A central force exerts zero torque about its centre, so angular momentum is conserved and the motion is planar. In plane polar coordinates, $J=m{r}^2\dot{\theta }$ andUse $F=-U^{\prime }$ and either the radial Euler-Lagrange equation or differentiation of the energy away from turning points to getThe radial acceleration is $\ddot{r}-r{\dot{\theta }}^2$, so this is exactly $m(\ddot{r}-r{\dot{\theta }}^2)=F(r)$, not an additional physical outward force in an inertial frame. The variational derivation remains valid at turning points where dividing an energy derivative by $\dot{r}$ would be illegitimate.

The escaping sand initially has the bucket's velocity, so it produces no relative-velocity thrust. This distinction is essential in a leaking Atwood machine. With upward velocity $v$ for the leaky side, the two bucket equations areTheir sum gives $\dot{v}=g(M_0-M)/(M+M_0)$. Since $\dot{M}=-\alpha$, it follows thatIntegrate from $M_0$, where the velocity is zero:At emptying, $M=(1-ϵ)M_0$, givingThis is upward and positive for $0<ϵ<1$. For no initial sand, $ϵ=0$, the emptying time is zero and the speed is zero. As the dry bucket mass tends to zero,This finite limit refers to the idealized model; the endpoint itself would describe a massless empty bucket. For small $ϵ$, $v_{\mathrm{empty}}\sim g{M}_0{ϵ}^2/(4\alpha )$, consistent with initially zero acceleration that grows as mass is lost.

During $dt$, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the expelled positive mass $-dM$ has velocity $v-u$ to first order. Subtract the initial momentum $Mv$ from their final momentum to obtain $Mdv+udM$. Equating this to the external impulse gives the rocket equationWith $F=-Mg-kMv$ and $\dot{M}/M=-b$, this reduces to $\dot{v}+kv=ub-g$. Therefore, for initial velocity $v_0$ and $k>0$,For a launch from rest,The inequality makes the initial upward acceleration positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.

The parallel axis through the centre is the rod's own long axis, with zero moment of inertia in the thin-rod approximation. By the parallel axis theorem, the displaced parallel axis hasEquivalently, every mass element is the same perpendicular distance $D$ from that axis, so the defining integral immediately gives this result.

Measure position $x$ along the ideal thin rod from its centre of mass. Its perpendicular distance from the inclined axis is $|x|\sin \alpha$, and its linear mass density is $M/L$. The moment of inertia is thereforeThis tends to zero for an axis along an infinitely thin rod and to $M{L}^2/12$ for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.

At threshold, the pion is at rest in the final centre-of-momentum frame, which moves in the laboratory withIn the pion's rest frame, its two-photon decay produces opposite photons of equal energy $m_{\pi }{c}^2/2$. The specified forward and backward directions remain collinear under this boost, soAt threshold, $\gamma (1+\beta )=(m_p+m_{\pi })/m_p$ and $\gamma (1-\beta )=m_p/(m_p+m_{\pi })$. HenceThe forward photon is the more energetic one. Their product remains $(m_{\pi }{c}^2/2{)}^2$, as required by the invariant mass of an oppositely directed photon pair, while their sum equals the laboratory pion energy.

Use the squared total energy-momentum four-vector, expressed in energy-squared units. With the proton initially at rest,At the reaction threshold, the outgoing proton and pion have no relative kinetic energy in their centre-of-momentum frame, so $W=(m_p+m_{\pi })c^2$. Solving givesThe numerical value uses the supplied $938$ and $135\mathrm{MeV}$ rest energies. The threshold exceeds the pion's rest energy because the outgoing system must carry the incident photon's nonzero laboratory momentum.

For a boost along $+x$ at speed $v$, the energy-momentum four-vector transforms asThe total four-vector is the sum over the particles. Write $E={\sum }_i{E}_i$ and $\mathbf{P}={\sum }_i{\mathbf{p}}_i$. A centre-of-momentum rest frame exists when this total is timelike, with ${\mathbf{V}}_{\mathrm{CM}}=c^2\mathbf{P}/E$. Boost along $\mathbf{P}$ and use $P=E{V}_{\mathrm{CM}}/c^2$:ThereforeThe timelike qualification is needed: exactly parallel photons alone have null total four-momentum and no such rest frame.

