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Newton’s Second Law#
Newton’s second law says that \(F=ma\): the force \(F\) on a body with mass \(m\) causes an acceleration \(a\) of that body. We will explore how this law allows us to describe a range of different types of motion, but in order to do that systematically, we have to introduce the mathematical notions of position, velocity, and acceleration. Afterwards, we will link this force to the forces we discussed in the first lecture.
Position, velocity and acceleration as functions of time#
The position of a particle can change over time, and therefore we have to define the position as a function of time. We label the position function \(x(t)\). Since we have only one function \(x(t)\), this describes the position of the particle along a line, i.e., it is a one-dimensional description. This may or may not be enough for the problem you want to describe. We will return to this in the next lecture.
The velocity of a particle (in one dimension) can be thought of as “how fast the position changes”. Hence, the velocity \(v\) is the rate of change over time of the position. Mathematically, it is the time derivative of the position:
This is true regardless of the particulars of the position \(x(t)\). The particle can move fast at first, then slow down; or in some other complicated way. The velocity is defined as the rate of change of the position. It means that if a particle disappears at time \(t_0\) from one position and instantly reappears some distance away, the derivative at \(t_0\) does not exist — the function \(x(t)\) is discontinuous and not differentiable — and the particle does not have a velocity at \(t_0\). Luckily, this is not the kind of motion that we expect to encounter much in physics!
Similar to the velocity, the acceleration \(a(t)\) of a particle is defined as the rate of change of the velocity of the particle:
Again, you should treat \(a(t)\) as a mathematical function of time (\(t\)) and observe all the mathematical rules for functions. Note how we can just substitute equation above for \(v(t)\) and obtain the second derivative of position.
Treating position, velocity and acceleration as functions of time allows us to bring in the mathematical manipulations that are legal to perform on these functions. For example, we can leverage the fact that integration is the anti-derivative: in order to find the position of a particle when the velocity or acceleration is given, we have to integrate the velocity, and to find the velocity we have to integrate the acceleration:
Note that we need to add constants \(x(t_0)\) and \(v(t_0)\) to determine the starting value of the position and velocity. As you remember, these integration constants are normal for integration, and here you see they have a distinct physical meaning (the position and velocity at time \(t_0\)). These constants disappear when we take the derivative with respect to \(t\). You should verify that when you calculate the time derivatives of \(x(t)\) and \(v(t)\) in these integral forms, you retrieve \(v(t)\) and \(a(t)\).
Also note the change in integration variable \(t'\) to distinguish it from the integration limit \(t\). You cannot have the integration variable and the limit be the same symbol since these refer to two different things. This is an important property of functions: imagine you have a function \(x(t)\) but \(t\) is already used elsewhere in the problem. You then have to rename the variable to something else. Often people use primes (\(t'\)) as above, or people start labelling specific times with indices, like \(t_1\), \(t_2\), etc. Taking care with your notation is half the work of getting towards the solution of a problem.
In many ways, mechanics is the study of the position and velocity of bodies, so you should always take these seriously as mathematical functions, and treat them accordingly.
Worked Example: Position and acceleration from velocity#
Imagine a particle moves with velocity
where \(v_0 = 5.0 \,\text{m/s}\) and \(\omega = 1.3\) rad·s\(^{-1}\). We want to know the position and acceleration of the particle at arbitrary times \(t\).
To obtain the acceleration, we need to take the derivative of the velocity:
Note how we do not yet substitute the numerical values for \(v_0\) and \(\omega\), since that would gain us nothing and lose a lot of clarity in the expression. Only when we have a numerical value for \(t\) does it make sense to substitute the values.
To obtain the position, we need to take the anti-derivative of \(v(t)\):
Next, let \(t_0 = 0.0\) s, and \(x(t_0) = x(0.0) = 2.5\) cm. Now we can calculate the position, velocity and acceleration of the particle for any time \(t\). For this example we take \(t=4.0\) s:
Note how the integration constant tells us where the particle starts at \(t=0\).
We will often stop shy of substituting actual numbers into our equations. In some sense the problem is “solved” when we have the equations of motion. To find any specific value, for example for comparison with an experiment, we can just substitute the given values into the equation of motion. You have already been well-trained in this aspect.
Equation of motion for problems in one dimension#
We can now create the mathematical description of the motion of the particles in the examples from the previous lecture. Newton’s second law tells us that \(F = ma\), and for conservative forces we established that the force is equal to the space derivative of a potential \(V\). If we write the acceleration as the second time derivative of position, we have the general relation
This is a so-called differential equation in \(x\), since it involves \(x(t)\) and its derivatives. Note that this is a higher-level description of physical processes than you are used to. You cannot substitute numbers into these differential equations. Instead, we use these equations to find functions, like the position \(x(t)\), into which we can substitute our numbers. This higher level of abstraction allows us to cover a large family of problems with a single simple unified mathematical description. The downside is that \(x\) appears both as a function (left-hand side) and a variable (right-hand side), so we need new techniques to solve this equation. If this approach works, i.e., if the numbers we get match our experiments, then we have strong evidence that the conceptual framework of forces and potentials that led to this equation is correct, and we have a coherent theory that explains the natural world in terms of forces and potentials. We will look at a number of examples / applications of this equation in the remainder of this lecture.
