Hints & solutions

Yes. A particle can move freely in three-dimensional space while gravity acts. Gravity determines its acceleration; it does not by itself constrain all possible configurations to a particular surface. A particular trajectory selected by initial data is not the same thing as an imposed constraint.

For a circle of radius $a$ in the $xy$ plane, $z=0$ and $x^2+y^2=a^2$. These position-only equalities are holonomic. They have no explicit time dependence, so they are scleronomous. A threaded bead remains on the wire, so the model is bilateral. Smoothness is a separate assumption about friction, not part of this geometrical classification.

For constant $c$, $\dot x=c$ integrates to $x-ct=C$. On a branch fixed by the initial value $C$, it is a holonomic relation with explicit time dependence when $c\ne0$. Its velocity form alone does not make it non-holonomic. If the initial position is unrestricted, the family includes different values of $C$; each fixed initial branch has its own integrated relation.

A rigid rod can transmit tension or compression, so its axial force may pull or push the endpoints. A flexible string can pull while taut but cannot transmit compression. This is why a rod's equality constraint can remain valid in a situation where a string would go slack.

For a planar pendulum, $x^2+y^2=a^2$ restricts the bob to a circle. It does not tell us how quickly the bob moves around it. Under ideal gravity-driven motion, $\ddot\theta+(g/a)\sin\theta=0$ determines angular evolution once initial angle and velocity are specified. The first equation describes geometry; the second describes dynamics within that geometry.

Holonomic means reducible to an equality involving positions and possibly time. Rheonomous means explicit time dependence. The expanding circle $x^2+y^2-a(t)^2=0$ satisfies both descriptions: it contains no velocities, but its prescribed radius changes with time. The categories therefore do not exclude each other.

The restriction is $x^2+y^2+z^2\le a^2$. It permits the interior and surface, so it is unilateral. The equality $x^2+y^2+z^2=a^2$ would describe a different model in which the particle always remains on the spherical surface.

From $x^2+y^2=a^2$, with constant $a$, differentiation gives $2x\dot x+2y\dot y=0$. Dividing by two yields $x\dot x+y\dot y=0$, or $\mathbf r\cdot\mathbf v=0$. Thus velocity is perpendicular to radius and tangent to the circle. The zero-speed case also satisfies the relation.

The chain rule gives $2(x-X)(\dot x-\dot X)+2(y-Y)(\dot y-\dot Y)=0$. Therefore

The relative radius is perpendicular to the bob's velocity relative to the support. The bob's laboratory-frame velocity alone need not be perpendicular to that radius.

Substitution gives $3(2)+4\dot y=0$, so $\dot y=-6/4=-1.5\,\mathrm{m\,s^{-1}}$. The bead moves right and down at that instant, tangent to the circle in the first quadrant. Also $3^2+4^2=25$, confirming that the supplied position lies on the radius-five circle.

The position relation gives $x_2=2.4-0.9=1.5\,\mathrm m$. Differentiating $x_1+x_2=2.4$ gives $\dot x_2=-\dot x_1=-0.3\,\mathrm{m\,s^{-1}}$. Because both segment coordinates increase downward, the negative value means mass 2 moves upward.

Differentiate $x=2t$ to obtain $\dot x=2\,\mathrm{m\,s^{-1}}$. The equation contains no $y$, so it does not fix $\dot y$. In a planar model, $y$ remains an independent coordinate along the moving wire. Additional forces determine its evolution.

Define a constraint as a restriction on admissible positions or velocities. Holonomic: $f(q,t)=0$, such as a pendulum length. Non-holonomic: non-integrable velocity restrictions, such as the steerable no-slip wheel. Scleronomous: no explicit time, such as a fixed circle. Rheonomous: explicit time, such as $x=Ut$. Bilateral: equality, such as a bead threaded on a wire. Unilateral: inequality, such as $y\ge0$ above a floor. Explain that a constraint may receive a label from each independent classification, and give the straight-line rolling counterexample to “velocity means non-holonomic.”

Write $m_a\mathbf a_a=\mathbf F_a+\mathbf R_a$. Applied forces are specified interactions such as gravity; reactions enforce a modeled restriction. A smooth surface supplies a normal reaction. An ideal taut string supplies tension, which cannot be compressive. A rigid rod can sustain tension or compression. A rough surface also supplies friction, generally tangential and not removable by a smooth-constraint assumption. Reactions may be unknown until motion is determined. Worklessness along a fixed smooth surface does not make a reaction dynamically unimportant: it can turn the velocity.

