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Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

For a particle in the scalar potential energy $U(x)$,The canonical commutation relation $[x,p]=i\hslash$ givesThe Ehrenfest theorem with $O=x$ therefore givesSince $[p^2,p]=0$ and direct action on a wavefunction gives $[U(x),p]=i\hslash U^{\prime }(x)$,Finally $H$ has no explicit time dependence and $[H,H]=0$, so

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that $⟨U^{\prime }(x)⟩\simeq U^{\prime }(⟨x⟩)$. These conclusions are summarized by the classical equations from Ehrenfest theorem.

By Newton's second law, the central force obeysDifferentiating the cross product givessince the acceleration is parallel to $\mathbf{r}$. Hence $\mathbf{n}=\mathbf{n}(0)$ is constant. Since $\mathbf{r}\cdot \mathbf{n}=0$, the entire motion lies in the fixed plane through the origin with nonzero normal vector $\mathbf{n}(0)$.

Choose polar coordinates in that plane, orienting $\theta$ so that its positive normal vector points along $\mathbf{n}(0)$. The velocity isThereforeMultiplication by $m$ identifies this with conservation of angular momentum; geometrically, the swept area per unit time is $h/2$.

Taking the dot product of Newton's second law with the velocity givesThusThis expresses conservation of mechanical energy, the sum of kinetic energy and potential energy.

Finally, for $u=1/r$ and a prime denoting a derivative with respect to $\theta$, the chain rule givesSubstitution into the conserved energy yields

In cylindrical coordinates,Differentiating the position vector $\mathbf{r}=r{\mathbf{e}}_r+z{\mathbf{e}}_z$ twice gives the velocity and accelerationThis derives acceleration in cylindrical coordinates rather than assuming it.

At a fixed time, the wire has parametrization $r\mapsto r{\mathbf{e}}_r+f(r){\mathbf{e}}_z$. Its derivative is the nonzero tangent vectorA smooth wire exerts a constraint force $\mathbf{R}$ with $\mathbf{R}\cdot \mathbf{t}=0$. Because $g$ is the gravitational acceleration, the gravitational force is $-mg{\mathbf{e}}_z$; the printed $-g{\mathbf{e}}_z$ is interpreted per unit mass. Projecting Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{R}-mg{\mathbf{e}}_z$, along $\mathbf{t}$ eliminates $\mathbf{R}$:The chain rule gives $\ddot{z}=f^{\prime }(r)\ddot{r}+f^{\prime \prime }(r){\dot{r}}^2$. Rearrangement yieldsThis is the governing equation for a bead on a uniformly rotating wire. Its azimuthal acceleration contributes nothing to the projection because ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=0$.

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

Put $C=q^2/(4\pi {ϵ}_0)>0$. The spring exerts the force $k(\ell -x)$ by Hooke's law, while Coulomb's law gives the attractive force $-C/x^2$. Newton's second law therefore gives the equation of motionFor $x>0$, the derivative is $f^{\prime }(x)=-k+2C/x^3$. Thus $f$ increases up to $x_{\ast }=(2C/k{)}^{1/3}$ and decreases thereafter. Also $f(x)\to -\infty$ at both ends of the positive half-axis. Its unique maximum isThere are therefore two equilibrium points precisely whenWriting them $x_{-}<x_{\ast }<x_{+}$, the linearization $m\ddot{\eta }=f^{\prime }(x_{\pm })\eta$ has exponentially growing solutions at $x_{-}$ and oscillatory solutions at $x_{+}$. Equivalently, the potential energy has a strict local maximum at $x_{-}$ and a strict local minimum at $x_{+}$. Hence the smaller equilibrium point is an unstable equilibrium; the larger is a stable equilibrium, with no damping to make it attracting.

At equality $\ell ={\ell }_1$ there is one degenerate equilibrium point $x_{\ast }$; it is unstable, since a displacement to its left gives a leftward force. For $\ell <{\ell }_1$ there are no equilibrium points.

Write $\mu =GM$. In polar coordinates, Newton's second law per unit mass gives the radial and transverse equations of motionThe transverse equation impliesThis is conservation of angular momentum per unit mass. For the nonradial trajectory in the next part, choose orientation so $h>0$. With $u=1/r$ and primes denoting derivatives with respect to $\theta$, the chain rule givesSubstituting into the radial equation yields $-h^2{u}^2(u^{\prime \prime }+u)=-\mu u^2$, henceThe change of independent variable requires $h\ne 0$; a purely radial trajectory instead has constant $\theta$.

In cylindrical coordinates, $\theta =\omega t$ and $z=r^2/(2a)$. ThusThe acceleration isA tangent vector along the wire is $\mathbf{t}={\mathbf{e}}_r+(r/a){\mathbf{e}}_z$. Because the wire is smooth, its constraint force is orthogonal to $\mathbf{t}$. Taking the dot product of Newton's second law with $\mathbf{t}$ eliminates that force; gravitational weight contributes $-mgr/a$. HenceMultiplying by $a^2/m$ gives

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

Use $g$ as the gravitational acceleration, so the weights are $mg$ and $\sqrt{2}mg$, consistently with the requested equations. The printed introductory phrase calling $g$ a force magnitude omits the factor $m$.