Set $q=1$ in the laboratory scattering circle formula. For $M_1>M_2$, a real outgoing speed requiresAlso $v_1\ge 0$ excludes the backward-angle interval: the square root is smaller than $|\cos {\theta }_1|$ when $M_2/M_1<1$, so both roots would be negative for $\cos {\theta }_1<0$. ThereforeAt this boundary the two possible speeds coincide, with $v_1/u_1=\sqrt{(M_1-M_2)/(M_1+M_2)}$. Geometrically the outgoing ray is tangent to the laboratory scattering circle. As $M_2/M_1\to 0$ the maximum tends to zero, while it approaches $\pi /2$ as the masses approach equality from $M_1>M_2$.

A totally inelastic collision in the sticking sense has zero final relative velocity, so $q=0$. Both velocities are the conserved centre-of-momentum frame velocity:Thus the bodies move together along the original incident direction; the moving projectile has ${\theta }_1=0$. The common outgoing state is stationary in the centre-of-momentum frame. The lost translational kinetic energy iswhich can become internal energy. As explained in part 8, setting only the normal coefficient of restitution to zero in a smooth oblique collision does not force sticking: tangential relative motion can remain.

For an elastic collision with $M_2>M_1$, the positive root in the laboratory scattering circle formula is the only nonnegative one. With $\epsilon =M_1/M_2\ll 1$,ConsequentlyThe heavy target can absorb a finite momentum with negligible velocity and negligible kinetic energy. Thus the projectile may change direction substantially while retaining its speed; a backward reflection also has $v_1\simeq u_1$.

For equal masses and an elastic collision, $q=1$ andThus the nonzero outgoing branch isand the other nonnegative algebraic root is $v_1=0$. The latter is the head-on transfer of the incoming momentum and kinetic energy to particle 2: $v_2=u_1$. A stationary particle has no defined outgoing direction, so this root is not a separate physical outcome for every chosen ${\theta }_1$.

On the nonzero branch, momentum conservation givesThe two nonzero outgoing velocities are perpendicular: squaring momentum conservation and comparing with kinetic energy conservation gives ${\mathbf{v}}_1^{\prime }\cdot {\mathbf{v}}_2^{\prime }=0$. For $\cos {\theta }_1<0$, the nonzero root is negative and must be rejected; backward scattering of a moving incident particle is impossible in this equal-mass, initially stationary-target case.

About a fixed origin in an inertial frame, the angular momentum and torque areFor a specified fixed axis with unit vector $\hat{\mathbf{n}}$, use the projections $L_n=\hat{\mathbf{n}}\cdot \mathbf{L}$ and ${\tau }_n=\hat{\mathbf{n}}\cdot \mathbf{\tau }$. The mass is constant. Differentiating and applying Newton's second law,Hence ${\dot{L}}_n={\tau }_n$ for the fixed axis. If the net torque vanishes, the corresponding angular momentum is conserved. Fixing the origin and axis is essential to this unmodified form of the law.

The radial and angular Euler-Lagrange equations areWriting the conserved angular momentum aswe obtainA positive constant radius requires zero radial acceleration. Hence the circular motion hasprovided $J\ne 0$ and $0<r_0<l$. Its angular velocity is ${\dot{\theta }}_0=J/(m{r}_0^2)$. No positive circular radius exists for $J=0$ with $M,g>0$.

For later use, the conserved radial energy isThe effective potential has $U_J^{\prime \prime }(r)=3J^2/(m{r}^4)>0$, so its stationary radius is a stable minimum. The tension remains positive in the ideal allowed region: the hanging Newton's second law gives $T_s=M(g+\ddot{r})=Mm(g+r{\dot{\theta }}^2)/(M+m)>0$.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.