Along a fixed line, $dx-Rd\phi=0$ integrates to $x-R\phi=C$. For a steerable wheel, $dx=R\cos\psi\,d\phi$ and $dy=R\sin\psi\,d\phi$. Changes in heading alter the integrals. Motion can return the angular variables to their initial values while changing position, so endpoint angles alone do not impose the position-only relation of straight-line rolling. The path-dependent restriction is non-holonomic.

No, the circle is not a constraint in the stated spring model. It is one dynamically possible trajectory selected by special initial conditions. Different initial speed or radial motion generally changes the orbit. A true circular-wire constraint would require every admissible motion to remain on the same circle, supported by a reaction force as needed.

There are two degrees of freedom, represented by $(x,y)$. Velocity requires two additional numbers to specify a state, but these do not increase the configuration-space dimension.

No. At any chosen time, $a(t)$ is known. One angle specifies a point on the prescribed circle. If radius instead evolves as an independent unknown, then radial position can be an additional degree of freedom.

Yes. The equations $x+y=0$ and $2x+2y=0$ are equivalent. The second imposes nothing new, so they count as one independent restriction. Degrees of freedom depend on independent information, not the number of written lines.

They describe the same pendulum configuration: sine is zero and cosine is one for both. Configuration space identifies angles differing by $2\pi$. The number of completed rotations may matter for a history, but not for the instantaneous ideal pendulum position.

Configuration space is the set of all admissible arrangements of the entire system. Physical space is the space in which the particles lie. Two free particles occupy ordinary three-dimensional space, but their joint configuration requires six coordinates, so their configuration space is six-dimensional.

For two degrees of freedom, the coordinates are $q_1,q_2$. A regular second-order initial-value problem normally requires $q_1(0),q_2(0),\dot q_1(0),\dot q_2(0)$. The first pair specifies configuration; the second pair specifies how it starts moving. Counting both pairs as configuration freedoms confuses configuration with dynamical state.

Nothing: all azimuth values locate the same pole. The azimuth label loses uniqueness there. A local pair of coordinates in the tangent directions still describes nearby positions, so the physical configuration space remains two-dimensional. This is a coordinate singularity.

There are $3N$ Cartesian position coordinates for $N$ point particles. Each independent regular holonomic equality locally determines one coordinate from the remaining ones. After $k$ such eliminations, $n=3N-k$. The constraints must be independent, smooth and regular at the configuration, and position-level rather than arbitrary non-integrable velocity restrictions. For a moving constraint, count at fixed prescribed time.

The first pair has the single condition $y=-x$ and leaves one free coordinate. In the second pair, add $x+y=0$ and $x-y=0$ to get $2x=0$, then $y=0$. Two independent equations fix the point, leaving zero freedom in the plane. Their coefficient rows are respectively proportional and independent.

The assumptions permit direct use of $n=3N-k=3(4)-5=7$. Seven independent coordinates specify the configuration locally. No additional velocity restrictions were supplied.

Three bobs in the plane give six Cartesian coordinates. The three independent rod-length equalities remove three choices, leaving $n=3$. Suitable coordinates are the three rod angles, each measured relative to a fixed direction. The support is fixed, so it contributes no unknown position.

There are three Cartesian coordinates and two independent regular restrictions, giving one freedom. Choose $\theta$: $x=3\cos\theta$, $y=3\sin\theta$, $z=2$ in metres. Substitution gives $x^2+y^2=9$ and the specified height, verifying the parameterization.

A free particle's position gives a point in three-dimensional configuration space. A fixed-length planar pendulum needs one angle modulo $2\pi$, so its configuration space is a circle. A double pendulum needs two independent periodic angles; a square of angle values with opposite edges identified represents a torus. One point of this two-dimensional space gives the complete positions of both bobs. Do not confuse this abstract space with their common physical plane.

Independent constraints remove separate local choices: $z=0$ and $x^2+y^2=a^2$ with $a>0$ leave a circle. Redundant constraints repeat information: $f=0$ and $2f=0$ remove only one choice. Singular constraint descriptions need special care: $x^2+y^2=0$ in a plane has vanishing first derivatives at the only allowed point. Its set is a single point, even though naive subtraction of one written equation would suggest a curve. State the regularity assumption when applying the counting formula.

The sum $x^2+y^2$ is zero only when $x=y=0$. Therefore the configuration set is a point, with zero degrees of freedom. The derivative row $(2x,2y)$ vanishes there, so the regular “one equation removes one dimension” argument is inapplicable. Directly inspect the allowed set instead.

The configuration variables $(x,y,\psi,\phi)$ describe four independent positional/orientation labels. Two no-slip equations restrict their instantaneous velocities. Because these restrictions are non-integrable, they do not produce two independent position equalities that could be subtracted from the configuration count. The allowed velocity directions at a configuration and the dimension of configuration space answer different questions.

