version of the differential equation, where $z$ is complex

$$Re(z(t))=x_p(t)$$
$$\ddot{z}+2\beta \dot{z}+{\omega }_0^{2}z=f_0{e}^{i\omega t}$$

Trial Solution
$$z(t)=c{e}^{i\omega t}$$
$$(-{\omega }^2+2i\beta +{\omega }_0^2)c{e}^{i\omega t}=f_0{e}^{i\omega t}$$
$$(-{\omega }^2+2i\beta +{\omega }_0^2)c=f_0⟹c=\frac{f_0}{{\omega }_0^2-\omega +2i\beta \omega }$$

For any $D$ is a complex number, you can write it as follows
$$D=A{e}^{-i\delta }$$
where $A$ is the real magnitude of the complex number and $\delta$ is the phase (imagine complex plane where $\delta$ is an angle, sort of like plane polar coords)

So you can write $$c=A{e}^{-i\delta }$$

• $A^2=c{c}^{\ast }$, where $c^{\ast }$ is the complex conjugate

$$A^2=c{c}^{\ast }=\frac{f_0^2}{({\omega }_0^2-{\omega }^2{)}^2+4{\beta }^2{\omega }_0^2}$$
$$A{e}^{-i\delta }=\frac{f_0}{{\omega }_0^2-{\omega }^2+2i\beta \omega }$$

$$f_0{e}^{-i\delta }=A({\omega }_0^2-{\omega }^2+2i\beta \omega )$$

$$\delta =arctan(\frac{Im}{Re})=arctan(\frac{2\beta \omega }{{\omega }_0^2-{\omega }^2})$$

$$z(t)=c{e}^{i\omega t}=A{e}^{-i\delta }{e}^{i\omega t}=A{e}^{i(\omega t-\delta )}=A(\cos (\omega t-\delta )+i\sin (\omega t-\delta ))$$
$$x_h(t)=Re(z(t))=Re(A(\cos (\omega t-\delta )+i\sin (\omega t-\delta )))=A\cos (\omega t-\delta )$$

$$x(t)=x_p(t)+x_h(t)$$

$$x(t)=A_1\cos (\omega t-{\delta }_1)+(A_2)(\cos ({\omega }_{1}t-{\delta }_2)e^{-\beta t}$$
$$x(t)=A\cos (\omega t-{\delta }_1)+(c_1{e}^{i{\omega }_{1}t}+c_2{e}^{-{\omega }_{1}t})e^{-\beta t}$$

Chapter 8: Rigid Bodies

$$\overrightarrow{L}=\underset{3x3\text{ matrix }}{\underbrace{I}}\omega$$

$$I=\left[\begin{array}{ccc}{I}_{xx}& I_{xy}& I_{xz}\\ I_{yx}& I_{yy}& I_{yz}\\ I_{zx}& I_{zy}& I_{zz}\end{array}\right]$$

$I$ is a symmetric matrix, which guarantees it is diagonalizable, which means we can find the eigenvectors and thus the principal axes.

Inertia Tensor of a Solid Cone

Consider a cone with the vertex at the origin, with uniform density $\rho$, height $h$, mass $M$, and radius $R$
$$I_{zz}=\sum _{\alpha }{m}_{\alpha }(x_{\alpha }^2+y_{\alpha }^2)$$

• Choose cylindrical polar coordinates ($r,\varphi ,z$)

