# Differential Equations

2nd Order Equations# Newton’s Second Law of Motion

Theory

Problems 1-2

Newton’s second law establishes a relationship between the force \(\mathbf{F}\) acting on a body of mass \(m\) and the acceleration \(\mathbf{a}\) caused by this force.

The acceleration \(\mathbf{a}\) of a body is directly proportional to the acting force \(\mathbf{F}\) and inversely proportional to its mass \(m,\) that is

This formulation is valid for systems with constant mass. When the mass changes (for example, in the case of relativistic motion), Newton’s second law takes the form

where \(\mathbf{p}\) is the impulse (momentum) of the body.

In general, the force \(\mathbf{F}\) can depend on the coordinates of the body, i.e., the radius vector \(\mathbf{r},\) its velocity \(\mathbf{v},\) and time \(t:\)

Below we consider the special cases where the force \(\mathbf{F}\) depends only on one of these variables.

### Force Depends on Time: \(\mathbf{F} = \mathbf{F}\left( t \right)\)

Assuming that the motion is one-dimensional, Newton’s second law is written as the second order differential equation:

Integrating once, we find the velocity of the body \(v\left( t \right):\)

Here we assume that the body begins to move at time \(t = 0\) with the initial velocity \(v\left( {t = 0} \right) = {v_0}.\) Integrating again, we get the law of motion \(x\left( t \right):\)

where \({x_0}\) is the initial coordinate of the body, \(\tau\) is the variable of integration.

### Force Depends on the Velocity: \(\mathbf{F} = \mathbf{F}\left( {\mathbf{v}} \right)\)

When a solid body moves in a liquid or gaseous environment it experiences a drag force (or a frictional force). At low velocities \(\mathbf{v},\) this force is proportional to the velocity \(\mathbf{v}:\)

The coefficient \(k\) in turn is proportional to the viscosity \(\eta.\) In particular, if the body has a spherical shape, the drag force is described by the Stokes’ law:

where \(R\) is the radius of the ball, \(\eta\) is the viscosity of the environment.

In this mode of motion Newton’s second law is written (in one-dimensional approximation) as the following differential equation:

Integrating this equation with the initial condition \(v\left( {t = 0} \right) = {v_0}\) gives

{\frac{{dv}}{v} = – \frac{k}{m}dt,\;\;}\Rightarrow

{\int\limits_{{v_0}}^v {\frac{{du}}{u}} = – \frac{k}{m}\int\limits_0^t {d\tau } .}

\]

Here \(u\) and \(\tau\) are integration variables. The velocity of the body varies from \({v_0}\) to \(v\) as the time changes from \(0\) to \(t.\) Consequently,

{\ln v – \ln {v_0} = – \frac{k}{m}t,\;\;}\Rightarrow

{\ln \frac{v}{{{v_0}}} = – \frac{k}{m}t,\;\;}\Rightarrow

{v\left( t \right) = {v_0}{e^{ – {\large\frac{k}{m}\normalsize}t}}.}

\]

Thus, if the drag force is proportional to the velocity of the body, its speed will decrease exponentially.

The law of motion \(x\left( t \right)\) can be easily found by repeated integration:

{x\left( t \right) = {x_0} + \int\limits_0^t {v\left( \tau \right)d\tau } }

= {{x_0} + \int\limits_0^t {{v_0}{e^{ – {\large\frac{k}{m}\normalsize}\tau }}d\tau } }

= {{x_0} – \frac{{m{v_0}}}{k}\left( {{e^{ – {\large\frac{k}{m}\normalsize}t}} – 1} \right) }

= {{x_0} + \frac{{m{v_0}}}{k}\left( {1 – {e^{ – {\large\frac{k}{m}\normalsize} t}}} \right).}

\]

The last formula shows that the path traversed by the body to a complete stop, is equal to \(\large\frac{{m{v_0}}}{k}\normalsize,\) i.e., proportional to the initial momentum of the body \(m{v_0}.\)

As the velocity of a body increases, the physics of the process changes. The kinetic energy of the body begins to be spent not only on the friction between the layers of liquid, but also on the movement of the fluid in front of the body. In this mode, the drag force becomes proportional to the square of the velocity:

where \(\mu\) is the coefficient of proportionality, \(S\) is the cross-sectional area of the body, \(\rho\) is the density of the medium.

The nonlinear regime desctribed above appears on the conditions

where \(\mathbf{\text{Re}}\) is the dimensionless Reynolds number, \(\eta\) is the viscosity of the medium, \(L\) is a characteristic cross-sectional size, for example, the radius of the body.

