Chapter 2

Equations of Motion and Variable-Mass Systems

Theory, worked examples and 5 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.

Theory

2.1 Equations of motion in one dimension

(a) The equation of motion and its integration

The relations between the acceleration, the velocity and the coordinates of a body are its equations of motion. For a particle of mass \(m\) moving along the \(x\)-axis under a force \(F\), Newton’s second law (Chapter 1) gives the second-order differential equation

\[m\frac{d^2x}{dt^2}=F,\]

and its integration, with two initial conditions such as the position and the velocity at \(t=0\), determines the function \(x(t)\) and hence the whole path of the particle. In this chapter we confine ourselves to motion in one dimension and study the forces that occur most often in practice: constant forces, forces that depend on the velocity, and forces that depend linearly on the position. In the second half of the chapter we turn to systems whose mass changes with time, such as rockets, conveyor belts and falling chains.

The acceleration can be written in several equivalent forms,

\[a=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{d^2x}{dt^2},\]

and, whenever the velocity can be regarded as a function of the position, the chain rule gives the useful alternative

\[\boxed{a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}}.\]

The choice between these forms is dictated by the force. If \(F=F(t)\), the equation \(m\,dv/dt=F(t)\) is integrated directly in time. If \(F=F(v)\), the variables separate as \(m\,dv/F(v)=dt\), which gives \(v(t)\); alternatively \(m\,v\,dv/F(v)=dx\) gives \(v(x)\) and so the distance travelled. If \(F=F(x)\), the form \(mv\,dv/dx=F(x)\) integrates to \(\frac12mv^2-\frac12mv_0^2=\int_{x_0}^{x}F\,dx\), which is the work–energy theorem (Chapter 4).

(b) The free particle

A particle on which no force acts is a free particle. Its equation of motion is

\[m\frac{d^2x}{dt^2}=0\quad\Rightarrow\quad\frac{d\dot x}{dt}=0\quad\Rightarrow\quad\dot x=c_1,\]

and a second integration gives \(x=c_1t+c_2\). The velocity is constant and the position varies linearly with time; the momentum \(m\dot x\) is conserved. The two graphs are shown below, the dashed part of the second being the extrapolation of the straight line to earlier times.

(c) Motion under a constant force

When the force \(F\) is constant, as for a particle near the surface of the Earth or a charge in a uniform electric field,

\[\begin{aligned}&m\frac{d^2x}{dt^2}=F\quad\Rightarrow\quad\frac{d\dot x}{dt}=\frac Fm\equiv c_1\\&\Rightarrow\;\dot x=c_1t+c_2,\end{aligned}\]

and a second integration gives

\[x(t)=\frac{c_1}{2}t^2+c_2t+c_3.\]

Here \(c_1=F/m\) is the constant acceleration, \(c_2\) the initial velocity and \(c_3\) the initial position. The velocity grows linearly and the position quadratically with time, while the acceleration is constant.

2.2 Velocity-dependent resistive forces

A body moving through a fluid experiences a drag force that opposes its velocity. At low speeds the drag is nearly proportional to the speed (viscous or Stokes drag), while at higher speeds it is nearly proportional to the square of the speed. We treat the two cases in turn.

(a) Linear drag on a horizontal motion

Let a particle with initial velocity \(v_0\) at \(t=0\), and initially at the origin, be subject only to the force \(F=-bv\) with \(b>0\). Then

\[\begin{aligned}&m\frac{dv}{dt}=-bv\quad\Rightarrow\quad\frac{dv}{v}=-\frac bm\,dt\\&\Rightarrow\;\ln v=-\frac{bt}{m}+\ln v_0,\end{aligned}\]

so that

\[\boxed{v(t)=v_0\exp\left(-\frac{bt}{m}\right)}.\]

The velocity decays exponentially with time constant \(m/b\). A further integration with \(x(0)=0\) gives

\[x(t)=\frac{mv_0}{b}\left[1-\exp\left(-\frac{bt}{m}\right)\right].\]

The particle never comes to rest in a finite time, yet the total distance it travels is finite and equal to \(mv_0/b\). The same result follows from the form \(mv\,dv/dx=-bv\), which gives \(v=v_0-bx/m\), so that \(v\) falls to zero at \(x=mv_0/b\).