No. A coordinate may be a length, an angle, or another independent parameter. Its dimensions depend on its definition. A pendulum angle is dimensionless in SI dimensional analysis, although the radian label is useful for clarity.

If $q=\theta$, then $\dot q=\dot\theta$ is angular velocity in radians per second. It is not the bob's linear speed; for fixed pendulum length $a$, the speed is $a|\dot\theta|$.

Yes. Changing $\theta_1$ changes the first bob's position and the position of the support point for the second rod, so both bobs move. A generalized coordinate describes an independent configuration change, not necessarily a single particle.

Every $\theta$ gives $(x,y)=(0,0)$, so angle is not uniquely defined at the origin. Polar coordinates are singular there. Cartesian coordinates provide a regular description.

In $x=Ut+a\sin\theta$, the term $Ut$ changes even if $\theta$ is held fixed; this is explicit time dependence. The angle also changes along a trajectory, producing implicit time dependence through $\theta(t)$. Consequently, $dx/dt=U+a\cos\theta\dot\theta$, while $\partial x/\partial t=U$ at fixed angle.

The equation $x^2+y^2=a^2$ relates the two coordinates. On a chosen local branch, specifying one fixes the other. Treating both as unrestricted independent generalized coordinates would include positions off the circle unless a constraint equation were retained. One angle incorporates the restriction automatically.

The vector $\partial\mathbf r/\partial q_i$ tells how physical position changes per unit change in $q_i$, with all other coordinates and time fixed. Its dimensions are length divided by $[q_i]$. It need not have length one. Multiplying by $\delta q_i$ gives that coordinate's contribution to virtual displacement.

For bob 1, $x_1^2+y_1^2=a^2(\sin^2\theta_1+\cos^2\theta_1)=a^2$. For the second rod, $x_2-x_1=b\sin\theta_2$ and $y_2-y_1=-b\cos\theta_2$. Squaring and adding gives $(x_2-x_1)^2+(y_2-y_1)^2=b^2$. Thus both fixed lengths are satisfied for arbitrary independent angles.

The derivatives are $\partial\mathbf r/\partial r=(\cos\theta,\sin\theta)$ and $\partial\mathbf r/\partial\theta=(-r\sin\theta,r\cos\theta)$. Their dot product is $-r\cos\theta\sin\theta+r\sin\theta\cos\theta=0$. The first has unit length; the second has length $r$ for $r\ge0$. They correspond to radial and tangential changes.

The transformations give $x=1.2\sin60^\circ=1.039\,\mathrm m$ and $y=-1.2\cos60^\circ=-0.600\,\mathrm m$. The bob is to the right and below the pivot. The distance is $\sqrt{1.039^2+0.600^2}\approx1.2\,\mathrm m$, allowing for rounding.

From the derivative magnitude, $ds/d\phi=\sqrt{R^2+b^2}=\sqrt{9+16}=5\,\mathrm m$ per radian. Since this factor is constant and $\phi$ increases monotonically, the finite arc distance is exactly $5(0.2)=1.0\,\mathrm m$.

Choose $\mathbf r(q)=(q,cq^2)$. Differentiate: $\partial\mathbf r/\partial q=(1,2cq)$. It is tangent to the parabola and generally not a unit vector. Since $cq$ is dimensionless, $c$ has dimensions of inverse length if both Cartesian coordinates are lengths.

Define generalized coordinates as independent parameters specifying the configuration locally. Write $\mathbf r_a=\mathbf r_a(q,t)$. A pendulum uses one angle: $(a\sin\theta,-a\cos\theta)$. A free planar particle in polar coordinates uses $(r,\theta)$: $(r\cos\theta,r\sin\theta)$. A double pendulum uses two angles, with the second bob obtained by adding rod vectors. Define axes and angle conventions and verify each constraint by substitution. Explain that generalized velocities are time derivatives of the chosen variables.

Advantages include fewer independent variables, automatic satisfaction of holonomic constraints, natural geometry, and simpler work or energy expressions. Coordinates may be angles or lengths. Limitations include branch choices, periodic identifications, and singularities such as polar coordinates at the origin. A poor choice may complicate equations without changing physics. Non-holonomic restrictions cannot generally be eliminated by merely reducing the position-coordinate count.

The position is $(a\sin(s/a),-a\cos(s/a))$. Applying the chain rule gives $\partial\mathbf r/\partial s=(\cos(s/a),\sin(s/a))$. Its magnitude is one, so it is the tangent unit vector in the direction of increasing arc length. The length factor present in $\partial\mathbf r/\partial\theta$ has been absorbed into $s$.