Considering one-dimensional motion, we write Newton’s second law for this case in the form

Integrating, we find the velocity of the body:

{\frac{{dv}}{{{v^2}}} = – \frac{{\mu \rho S}}{m}dt,\;\;}\Rightarrow

{\int\limits_{{v_0}}^v {\frac{{du}}{{{u^2}}}} = – \frac{{\mu \rho S}}{m}\int\limits_0^t {d\tau } .}

\]

Here \(u\) and \(\tau\) again denote the integration variables. For the time \(t,\) the velocity of the body will decrease from an initial value \({v_0}\) to the final value \(v.\) As a result, we obtain

{- \left( {\frac{1}{v} – \frac{1}{{{v_0}}}} \right) = – \frac{{\mu \rho S}}{m}t,\;\;}\Rightarrow

{\frac{1}{v} = \frac{1}{{{v_0}}} + \frac{{\mu \rho S}}{m}t,\;\;}\Rightarrow

{v\left( t \right) = \frac{1}{{\frac{1}{{{v_0}}} + \frac{{\mu \rho S}}{m}t}} }={ \frac{{{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}t}}.}

\]

Integrate again to find the law of motion \(x\left( t \right):\)

{x\left( t \right) = \int\limits_0^t {\frac{{{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}d\tau } }

= {\int\limits_0^t {\frac{{\cancel{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}\frac{{d\left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)}}{{\frac{{\mu \rho S\cancel{v_0}}}{m}}}} }

= {\frac{m}{{\mu \rho S}}\int\limits_0^t {\frac{{d\left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}} }

= {\frac{m}{{\mu \rho S}}\left[ {\left. {\ln \left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)} \right|_0^t} \right] }

= {\frac{m}{{\mu \rho S}}\ln \left( {1 + \frac{{\mu \rho S{v_0}}}{m}t} \right).}

\]

It is important to bear in mind that these formulas are valid for sufficiently large values of the velocity: at lower velocities, this model is physically incorrect, since the drag force begins to depend on the velocity linearly (this case was considered previously).

### Force Depends on the Position: \(\mathbf{F} = \mathbf{F}\left( x \right)\)

Examples of forces that depend only on the coordinate are, in particular:

- Elastic force \(F = -kx;\)
- Force of gravitational attraction \(F = – G\large\frac{{{m_1}{m_2}}}{{{x^2}}}\normalsize.\)

The motion of a body of mass \(m\) connected to a spring under the force of elasticity is determined by the differential equation

This equation describes the undamped periodic oscillations with a period

In the case of gravitational attraction, the body motion is described by the nonlinear differential equation

where \(M\) is the mass of the attracting body (for example, the mass of the Earth or the Sun), \(G\) is the universal gravitational constant.

The solution of this equation is given on the page Newton’s Law of Universal Gravitation.

In the case where the force depends on the coordinate, the acceleration is conveniently represented in the form:

Then the differential equation can be written as

Separating the variables \(v\) and \(x,\) we have

{mvdv = F\left( x \right)dx,\;\;}\Rightarrow

{m\int\limits_{{v_0}}^v {udu} = \int\limits_0^L {F\left( x \right)dx} ,\;\;}\Rightarrow

{\frac{{m{v^2}}}{2} – \frac{{mv_0^2}}{2} }={ \int\limits_0^L {F\left( x \right)dx} .}

\]

The last equation expresses the law of conservation of energy. The left side describes the change in kinetic energy, and the right side corresponds to the work of a variable force \({F\left( x \right)}\) when the body is moved by a distance \(L.\)

The subsequent integration of the function \({v\left( t \right)}\) allows to find the law of motion \({x\left( t \right)}.\) Unfortunately, this is not always possible because of the cumbersome analytical expressions for \({v\left( t \right)}.\)

## Solved Problems

Click on problem description to see solution.

### ✓ Example 1

A body begins to fall from a height \(H\) under the action of gravity. While falling it experiences resistance proportional to the velocity. Determine the time of the drop.

### ✓ Example 2

At the initial moment, a chain of length \(L\) hangs over the edge of the table so that the force of gravity is balanced by the friction force (Figure \(3\)). As a result of a small shift \(\varepsilon\) the chain starts sliding. Determine the time \(T,\) for which the chain completely slips off the table. The coefficient of friction between the chain and the surface of the table is equal to \(\mu.\)

Theory

Problems 1-2

### Related Pages

- The Indefinite Integral and Basic Formulas of Integration. Table of Integrals.
- The Definite Integral and Fundamental Theorem of Calculus
- Separable Equations
- Second Order Linear Homogeneous Differential Equations with Constant Coefficients
- Second Order Linear Nonhomogeneous Differential Equations with Constant Coefficients
- Newton’s Law of Universal Gravitation