3.5 - Variable forces depending on time or velocity

3.5 - Variable forces depending on time or velocity

A force that changes with time or velocity usually produces changing acceleration. You will form Newton’s second-law equations with consistent signs, solve separable equations, and distinguish a speed approached indefinitely from a speed reached at a finite time.

Forming the differential equation

Can a particle keep slowing down without ever stopping? To answer that, first replace a constant acceleration by a rate of change of velocity.

Suppose a particle of mass 3 kg3\,\mathrm{kg} moves to the right at 3 m s−13\,\mathrm{m\,s^{-1}}. A driving force of 12 N12\,\mathrm{N} acts rightwards and resistance of 6 N6\,\mathrm{N} acts leftwards. The resultant is 6 N6\,\mathrm{N} rightwards, so its acceleration is 2 m s−22\,\mathrm{m\,s^{-2}} rightwards. If the resistance changes as the velocity changes, this calculation becomes an equation involving dv/dt\mathrm{d}v/\mathrm{d}t.

Newton’s second law along a line

Choose a positive direction and use signed force components and signed velocity vv:

mdvdt=resultant force in the positive direction.m\frac{\mathrm{d}v}{\mathrm{d}t}=\text{resultant force in the positive direction}.

The mass mm is constant. Not in the formula booklet: learn it.

Velocity includes direction; speed is ∣v∣|v|. A resistance acts opposite to the motion. A resistance whose magnitude is proportional to speed can therefore have signed component −kv-kv, with k>0k>0: it is negative when v>0v>0 and positive when v<0v<0. The direction of the driving force need not change when the particle reverses.

Worked example 1: resistance during rightward motion

Question. A particle of mass 3 kg3\,\mathrm{kg} moves horizontally. A driving force of 12 N12\,\mathrm{N} acts rightwards and resistance has magnitude 2∣v∣ N2|v|\,\mathrm{N}, where vv is the numerical value of rightward velocity in m s−1\mathrm{m\,s^{-1}}. Form the differential equation while v>0v>0, and find the acceleration when v=3v=3.

Plan. Take rightwards as positive and subtract the leftward resistance.

For rightward motion, the resistance is 2v N2v\,\mathrm{N} leftwards. Hence

3dvdt=12−2v.3\frac{\mathrm{d}v}{\mathrm{d}t}=12-2v.

At the stated velocity,

dvdt=12−2(3)3=2.\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{12-2(3)}{3}=2.

Answer. The equation is 3dv/dt=12−2v3\mathrm{d}v/\mathrm{d}t=12-2v; the acceleration is 2 m s−22\,\mathrm{m\,s^{-2}} rightwards.

Check. In the next very short interval, the positive acceleration increases the positive velocity. That increases the opposing resistance and reduces the resultant, as the equation predicts.

Keep the same forces and positive direction, but change the direction of motion. This tests whether the resistance sign follows the motion rather than the coordinate axis.

Worked example 2: the same resistance during leftward motion

Question. For the same particle and force laws, form the equation while v<0v<0, and find the acceleration when v=−3v=-3.

Plan. The particle now moves leftwards, so both the drive and the resistance act rightwards.

The resistance magnitude is 2∣v∣=−2v2|v|=-2v, since v<0v<0. Therefore

3dvdt=12+(−2v)=12−2v\begin{aligned} 3\frac{\mathrm{d}v}{\mathrm{d}t}&=12+(-2v) \\ &=12-2v \end{aligned}

The same signed equation covers both directions. At v=−3v=-3,

dvdt=12−2(−3)3=6.\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{12-2(-3)}{3}=6.

Answer. The equation remains 3dv/dt=12−2v3\mathrm{d}v/\mathrm{d}t=12-2v; the acceleration is 6 m s−26\,\mathrm{m\,s^{-2}} rightwards.

Check. Directly, the drive is 12 N12\,\mathrm{N} rightwards and the resistance is 6 N6\,\mathrm{N} rightwards. Their resultant 18 N18\,\mathrm{N} gives 18/3=6 m s−218/3=6\,\mathrm{m\,s^{-2}}. The negative velocity becomes less negative, so the speed decreases.

For vertical motion, include weight as another signed force. With downwards positive during a fall, weight contributes +mg+mg and an upward resistance contributes a negative term. For a force specified as a function of time, put that expression directly on the force side; for example, a rightward force F(t)F(t) acting alone gives mdv/dt=F(t)m\mathrm{d}v/\mathrm{d}t=F(t).

Watch out

Positive acceleration does not always mean increasing speed. If v<0v<0 and dv/dt>0\mathrm{d}v/\mathrm{d}t>0, the velocity moves towards zero and the particle slows down.

Use the force directions before substituting the negative velocity.

Separating variables

A differential equation is separable when its velocity terms and time terms can be collected on opposite sides before integrating.

For example, dv/dt=−2v/(1+t)\mathrm{d}v/\mathrm{d}t=-2v/(1+t) can be written as dv/v=−2 dt/(1+t)\mathrm{d}v/v=-2\,\mathrm{d}t/(1+t) on a branch where v≠0v\ne0. Integrating changes the equation into a relation between vv and tt. Differentiating the force expression would answer a different question.

Separation and initial data

For

dvdt=f(t)g(v),∫1g(v) dv=∫f(t) dt+C\frac{\mathrm{d}v}{\mathrm{d}t}=f(t)g(v), \qquad \int\frac{1}{g(v)}\,\mathrm{d}v=\int f(t)\,\mathrm{d}t+C

on an interval where g(v)≠0g(v)\ne0. Use an initial condition to determine CC. Check separately any constant velocity satisfying g(v)=0g(v)=0, because division can remove it.

The separation procedure is not in the formula booklet: learn it. The standard power, exponential and logarithmic integrals are in the formula booklet.

If the force depends only on time, g(v)=1g(v)=1, so integration is direct. The first example uses that case.

Worked example 3: a force depending only on time

Question. A particle of mass 2 kg2\,\mathrm{kg} initially moves rightwards at 2 m s−12\,\mathrm{m\,s^{-1}}. While it moves rightwards, its only horizontal force is 4/(1+t) N4/(1+t)\,\mathrm{N} leftwards. Find vv in terms of tt, and state the interval before its first stop.

Plan. Integrate the signed acceleration and use v=2v=2 when t=0t=0.

Taking rightwards as positive,

2dvdt=−41+tv=−2ln⁡(1+t)+C\begin{aligned} 2\frac{\mathrm{d}v}{\mathrm{d}t}&=-\frac{4}{1+t} \\ v&=-2\ln(1+t)+C \end{aligned}

Since 2=−2ln⁡1+C2=-2\ln1+C, we obtain C=2C=2. The first zero of the velocity satisfies

2−2ln⁡(1+t)=0⟹t=e−1.2-2\ln(1+t)=0 \quad\Longrightarrow\quad t=e-1.

Answer. v=2−2ln⁡(1+t)v=2-2\ln(1+t), valid for the stated rightward phase 0≤t<e−10\le t<e-1, with first rest at t=e−1 st=e-1\,\mathrm{s}.

Check. The accumulated leftward impulse up to e−1e-1 is ∫0e−14/(1+t) dt=4 N s\int_0^{e-1}4/(1+t)\,\mathrm{d}t=4\,\mathrm{N\,s}. It exactly removes the initial rightward momentum 2×2=4 kg m s−12\times2=4\,\mathrm{kg\,m\,s^{-1}}.

Now change the force so that its magnitude also contains a factor of speed. The factor 1/(1+t)1/(1+t) is unchanged, but the velocity must join the left side before integration.

Continue this lesson.

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