Although $x$ and $y$ repeat when $\phi$ increases by $2\pi$, $z=b\phi$ increases by $2\pi b$. For $b\ne0$ the new point is different. Thus an unbounded helix requires an unwrapped real-valued parameter. Restricting it to one interval of length $2\pi$ would describe only one turn.

No. A virtual displacement compares neighboring admissible configurations at the same specified time. It assigns no speed and does not describe a journey completed instantaneously. Therefore nonzero $\delta\mathbf r$ with $\delta t=0$ is not a statement about infinite velocity.

Hold time and all generalized coordinates other than $q_i$ fixed. Then change $q_i$ infinitesimally. The resulting derivative describes only that coordinate's contribution to the particle's positional change.

Yes. If $\mathbf r=(q,Ut)$ and $\dot q=0$, the velocity is still $(0,U)$. It comes entirely from $\partial\mathbf r/\partial t$. A coordinate can remain constant while the geometry carrying it moves.

They remain different concepts. For stationary transformations, actual and virtual displacements lie in the same allowed tangent directions, but the actual change is determined by the evolving trajectory. A virtual variation is any permissible infinitesimal fixed-time comparison. Their formal coordinate expressions may look alike without making the operations identical.

The fixed-time first-order condition is $\nabla f\cdot\delta\mathbf r=0$. At a regular point, $\nabla f$ is normal to the constraint surface. Its zero dot product with the virtual displacement therefore says the displacement is tangent. Regularity matters because a zero gradient provides no normal direction.

$\dot q_i=dq_i/dt$ is a rate along a motion. The actual infinitesimal change is $dq_i=\dot q_i\,dt$. The virtual change $\delta q_i$ is an independently chosen admissible comparison with time fixed. It need not equal the actual $dq_i$ and is not defined by multiplying a velocity by $\delta t$.

For a rising horizontal wire, $d\mathbf r=(dq,Udt)$ while $\delta\mathbf r=(\delta q,0)$. With vertical reaction $(0,N)$, virtual work is zero but actual work is $NUdt$. The support's vertical transport can transfer energy even though the reaction has no component along the instantaneous allowed tangent variation.

From $\mathbf r_a(q,t)$, total differentiation gives $\mathbf v_a=\sum_j\mathbf r_{a,j}\dot q_j+\mathbf r_{a,t}$. Partial differentiation with respect to $\dot q_i$ selects the coefficient $\mathbf r_{a,i}$ because the response vectors contain no velocities. For the second identity,

Smoothness permits interchange of mixed derivatives, so these expressions agree. Define the comma notation if you use it: $\mathbf r_{a,i}=\partial\mathbf r_a/\partial q_i$.

Along a trajectory, the total differential of $f=0$ is $\nabla f\cdot d\mathbf r+f_tdt=0$. For a fixed-time variation, $\delta t=0$, so $\nabla f\cdot\delta\mathbf r=0$. In a scleronomous description, $f_t=0$ and both satisfy the same tangent restriction. In a rheonomous description, actual displacement generally includes the motion of the surface through $f_tdt$.

$\mathbf v=(\dot q,3)=(4,3)\,\mathrm{m\,s^{-1}}$, so speed is $\sqrt{4^2+3^2}=5\,\mathrm{m\,s^{-1}}$. At fixed time, $\delta\mathbf r=(\delta q,0)$. The upward component of actual velocity has no counterpart in the virtual displacement.

At $\theta=0$, $\partial\mathbf r/\partial\theta=(a,0)=(0.50,0)\,\mathrm m$ per radian. Thus $\mathbf v=(0.50,0)(2)=(1.0,0)\,\mathrm{m\,s^{-1}}$. The first-order virtual displacement is $(0.50,0)(0.01)=(0.005,0)\,\mathrm m$. A finite angular change would include a small second-order vertical change, which the infinitesimal tangent approximation omits.

The tangential speed is $r\dot\theta=2(2)=4\,\mathrm{m\,s^{-1}}$. Hence $v^2=\dot r^2+r^2\dot\theta^2=9+16=25$, giving $v=5\,\mathrm{m\,s^{-1}}$. Kinetic energy is $T=\tfrac12(2)(25)=25\,\mathrm J$.

Define actual displacement as change along a trajectory over $dt$, and virtual displacement as an admissible fixed-time comparison. For a fixed circle, $d\mathbf r=\mathbf r_{,\theta}d\theta$ and $\delta\mathbf r=\mathbf r_{,\theta}\delta\theta$; both are tangent. For $y=Ut$, actual displacement is $(dq,Udt)$ and virtual displacement $(\delta q,0)$. Draw the horizontal instantaneous wire, a tangent comparison arrow, and a slanted actual arrow to a later wire. Explain that a virtual displacement is not instantaneous physical motion.