(b) Fall under gravity with linear drag

Let a particle be released from rest at \(t=0\) and fall under gravity against the drag \(-bv\). Taking \(x\) downwards,

\[m\frac{dv}{dt}=mg-bv\quad\Rightarrow\quad\frac{dv}{g-\dfrac bmv}=dt.\]

Putting \(y=g-bv/m\), so that \(dv=-(m/b)\,dy\), we find \(\ln y=-bt/m+\text{const}\); the condition \(v=0\) at \(t=0\) fixes the constant as \(\ln g\), and

\[\begin{aligned}&\ln\frac{g-\dfrac bmv}{g}=-\frac{bt}{m}\\&\Rightarrow\;\boxed{v(t)=\frac{mg}{b}\left[1-\exp\left(-\frac{bt}{m}\right)\right]}.\end{aligned}\]

The velocity rises from zero and approaches the terminal velocity \(v_t=mg/b\), at which the drag balances the weight. Integrating once more with \(x(0)=0\),

\[x(t)=\frac{mg}{b}t-\frac{m^2g}{b^2}\left[1-\exp\left(-\frac{bt}{m}\right)\right].\]

For \(t\ll m/b\) the expansion of the exponential gives \(x\approx\frac12gt^2\), the free fall, while for \(t\gg m/b\) the graph of \(x(t)\) approaches the straight line \(x=(mg/b)(t-m/b)\), which meets the time axis at \(t=m/b\).

(c) Quadratic drag

For a drag force of magnitude \(bv^2\) the terminal speed in a fall is fixed by \(bv_t^2=mg\), that is \(v_t=\sqrt{mg/b}\). A body released from rest and falling with \(x\) measured downwards obeys

\[m\frac{dv}{dt}=mg-bv^2\quad\Rightarrow\quad\frac{dv}{v_t^2-v^2}=\frac{g}{v_t^2}\,dt.\]

With the partial fractions \(\dfrac{1}{v_t^2-v^2}=\dfrac{1}{2v_t}\left(\dfrac{1}{v_t+v}+\dfrac{1}{v_t-v}\right)\) the integral from \(0\) to \(t\) is

\[\begin{aligned}&\frac{1}{2v_t}\ln\frac{v_t+v}{v_t-v}=\frac{gt}{v_t},\\&\text{so}\ \boxed{v=v_t\tanh\frac{gt}{v_t}},\qquad v_t=\sqrt{\frac{mg}{b}}.\end{aligned}\]

A second integration gives the distance fallen, \(x=(v_t^2/g)\ln\cosh(gt/v_t)\).

For a body thrown upwards with speed \(v_0\), gravity and drag both act downwards. With \(v\) now measured upwards,

\[\begin{aligned}&m\frac{dv}{dt}=-mg-bv^2\quad\Rightarrow\quad\frac{dv}{v_t^2+v^2}=-\frac{g}{v_t^2}\,dt\\&\Rightarrow\;\tan^{-1}\frac{v}{v_t}=\tan^{-1}\frac{v_0}{v_t}-\frac{gt}{v_t}.\end{aligned}\]

The body reaches its highest point, \(v=0\), at the time \(t_m\) given by

\[\boxed{v_0=v_t\tan\frac{gt_m}{v_t}}\qquad\text{or}\qquad t_m=\frac{v_t}{g}\tan^{-1}\frac{v_0}{v_t}.\]

Using \(mv\,dv/dx=-mg-bv^2\) instead, the maximum height is \(h=\dfrac{v_t^2}{2g}\ln\left(1+\dfrac{v_0^2}{v_t^2}\right)\), which reduces to \(v_0^2/2g\) when the drag is negligible (\(v_0\ll v_t\)).

Finally, in horizontal motion with quadratic drag alone, \(m\,dv/dt=-bv^2\) integrates to \(1/v-1/v_0=bt/m\), so that

\[v=\frac{mv_0}{m+bv_0t},\qquad x=\frac mb\ln\left(1+\frac{bv_0t}{m}\right).\]

The speed decreases only as \(1/t\) at long times, and the distance travelled grows without limit.