Type of motion 1: Inertial motion#
Let’s start with the simplest case, when there is no force acting on the particle: \(F=0\). Using Newton’s second law, we find that \(F = ma = 0\), which means that the acceleration \(a\) is equal to zero (we are not considering massless particles in this course). Qualitatively, you know what it means for a particle to have zero acceleration: it has a constant velocity. However, in preparation for the more involved examples, we will see how we treat this mathematically. In order to find the velocity of a particle, we have to integrate the acceleration over time. From the earlier integral form we find:
since \(a(t')=0\). In other words, the velocity \(v(t)\) at any time \(t\) is equal to the velocity \(v(t_0)\) at some time \(t_0\). That is just a fancy way of saying that the velocity is constant, and its value is \(v(t_0)\).
We can go further and integrate the velocity to get the position as a function of time:
Often we use the special value \(t_0 = 0\), and the constants \(x(t_0)\) and \(v(t_0)\) are given the special symbols \(x_0\) and \(v_0\), respectively. With these substitutions, we obtain the very familiar result
This is called inertial motion: in the absence of forces, a particle will move with a constant velocity. You recognise this as Newton’s first law of motion: “a uniformly moving body continues to move uniformly unless acted on by a force”.
By integrating the acceleration twice we have solved the differential equation of motion for \(V\) equal to a constant (which leads to a zero force). Let’s solve the differential equation for a constant force next.
Type of motion 2: Uniformly accelerated motion#
As soon as a force acts on our particle, it no longer moves uniformly, that is, with constant velocity. Let’s consider the next simplest case, where a constant force is acting on our particle: \(F = k\), where \(k\) is a constant[1]. Again, we use Newton’s second law to write \(F = ma = k\) and integrate twice to find the velocity and the position of the particle:
where we have written \(a_0 = k/m\) to tidy up our equations. We also explicitly wrote \(a_0 \times (t-t_0)\) to avoid mistaking it for a function \(a\) at time \(t-t_0\).
Next, we integrate again to obtain the position of the particle:
Using again the special values \(t_0 = 0\), \(x_0\), and \(v_0\), these are easily recognised as the kinematic equations (“suvat”):
Hence, the familiar equations for uniformly accelerated motion follow directly from Newton’s second law and a constant force. …
Example 1#
We want to find the position as a function of time of an object with mass \(m\) sliding along a ramp without friction. The potential energy of the object along the ramp is \(V=mgy\), with \(g\) the gravitational acceleration and \(y\) the vertical coordinate. The equation for a ramp sloping down towards the right is \(y = -\tan\alpha\, x + h\), with \(\alpha\) the slope of the ramp. Since we want the horizontal position \(x(t)\), we substitute the ramp equation into \(V\) to obtain
The force in the \(x\) direction on the object is the negative derivative of \(V\) with respect to \(x\):
However, we want the force along the ramp, so we need to make an adjustment. From trigonometry, we can see that we need to include a factor \(\cos\alpha\), which means that \(F = mg \tan\alpha \cos\alpha = mg\sin\alpha\). Setting this force equal to \(F=ma\), we obtain
and we have motion with a constant acceleration \(a_0 = g\sin\alpha\). The equations of motion become:
At the bottom of the ramp, \(y=0\), the horizontal position is \(x = h/\sin\alpha\). From this we can find the time it takes for the object to slide to the bottom:
You should check that this expression makes sense for various values of \(\alpha\), especially \(\alpha=90^\circ\) and \(\alpha=0^\circ\). Doing this check after every calculation helps you find errors that you may have made.
Type of motion 3: Simple harmonic motion#
Let’s see how this works when we have a quadratic potential \(V=\tfrac12 k x^2\), for example a block with mass \(m\) on a spring with spring constant \(k\) and displacement from the equilibrium position \(x\). The force on the block due to the spring is given by
This is Hooke’s law. Equating this to \(F=ma\), we obtain
At this point, we can write \(a\) as the second derivative of \(x\) with respect to time, and reorganise as
with \(\omega^2 = k/m\).
This is a differential equation in \(x(t)\), an equation that cannot be solved algebraically but must be integrated. … Fortunately, it has a known solution:
with \(A\), \(B\) constants determined by initial conditions. You will study these equations of motion in detail in the Oscillations and Waves part of this course.
Type of motion 4: Motion from a central force#
As a final example of a potential energy that leads to equations of motion, consider the central potential energy \(V(r) = -k/r\). One example is a body of mass \(m\) in the gravitational potential of another body of mass \(M\), where \(k = GMm\) and \(G\) is the gravitational constant. Another example is the electric potential energy between two charges \(q_1\) and \(q_2\), where \(k= -q_1 q_2/4\pi\epsilon_0\) with \(\epsilon_0\) the electric permittivity. We use the radial coordinate \(r\). The force is:
This is the inverse-square force law that Newton identified as describing gravity. When \(k>0\) the force is attractive, and when \(k<0\) the force is repulsive. The negative sign denotes that the force is always towards (or away from) the other body. Substituting into \(F=ma\), we obtain
This is again a differential equation that requires new techniques to solve. It is not an easy equation to solve, but it will yield a full description of planetary orbits and other motions. In fact, solving this equation shows that Kepler’s laws are a direct consequence of Newton’s inverse-square law.