Let $a=1,\ldots,N$ label particles and $i=1,\ldots,n$ label independent coordinates. Define smooth transformations $\mathbf r_a(q,t)$. Derive $\mathbf v_a=\sum_i\mathbf r_{a,i}\dot q_i+\mathbf r_{a,t}$ and $\delta\mathbf r_a=\sum_i\mathbf r_{a,i}\delta q_i$. Explain that partial derivatives hold the other arguments fixed and that generalized velocities are separate arguments in partial differentiation. Prove both identities as in Solution 4.8. State that independent holonomic coordinates underlie the unrestricted independent variations used later.

At fixed time the radius is fixed, so $2\mathbf r\cdot\delta\mathbf r=0$. Along actual motion, $2\mathbf r\cdot\mathbf v=2a\dot a$. Therefore $\mathbf r\cdot\delta\mathbf r=0$ while $\mathbf r\cdot\mathbf v=a\dot a$. The actual radial speed is $\dot a$, but virtual displacements remain tangent to the instantaneous circle.

At fixed angle, $\partial\mathbf v/\partial\dot\theta=(a\cos\theta,a\sin\theta)=\partial\mathbf r/\partial\theta$. At fixed generalized velocity, $\partial\mathbf v/\partial\theta=(-a\sin\theta\dot\theta,a\cos\theta\dot\theta)$. The total time derivative of $\partial\mathbf r/\partial\theta=(a\cos\theta,a\sin\theta)$ is exactly the same. The constant support speed $U$ contributes neither derivative, confirming the general relations in this example.

No. Virtual work has dimensions of energy but uses an admissible fixed-time comparison. It measures the force projection relevant to that comparison. Actual energy transfer uses force integrated along actual displacement over an actual motion.

Not necessarily. The ideality condition concerns $\sum_a\mathbf R_a\cdot\delta\mathbf r_a$. In the Atwood system, tension contributes $-\mathcal T\delta x_1$ and $-\mathcal T\delta x_2$; these cancel because $\delta x_2=-\delta x_1$, although each can be nonzero.

For a length coordinate, $[Q_i]=\mathrm{J/m}=\mathrm N$. For an angle measured in radians, the generalized force is torque-like, conventionally written in $\mathrm{N\,m}$ or joules per radian. Its physical interpretation differs from a Cartesian force even though radians are dimensionless in SI.

No. A nonzero force can have zero dot product with a particular displacement. Equilibrium requires zero total applied virtual work for every admissible variation. Only then can independence of the variations force every generalized-force coefficient to vanish.

Ideal constraints have zero total reaction virtual work for every admissible virtual displacement. On a smooth surface the reaction is normal, while the virtual displacement is tangent to the instantaneous surface. Their dot product is zero. Smoothness excludes a tangential sliding-friction force from that eliminated reaction.

Changing $q_i$ may displace many particles, as changing one rod angle moves both bobs of a double pendulum. Each applied force then contributes $\mathbf F_a\cdot(\partial\mathbf r_a/\partial q_i)$ to the coefficient of $\delta q_i$. Summing those contributions produces the one generalized force $Q_i$.

Stationarity requires the first derivative to vanish. Both a maximum and a minimum satisfy that. Near a minimum, a small displacement raises potential and gives a restoring tendency. Near a maximum, it lowers potential and the tendency is away from equilibrium. Therefore $V'=0$ alone does not establish stability.

Start from $\delta W=\sum_a\mathbf F_a\cdot\delta\mathbf r_a$. Substitute $\delta\mathbf r_a=\sum_i\mathbf r_{a,i}\delta q_i$ to get $\delta W=\sum_a\sum_i(\mathbf F_a\cdot\mathbf r_{a,i})\delta q_i$. Interchange finite sums and collect: $\delta W=\sum_i[\sum_a\mathbf F_a\cdot\mathbf r_{a,i}]\delta q_i$. Define the bracket as $Q_i$. Its dimensions must satisfy $[Q_i][q_i]=\mathrm J$.

At equilibrium, $\mathbf F_a+\mathbf R_a=0$. Dot with each admissible $\delta\mathbf r_a$ and sum. Ideality removes $\sum_a\mathbf R_a\cdot\delta\mathbf r_a$, leaving $\sum_a\mathbf F_a\cdot\delta\mathbf r_a=0$ for every admissible variation. With independent regular holonomic coordinates this is $\sum_iQ_i\delta q_i=0$. Set all variations except one to zero to infer each $Q_i=0$. The usual static equivalence assumes stationary bilateral geometry and physically available reactions; unilateral contacts require additional sign checks.