(d) Power-law resistance

Consider more generally a resistive force of magnitude \(\eta v^\alpha\). Separating the variables in \(m\,dv/dt=-\eta v^\alpha\) gives the time to stop,

\[t_s=\frac m\eta\int_0^{v_0}\frac{dv}{v^\alpha}=\frac m\eta\frac{v_0^{1-\alpha}}{1-\alpha}\qquad(\alpha<1),\]

while the form \(mv\,dv/dx=-\eta v^\alpha\) gives the stopping distance

\[x_s=\frac m\eta\int_0^{v_0}v^{1-\alpha}\,dv=\frac m\eta\frac{v_0^{2-\alpha}}{2-\alpha}\qquad(\alpha<2).\]

Hence the body stops in a finite time only if \(\alpha<1\), and within a finite distance only if \(\alpha<2\). The linear law \(\alpha=1\) gives an infinite time but a finite distance, and the quadratic law \(\alpha=2\) gives both infinite, in agreement with Sections 2.2(a) and (c).

Exam tip. For a force that depends only on the velocity, use \(m\,dv/dt=F(v)\) when time is asked for and \(mv\,dv/dx=F(v)\) when distance is asked for; each separates at once.

2.3 Linear forces and the damped oscillator

(a) A linear restoring force

A particle of mass \(m\) acted on by the restoring force \(F=-kx\) (\(k>0\)) obeys

\[m\frac{d^2x}{dt^2}=-kx\quad\Rightarrow\quad\frac{d^2x}{dt^2}+\omega^2x=0,\qquad\omega^2=\frac km.\]

The general solution is \(x=Ae^{i\omega t}+Be^{-i\omega t}\), or equivalently \(x=C\sin\omega t+D\cos\omega t\). Suppose that \(x=0\) and \(\dot x=a\) at \(t=0\). The first condition gives \(B=-A\) and hence \(x=c\sin\omega t\) with \(c=2iA\); the second gives \(\dot x(0)=c\omega=a\). Therefore

\[x=\frac a\omega\sin\omega t,\]

which is simple harmonic motion of angular frequency \(\omega\) and amplitude \(a/\omega\). Oscillations are studied in detail in Chapter 12.

(b) A linear repulsive force

If instead \(F=+kx\) (\(k>0\)), the force pushes the particle away from the origin. Writing \(\gamma^2=k/m\),

\[\frac{d^2x}{dt^2}-\gamma^2x=0\quad\Rightarrow\quad x=Ae^{\gamma t}+Be^{-\gamma t}.\]

With \(x=0\) and \(\dot x=a\) at \(t=0\), \(B=-A\), so \(x=c\sinh\gamma t\) with \(c=2A\), and \(\dot x(0)=c\gamma=a\). Hence

\[x=\frac a\gamma\sinh\gamma t,\qquad\dot x=a\cosh\gamma t.\]

The motion is not oscillatory; the particle runs away exponentially, as near the top of a potential hill (Chapter 5).

(c) The damped harmonic oscillator

Let the particle be subject both to the restoring force \(-kx\) and to the damping force \(-bv\), with \(k,b>0\). Then

\[\begin{aligned}&m\frac{d^2x}{dt^2}=-kx-b\frac{dx}{dt}\quad\Rightarrow\quad\boxed{\frac{d^2x}{dt^2}+2\gamma\frac{dx}{dt}+\omega_0^2x=0},\\&\omega_0^2=\frac km,\quad 2\gamma=\frac bm,\end{aligned}\]

where \(\omega_0\) is the natural frequency of the undamped oscillator. The equation is linear with constant coefficients, so we try \(x=e^{st}\), which gives the characteristic equation

\[s^2+2\gamma s+\omega_0^2=0\quad\Rightarrow\quad s=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}.\]

Three cases arise.

Underdamping, \(\gamma<\omega_0\). The roots are complex, \(s=-\gamma\pm i\omega_d\) with \(\omega_d=\sqrt{\omega_0^2-\gamma^2}\), and the real solution is

\[x=Ae^{-\gamma t}\cos\left(\omega_dt+\delta\right).\]

The motion is oscillatory with a constant angular frequency \(\omega_d<\omega_0\), but the amplitude \(Ae^{-\gamma t}\)

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Chapter 2 Equations of Motion and Variable-Mass Systems