Mv8–Mv10 - Differential equations for variable forces

Mv8–Mv10 - Differential equations for variable forces

Translate a variable-force model into differential equations, verify a supplied solution, and apply initial or boundary conditions. You will choose between time and displacement as the independent variable and keep track of the physical branch of the motion.

From force models to time equations

A car starts below its terminal speed. Can it keep accelerating forever without ever exceeding that speed?

A force law can make the acceleration shrink as speed increases. A differential equation records that feedback. Its limiting speed can be approached without being reached at any finite time.

You already know

Choose a positive direction and write resultant force =ma=m\mathbf{a}. With velocity depending on time, a=dv/dt\mathbf{a}=d\mathbf{v}/dt. Core Pure supplies the calculus needed to differentiate a proposed solution.

A resistance opposes motion. If its magnitude is given in terms of speed, state the direction of travel before assigning a sign. A model written for v≥0v\ge0 may need a different signed expression after reversal.

Time-dependent motion equations

For constant mass,

mdvdt=Fresultant,drdt=vm\frac{d\mathbf{v}}{dt}=\mathbf{F}_{\mathrm{resultant}}, \qquad \frac{d\mathbf{r}}{dt}=\mathbf{v}

In components, each direction has its own Newton's-second-law equation. Forces may depend on time, position or velocity.

Not in the formula booklet: learn it.

Formulating an equation, verifying a supplied solution and finding constants from conditions are different jobs. A supplied solution can be checked directly without finding it afresh.

Worked example 1: resistance proportional to speed

Question. A 2 kg2\,\mathrm{kg} particle moves right on a smooth horizontal track. A driving force of 20 N20\,\mathrm{N} acts right, and resistance has magnitude 2v N2v\,\mathrm{N}, where v≥0v\ge0 is its speed in m s−1\mathrm{m\,s^{-1}}. Form its velocity equation and find its terminal speed.

Plan. Take right positive, subtract resistance and set acceleration to zero for a possible steady speed.

2dvdt=20−2vresultant forcedvdt+v=10divide by massdvdt=0  ⟹  v=10steady-speed condition\begin{aligned} 2\frac{dv}{dt}&=20-2v && \text{resultant force}\\ \frac{dv}{dt}+v&=10 && \text{divide by mass}\\ \frac{dv}{dt}=0&\implies v=10 && \text{steady-speed condition} \end{aligned}

Why this step? Resistance is leftwards while the particle moves right, so its force contribution is negative.

Answer. dv/dt=10−vdv/dt=10-v, with terminal speed 10 m s−110\,\mathrm{m\,s^{-1}}.

Check. For 0≤v<100\le v<10, acceleration is positive; for v>10v>10, it is negative.

Extend only the number of directions: retain the first force component and add resistance in a perpendicular horizontal direction.

Worked example 2: a two-dimensional model

Question. A 2 kg2\,\mathrm{kg} particle moves in a horizontal plane. Its resultant force components, in newtons, are 20−2vx20-2v_x and −4vy-4v_y, where the velocity components are in SI units. Form its component equations.

Plan. Apply Newton's second law separately in the two perpendicular directions.

2dvxdt=20−2vxfirst force component2dvydt=−4vysecond force componentdvxdt=10−vx,dvydt=−2vydivide by mass\begin{aligned} 2\frac{dv_x}{dt}&=20-2v_x && \text{first force component}\\ 2\frac{dv_y}{dt}&=-4v_y && \text{second force component}\\ \frac{dv_x}{dt}&=10-v_x,\quad \frac{dv_y}{dt}=-2v_y && \text{divide by mass} \end{aligned}

Why this step? A component resistance such as −4vy-4v_y automatically opposes either sign of vyv_y; it is a signed force law, not an unsigned magnitude.

Answer. The two equations displayed above, together with dx/dt=vxdx/dt=v_x, dy/dt=vydy/dt=v_y if positions are required.

Check. At vy=0v_y=0, the transverse force and acceleration both vanish.

The numerical coefficients in a force law carry the units needed to produce newtons; do not treat them as dimensionless when checking consistency.

Watch out

Wrong: use constant-acceleration formulae because the driving force is constant. Right: include resistance in the resultant. If that changes with velocity, acceleration generally changes too.

Change the mass and driving component.

Using displacement instead of time

When force or a supplied relation is expressed in terms of displacement, use a=v dv/dsa=v\,dv/ds on a branch where velocity is a function of position.

The chain rule gives

dvdt=dvdsdsdt=vdvds\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}

This can be useful even when a time-based solution would be awkward.

Acceleration on a monotone branch

a=vdvds,mvdvds=Fresultanta=v\frac{dv}{ds},\qquad mv\frac{dv}{ds}=F_{\mathrm{resultant}}

Work on a specified direction-of-motion branch. At a turning point, avoid dividing by v=0v=0; use a supplied v2v^2 relation or a limiting argument.

Not in the formula booklet: learn it.

For a supplied squared-speed relation, differentiating v2v^2 gives 2v dv/ds2v\,dv/ds. This verifies the equation without choosing an incorrect velocity sign.

Worked example 3: a supplied position–speed relation

Question. A 2 kg2\,\mathrm{kg} particle moves outward from s=0s=0 with speed 6 m s−16\,\mathrm{m\,s^{-1}}. Its signed resultant force is −4s N-4s\,\mathrm{N}. Verify v2=36−2s2v^2=36-2s^2 on the outward branch, and find the first turning displacement.

Plan. Form the displacement equation, differentiate the supplied relation and check the initial value.

2vdvds=−4sNewton’s second lawv2=36−2s2  ⟹  2vdvds=−4sdifferentiates=0  ⟹  v2=36,v=6outward initial conditionv=0  ⟹  s=32first positive turning point\begin{aligned} 2v\frac{dv}{ds}&=-4s && \text{Newton's second law}\\ v^2=36-2s^2&\implies2v\frac{dv}{ds}=-4s && \text{differentiate}\\ s=0&\implies v^2=36,\quad v=6 && \text{outward initial condition}\\ v=0&\implies s=3\sqrt2 && \text{first positive turning point} \end{aligned}

Why this step? The initial outward motion selects v=+36−2s2v=+\sqrt{36-2s^2} before the turn, not the negative square root.

Answer. The relation is verified; the outward branch has 0≤s≤32 m0\le s\le3\sqrt2\,\mathrm{m}.

Check. The squared speed is positive inside that interval and zero at its endpoint.

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