Choose $q$ up the incline. Applied virtual work is $(P-mg\sin30^\circ)\delta q$. Equilibrium for all $\delta q$ gives $P=3(9.8)(0.5)=14.7\,\mathrm N$ uphill. The normal reaction contributes no work along the plane.

For a positive counterclockwise variation, $\delta W=[60(0.20)-F(0.50)]\delta\theta$. Its coefficient vanishes at equilibrium, so $F=12/0.50=24\,\mathrm N$ downward on the right. Both moments then have magnitude $12\,\mathrm{N\,m}$ and opposite signs.

On the lower branch, $\tan\theta=P/(mg)=1/\sqrt3$. Thus $\theta=30^\circ$ from the downward vertical toward the applied force. Specifying the branch matters because tangent repeats every $\pi$.

Define virtual work and derive the generalized-force formula as in Solution 5.8. Define ideal constraints by zero total reaction work. For the pendulum use $(x,y)=(a\sin\theta,-a\cos\theta)$, so $\delta\mathbf r=(a\cos\theta,a\sin\theta)\delta\theta$. Tension is radial and contributes zero. Gravity plus horizontal force gives $\delta W=a(P\cos\theta-mg\sin\theta)\delta\theta$. For equilibrium, its coefficient vanishes, yielding the lower-branch relation $\tan\theta=P/(mg)$. Finish by checking tension/contact assumptions.

Newtonian equilibrium uses force balances including unknown reactions. Virtual-work equilibrium dots those balances with admissible variations, so ideal reactions disappear. In an Atwood machine, Newtonian balance gives $m_1g=\mathcal T$ and $m_2g=\mathcal T$. Virtual work instead uses $\delta x_2=-\delta x_1$ to cancel both tension terms at once, leaving $(m_1-m_2)g\delta x_1=0$. Both yield $m_1=m_2$. The second method is efficient when reactions are not requested; they can be recovered afterward.

The applied generalized force is $Q=mg-kq$. Thus $(mg-kq)\delta q=0$ for every $\delta q$, giving $q_0=mg/k$. The potential is $V=\tfrac12kq^2-mgq$, so $V''=k>0$ for an ordinary spring. This is a stable minimum. Equivalently, writing $q=q_0+\eta$ gives force $Q=-k\eta$, directed back toward equilibrium.

The rough surface supplies static friction $f$ opposing $P$. It cannot be discarded as a workless normal reaction. The correct virtual-work coefficient is $P-f$, so equilibrium gives $f=P$, provided the required value satisfies $|f|\le\mu_sN$. If $P$ exceeds the maximum available static friction, the assumed static equilibrium is impossible. Setting $P\delta x=0$ incorrectly removes the balancing force.

No. In this inertial-frame derivation, $-m\mathbf a$ is the acceleration term of Newton's law written on the force side. It is useful bookkeeping, not a new interaction or a third-law reaction on the same particle. It does not imply actual static equilibrium.

The expression becomes $\sum_a\mathbf F_a\cdot\delta\mathbf r_a=0$, the principle of virtual work for equilibrium under ideal constraints. The same ideal-reaction assumption supports both results.

Yes. A reaction can supply the radial or normal component needed for curved motion while doing zero work along tangent variations. For a pendulum, the acceleration includes an inward term proportional to $a\dot\theta^2$. The virtual-work projection removes that component from the tangent equation; it does not remove it physically.

Only when the generalized variations are independent within the admissible set. Then choosing one nonzero variation at a time proves its coefficient is zero. If variations remain linked by constraints, the sum may vanish through cancellation without each individual coefficient vanishing.

The derivation begins with Newton's law for every particle. It rearranges the equations and projects them onto admissible displacements. Ideality then eliminates total reaction work. No new force law is introduced; the method uses the geometry and the reaction model to reorganize the existing dynamics efficiently.

The driving generalized force is the difference of their weights, but both masses resist the shared acceleration. Since $\ddot x_2=-\ddot q$ and $\delta x_2=-\delta q$, the two minus signs in its inertial contribution cancel, making the contributions add. The effective coefficient is $m_1+m_2$, also visible in $T=\tfrac12(m_1+m_2)\dot q^2$.

Once the motion and acceleration are known, substitute them into the original force balance. For mass 1 of an Atwood machine, $m_1g-\mathcal T=m_1\ddot q$ determines tension. For a pendulum, radial force balance determines the rod reaction or string tension. Eliminating a reaction from the motion equation does not make it unknowable.