$$I_{zz}=\sum _{\alpha }{m}_{\alpha }(x_{\alpha }^2+y_{\alpha }^2)={\int }_V{r}^{2}dm$$
$$dm=\rho dV=\rho rd\varphi drdz$$
$$I_{zz}=\rho {\int }_V{r}^{3}drd\varphi dz$$
Physics integration notation is weird
$$I_{zz}=\rho {\int }_0^{h}dz{\int }_0^{2\pi }d\varphi {\int }_0^{z\frac{R}{h}}dr{r}^3=\rho \int dz\int d\varphi$$
Note the bounds for $r$ ends at $r=z\frac{R}{h}$ since every $z$ has a differnet $r$. Note the following relation is always true for any value of $r$ and $z$:
$$\frac{r}{z}=\frac{R}{h}$$
Now continue with the integration
$$I_{zz}=\rho \int dz\int d\varphi {\frac{r^4}{4}|}_0^{\frac{zR}{h}}=\rho \int dz\int d\varphi {\left(\frac{R}{h}\right)}^4\frac{z^4}{4}=\frac{3}{10}M{R}^2$$
Now find $I_{xx}$
$$I_{xx}=\sum _{\alpha }{m}_{\alpha }(y_{\alpha }^2+z_{\alpha }^2)$$
$$I_{xx}={\int }_{V}dm(y^2+z^2)=\rho \int dV(y^2+z^2)=\int dz\int d\varphi \int rdr(y^2+z^2)$$
$$I_{xx}=\int dz\int d\varphi \int rdr{y}^2+\int dz\int d\varphi \int rdr{z}^2)$$
Note that
$$I_{zz}=\rho {\int }_V(x^2+y^2)dV=\frac{3}{10}M{R}^2$$
$$I_{zz}=\rho {\int }_V(y^2)dV+\rho {\int }_V(y^2)=\frac{3}{10}M{R}^{2}dV⟹\rho {\int }_V(y^2)dV=\frac{3}{20}M{R}^2$$
since x and y are symmetric since we have a circle
$$I_{xx}=\int dz\int d\varphi \int rdr{y}^2+\int dz\int d\varphi \int rdr{z}^2$$

$$I_{xy}=-\sum _{\alpha }{m}_{\alpha }{x}_{\alpha }y\alpha =-\rho {\int }_{V}dVxy=0$$
Since along the x axis, the cone is symmetric, and along the $y$-axis, the cone is symmetric.

$$I_{cone}=\left[\begin{array}{ccc}\frac{3}{20}M(R^2+4h^2)& 0& 0\\ 0& \frac{3}{20}M(R^2+4h^2)& 0\\ 0& 0& \frac{3}{10}M{R}^2\end{array}\right]$$

Principal Axes of Inertia

$$\overrightarrow{L}=I\overrightarrow{\omega }$$

• So $\overrightarrow{L}\ne \overrightarrow{\omega }$ in general

• Can we find particular directions of $\overrightarrow{\omega }$ so that the resulting $\overrightarrow{L}$ is in the same direction of $\overrightarrow{\omega }$?

  • If so, then we get
    $$\overrightarrow{L}=\lambda \overrightarrow{\omega },\lambda \in \mathbb{R}$$
  • It turns out we can find principal axes for any rigid body, and these directions are called principal axes
  • Note in the above, $\lambda$ is an eigenvalue!

• Statement of Existence of Principal axes: For any rigid body at any point $\underset{\text{origin}}{\underbrace{O}}$, there are at least three principal axes and the inertial tensor, $I$, is diagonal. When $\overrightarrow{\omega }$ is along any of the three axes, the $\overrightarrow{L}$ will be along the same direction of $\overrightarrow{\omega }$

$$\overrightarrow{L}=I\overrightarrow{\omega }=\lambda \overrightarrow{\omega }$$
Eigenvalue problem!

$$\hat{I}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$
$$I\omega =\lambda \hat{I}\overrightarrow{\omega }⟹$$
$$(I-\lambda \hat{I})\overrightarrow{\omega }=0$$

Define $A=I-\lambda \hat{I}$
$$A\overrightarrow{\omega }=0$$

The above is satisfied only if $\overrightarrow{\omega }=0$ or $det(A)=0$. The former is when there is no rotation at all.

We are guaranteed to have ${\lambda }_1,{\lambda }_2,{\lambda }_3$ and ${\omega }_1,{\omega }_2,{\omega }_3$

Finding Principle Axes

Find Eigenvectors