For the particle on a cone coupled to a hanging mass, write $z=-r$ and use the cylindrical coordinates unit vectors ${\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_z$. The ideal string segment from the cone particle to the hole has length $\sqrt{2}r$ and direction $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. If $h$ is the downward length of the hanging segment and $s_0$ the fixed total length, thenThe light ideal string and smooth hole give the same tension $T$ on both sides. Applying the vertical component of Newton's second law to the hanging particle givesWrite the normal reaction on the cone particle as $N({\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$, allowing $N$ to be signed according to the side of the ideal constraining surface. The acceleration in cylindrical coordinates and Newton's second law giveThere is no azimuthal force. Multiplying the middle equation by $r/m$ proveswhere the positive angular direction is chosen along the initial velocity. This is conservation of angular momentum about the vertical axis.

Subtract the radial equation from the vertical equation to eliminate the normal reaction. Since $\ddot{r}=-\ddot{z}$, this givesSubstitute the tension and the conserved angular momentum, using $r=-z$:Multiplication by $\dot{z}$ now gives an exact time derivative:Initially $z=-a$ and $\dot{z}=0$, because the initial velocity is horizontal. ThereforeThis is also total mechanical energy divided by $m$, with an irrelevant constant removed: the two particles contribute $(1+\sqrt{2})m{\dot{z}}^2$ to the radial and vertical kinetic energy, while their combined gravitational potential energy is $-mgz$ plus a constant.

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

Let the uniform string have constant tension $\tau >0$ and linear mass density $\mu >0$. Assume small transverse motion with $|u_x|\ll 1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan \beta =u_x$, and the small-angle approximation gives $\sin \beta \simeq u_x$ and $ds\simeq dx$ to first order.

The net upward force on the segment from $x$ to $x+\Delta x$ is thereforeBy Newton's second law, its mass $\mu \Delta x$ times its transverse acceleration equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to 0$ to obtainThus the wave equation on a string iswhere $c$ is the wave speed. The horizontal force balance makes the horizontal component of tension constant; to first order it equals $\tau$.

Write $q=|\dot{\mathbf{r}}|=\sqrt{{\dot{r}}^2+r^2{\dot{\theta }}^2}$. Differentiating the angular momentum and applying Newton's second law givesThe central force contributes no torque, because it is parallel to $\mathbf{r}$. The angular momentum direction is fixed whenever it is nonzero; the motion is consequently planar. In the plane's polar coordinates, $\mathbf{L}=mh{\mathbf{e}}_z$, soFor the mechanical energy, use the chain rule, Newton's second law, and $V^{\prime }(r)=-F_c(r)$:ThusFor a trajectory on a time interval, both $h$ and $E$ are each a constant of motion precisely when $\alpha =0$ or the velocity vanishes throughout that interval. Under the stated strictly attractive central force at a regular point $r>0$, a permanently stationary trajectory is impossible by Newton's second law. For the stated attractive motion, both are conserved if and only if $\alpha =0$. An isolated instant of zero speed only makes the derivatives vanish at that instant. A radial trajectory has $h=0$ even with quadratic drag, but its mechanical energy still decreases whenever it moves.

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

For a smooth surface the friction force vanishes. The ${\mathbf{e}}_{\theta }$ component of Newton's second law then gives $r^2\dot{\theta }=h$, constant. Along the constraint $z=H(r)$, the chain rule givesThe acceleration in cylindrical coordinates isTake the dot product of Newton's second law with the tangent vector ${\mathbf{e}}_r+H^{\prime }{\mathbf{e}}_z$. The normal force disappears, while the gravity term is $-mg{H}^{\prime }$. Dividing by $m$ givesUsing $\dot{\theta }=h/r^2$ yields

Continue with $f=p\sin \varphi$ and $q=p\cos \varphi$. The tangent vector to the hoop is ${\mathbf{r}}_{\varphi }=f^{\prime }{\mathbf{e}}_1-q^{\prime }{\mathbf{e}}_3$. Smooth contact means that the hoop's normal force has zero dot product with this tangent vector. The remaining applied force is $-mg{\mathbf{e}}_3$, whose dot product with ${\mathbf{r}}_{\varphi }$ is $mg{q}^{\prime }$.

Since $\mathbf{\omega }=\omega {\mathbf{e}}_3$, the Coriolis force is perpendicular to the hoop's plane and hence to ${\mathbf{r}}_{\varphi }$. AlsoTake the dot product of the rotating-frame Newton's second law with ${\mathbf{r}}_{\varphi }$, substitute a, and divide by $m$:DefineThen $K^{\prime }/2=f^{\prime }{f}^{\prime \prime }+q^{\prime }{q}^{\prime \prime }$ and $J^{\prime }=-{\omega }^{2}f{f}^{\prime }-g{q}^{\prime }$, soExpanding the squares makes the cancellation of the mixed terms explicit:The strict inequality follows from $p>0$. Here $J$ is the rotating-frame effective potential per unit mass. Multiplication of the equation by $\dot{\varphi }$ gives the useful first integral $\frac{1}{2}K(\varphi ){\dot{\varphi }}^2+J(\varphi )=\text{constant}$.

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation:

The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.