Start with $\mathbf F_a+\mathbf R_a=m_a\mathbf a_a$. Rearrange, dot with admissible $\delta\mathbf r_a$, and sum. Ideality sets the reaction-work sum to zero, giving $\sum_a(\mathbf F_a-m_a\mathbf a_a)\cdot\delta\mathbf r_a=0$. Insert $\delta\mathbf r_a=\sum_i\mathbf r_{a,i}\delta q_i$ and group terms to obtain $\sum_i[Q_i-\sum_am_a\mathbf a_a\cdot\mathbf r_{a,i}]\delta q_i=0$. For independent variations, $Q_i=\sum_am_a\mathbf a_a\cdot\mathbf r_{a,i}$. State constant masses, inertial frame, ideal reactions, smooth holonomic transformations, and regular independent coordinates for the final step.

Use $\mathbf r=(a\sin\theta,-a\cos\theta)$ and $\mathbf r_{,\theta}=(a\cos\theta,a\sin\theta)$. Twice differentiating position gives $\mathbf a=(a\cos\theta\ddot\theta-a\sin\theta\dot\theta^2,\ a\sin\theta\ddot\theta+a\cos\theta\dot\theta^2)$. Dotting with $\mathbf r_{,\theta}$ cancels the opposite mixed terms and gives $a^2\ddot\theta$. Gravity has generalized force $-mga\sin\theta$. Thus $ma^2\ddot\theta=-mga\sin\theta$, or $\ddot\theta+(g/a)\sin\theta=0$. This is exact for the ideal model; replacing sine by angle is a separate small-angle approximation.

Acceleration is $\ddot q=(5-3)9.8/(5+3)=2.45\,\mathrm{m\,s^{-2}}$. The $5\,\mathrm{kg}$ mass goes down. Tension follows from its balance: $\mathcal T=5(9.8-2.45)=36.75\,\mathrm N$. Check with the lighter mass: $\mathcal T=3(9.8+2.45)=36.75\,\mathrm N$. Both methods agree.

D’Alembert gives $(mg\sin20^\circ-m\ddot q)\delta q=0$, so $\ddot q=9.8\sin20^\circ\approx3.35\,\mathrm{m\,s^{-2}}$ downhill. No mass value is needed because it cancels.

The equation is $0.25\ddot x+9x=0$, hence $\omega=\sqrt{36}=6\,\mathrm{s^{-1}}$. Write $x=A\cos6t+B\sin6t$. At $t=0$, $A=0.04\,\mathrm m$; differentiating gives $\dot x(0)=6B=0$, so $B=0$. Therefore $x(t)=0.04\cos(6t)\,\mathrm m$ for $t$ in seconds. Its period is $2\pi/6\approx1.047\,\mathrm s$.

Give the derivation in Solution 6.8. For the Atwood model state ideal string and pulley assumptions, take $x_1=q$ and $x_2=b-q$, and obtain variations $\delta q,-\delta q$. Gravity gives $(m_1-m_2)g\delta q$. Inertial terms give $-(m_1+m_2)\ddot q\delta q$. Set the total coefficient to zero to obtain $\ddot q=(m_1-m_2)g/(m_1+m_2)$. Check the equal-mass limit, acceleration direction, and $|\ddot q|<g$ for positive masses. Recover tension only if requested, using the original force balance.

Constraints restrict motion. Degrees of freedom identify how many independent labels remain. Generalized coordinates provide those labels while incorporating holonomic geometry. Transformation derivatives convert their changes into particle displacements. Virtual displacement freezes time and selects permissible directions. Virtual work projects equilibrium onto those directions and eliminates ideal reactions. D’Alembert applies the same projection to force minus mass–acceleration, producing motion equations. Lagrangian mechanics will organize the resulting inertial and conservative-force terms using $T$ and $V$, avoiding repeated Cartesian acceleration calculations.

The tangent generalized force is $Q_\theta=aP\cos\theta-mga\sin\theta$. The inertial projection remains $ma^2\ddot\theta$. Therefore

At a static equilibrium, $\ddot\theta=0$ and $\dot\theta=0$. The balance $P\cos\theta=mg\sin\theta$ gives $\tan\theta=P/(mg)$ on the lower branch, agreeing with Chapter 5. Setting $P=0$ recovers the ordinary pendulum equation.

With $\mathbf r=(q,cq^2)$, $\mathbf r_{,q}=(1,2cq)$ and $\mathbf a=(\ddot q,2c\dot q^2+2cq\ddot q)$. Therefore

Gravity $(0,-mg)$ gives $Q_q=-2mgcq$. Equating applied and inertial generalized terms and dividing by $m$ yields

The velocity-squared term arises from the curved transformation. For $c=0$ the wire is horizontal and the equation reduces to $\ddot q=0$. For $c>0$ and small motion near $q=0$, the leading equation is $\ddot q+2gcq=0$, showing a restoring tendency.

The constraints are $z=0$ and $(x-Ut)^2+y^2=a^2$. Both are holonomic and bilateral for a threaded bead. The circle equation is rheonomous for nonzero $U$; the plane equation remains scleronomous. Two independent constraints in three-dimensional space leave one freedom. Choose $\theta$ with $x=Ut+a\cos\theta$, $y=a\sin\theta$, $z=0$.

A free planar particle has two freedoms $(x,y)$. A pendulum with prescribed changing length has one, its angle, because length is known at each time. A pendulum with dynamically variable length generally has two, $(r,\theta)$. A planar double pendulum with two fixed rods has four Cartesian endpoint coordinates minus two independent length constraints, hence two angle freedoms. Initial velocities are additional state data, not configuration freedoms.

Differentiate R1's transformation:

Take the radial reaction to be $\mathbf R=N_r(\cos\theta,\sin\theta,0)$ with signed $N_r$. Its virtual work is $N_ra(-\cos\theta\sin\theta+\sin\theta\cos\theta)\delta\theta=0$. Its actual power is $\mathbf R\cdot\mathbf v=N_rU\cos\theta$: angular terms cancel, but the translating support term remains. This is generally nonzero, demonstrating the difference between virtual ideality and actual worklessness.

First give the force-balance derivation in Solution 5.9. In this application, gravity's generalized component is $mg\sin\alpha$ and the uphill spring force is $-kq$. The smooth plane's normal reaction has zero tangent work. Thus $\delta W=(mg\sin\alpha-kq)\delta q$, giving $q_0=mg\sin\alpha/k$. For $k>0$ the potential $V=\tfrac12kq^2-mgq\sin\alpha$ has positive second derivative, so this equilibrium is stable within the ideal model.

Derive D’Alembert from Newton's law as in Solution 6.8. With $m_1=4$ and $m_2=1$ in kilograms, the common coordinate equation is $5\ddot q=3g$. Therefore $\ddot q=5.88\,\mathrm{m\,s^{-2}}$ downward for the heavier mass. Tension is $4(9.8-5.88)=15.68\,\mathrm N$. The lighter mass check gives $1(9.8+5.88)=15.68\,\mathrm N$. Acceleration is less than $g$, as it should be with both positive masses connected.

The chain rule gives $\mathbf v_a=\sum_j\mathbf r_{a,j}\dot q_j+\mathbf r_{a,t}$. Taking $\partial/\partial\dot q_i$ at fixed coordinates selects $\mathbf r_{a,i}$. Taking $d\mathbf r_{a,i}/dt$ and $\partial\mathbf v_a/\partial q_i$ gives the two matching mixed-derivative sums shown in Solution 4.8. At one position, different initial velocities are possible. Therefore a function of position and velocity can regard these as independent arguments even though each particular physical solution supplies functions $q(t)$ and $\dot q(t)$.

The response vector is $\mathbf r_{,\phi}=(-R\sin\phi,R\cos\phi,b)$, with constant squared length $R^2+b^2$. Its second derivative is $\mathbf r_{,\phi\phi}=(-R\cos\phi,-R\sin\phi,0)$. Acceleration is $\mathbf a=\mathbf r_{,\phi}\ddot\phi+\mathbf r_{,\phi\phi}\dot\phi^2$. The second derivative has zero dot product with $\mathbf r_{,\phi}$ because its two horizontal mixed terms cancel. Hence $m\mathbf a\cdot\mathbf r_{,\phi}=m(R^2+b^2)\ddot\phi$. Gravity gives $Q_\phi=(0,0,-mg)\cdot\mathbf r_{,\phi}=-mgb$. Therefore

Integration gives $\phi(t)=\phi_0+\dot\phi_0t-gbt^2/[2(R^2+b^2)]$. For $b=0$ the helix becomes a horizontal circle: gravity has no tangential component and $\dot\phi$ stays constant. Normal acceleration and the necessary reaction may still be nonzero.

The straight-line rolling equation $\dot x-R\dot\phi=0$ integrates to $x-R\phi=C$, so a velocity form does not imply non-holonomy. A rising smooth wire has zero reaction virtual work but can have nonzero actual reaction power $NU$, so “never do work” is too broad. Finally, a nonzero force can be perpendicular to one chosen variation; equilibrium requires zero virtual work for every admissible variation, with the stated ideality and admissibility assumptions.