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## Newton’s 2nd law of motion, force and acceleration

Last week, I blogged about Newton’s 1st law of motion, and the concept of inertia. At the end I said that this week I would discuss what happens to an object if a force is applied. Or, to put it more correctly, an “external resultant force” is applied. This is what Newton’s 2nd law of motion is all about – the effect on a body of an applied force.

Newton’s three laws of motion appear in his masterpiece, The Principia, which was published in 1687.

If we apply a force to an object it will change its velocity, which means it will accelerate. As I have mentioned before, in physics acceleration has a more precise meaning than it does in everyday life. It doesn’t just mean an object is changing its speed, it can also be keeping a constant speed but changing its direction, such as an object moving at a constant speed in a circle. But, whether an object is changing its speed or changing its direction, it has to accelerate to do this, and so a force needs to be applied.

## The most important equation in physics

The relationship between force and acceleration is given by Newton’s 2nd law of motion, which states that

$\boxed{ F = m a }$

where $F$ is the force, $a$ is the acceleration, and $m$ is the mass of the body. From this equation, along with Newton’s 3rd law (which I will discuss next week), nearly all of mechanics can be derived. For example, this equation tells us that for the Moon to orbit the Earth, it must have a force acting upon it. That force is gravity, and Newton was also the first person to produce an equation to describe gravity. For example, in this blog I showed how we can derive the acceleration felt by a body travelling in a circle, which we call the centripetal acceleration.

Using calculus, this equation also allows us to derive the three equations of motion, equations like $v = u + at$ and $s=ut + \frac{1}{2}at^{2}$, as I did in this blog. It tells us that it is more difficult to accelerate a more massive object than it is a less massive one, which is why you need a more powerful engine in a large truck than you do in e.g. a small car. Along with Newton’s 3rd law, it explains why a bullet comes out at such a high speed from the nozzle of a rifle, but why the recoil of the gun moves much more slowly. As I said, the most important equation in physics.

Next week I will discuss the last of Newton’s laws of motion, his 3rd law.

## Derivation of centripetal acceleration using polar coordinates

In this blog, I introduced the idea of polar coordinates as a way to derive the area of a circle. I mentioned at the end of the blog that polar coordinates can also be used to derive the centripetal acceleration, the acceleration an object feels when it moves in a circle. There are a few ways to do this, but I will choose what I think is the easiest way.

## The position of a point on a circle in polar coordinates

Remember, as I mentioned in my blog on deriving the formula for the area of a circle, we can express any point on a circle, which is usually given in terms of its x and y coordinates, instead in terms of the radius vector $\vec{r}$ and the angle the radius vector makes with the x-axis $\theta$.

The point $(x,y)$ on the circle can be written in terms of the radius $r$ and the angle $\theta$. We can write that $x = r \cos \theta$ and $y = r \sin \theta$

When we do this, we can write the x-direction vector to this point as $\vec{x}=\vec{r} \cos \theta$ and the y-direction vector as $\vec{y}=\vec{r} \sin \theta$.

## The velocity of circular motion in polar coordinates

The velocity of an object undergoing circular motion is given by $\vec{v}$, the direction of which is tangential to the the radius vector $\vec{r}$, as I described in my blog on the direction of the angular velocity vector here, and shown in the figure below.

The velocity of an object moving in a circle is in a direction which is at right angles to the radius. We can split that velocity up into its x and y-components

We can write the x and y-coordinates of the point in terms of their polar coordinates $r \text{ and } \theta$; that is

$\vec{x} = \vec{r} \cos \theta \text{ and } \vec{y} = \vec{r} \sin \theta$

But, as I mentioned in this blog here, we can write $\theta$ in terms of the angular velocity $\omega$. From the definition of the angular velocity, which is the angle moved per unit time, we have $\omega = \theta / t$ and so rearranging we can write $\theta = \omega t$. We can then re-write the x and y-coordinates in terms of the polar coordinates, but using $\omega t$ instead of $\theta$.

$\vec{x} = \vec{r} \cos (\omega t) \text{ and } \vec{y} = \vec{r} \sin (\omega t)$

We can then use differentiation to find the x and y-components of the velocity $\vec{ v_{x} } \text{ and } \vec{ v_{y}}$.

$\vec{ v_{x} } = \frac{ d \vec{x} }{ dt } = \frac{ d }{ dt } \vec{r} \cos (\omega t) = -\omega \vec{r} \sin (\omega t)$

$\vec{ v_{y} } = \frac{ d\vec{y} }{ dt } = \frac{ d }{ dt } \vec{r} \sin (\omega t) = \omega \vec{r} \cos (\omega t)$

To find the components of the acceleration, $\vec{ a_{x} } \text{ and } \vec{ a_{y} }$ we differentiate again, so we have

$\vec{ a_{x} } = \frac{ d \vec{ v_{x} } }{ dt } = \frac{ d }{ dt } -\omega \vec{r} \sin (\omega t) = -\omega^{2} \vec{r} \cos (\omega t) = -\omega^{2}\vec{x}$

and

$\vec{ a_{y} } = \frac{ d \vec{ v_{y} } }{ dt } = \frac{ d }{ dt } -\omega \vec{r} \cos (\omega t) = -\omega^{2} \vec{r} \sin (\omega t) = -\omega^{2}\vec{y}$

To find the magnitude (size) of the acceleration $| \vec{a} |$ we remember that it is given by

$| \vec{a} | = \sqrt{ \vec{ a_{x}^{2} } + \vec{ a_{y}^{2} } }$

so

$| \vec{a} | = \sqrt{ (-\omega^{2} \vec{x} )^{2} + (-\omega^{2} \vec{y} )^{2} } = \sqrt{ \omega^{4}x^{2} + \omega^{4}y^{2} } = \omega^{2} \sqrt{ x^{2} + y^{2} } = \omega^{2}r$

(because $\sqrt{ x^{2} + y^{2} } = r$).
To find the direction of the acceleration, we can use a diagram of the two components $\vec{ a_{x} } \text{ and } \vec{ a_{y} }$.

The direction of the acceleration can be found from this vector diagram, and is found to be in the negative radial direction, which means towards the centre of the circle

As can be seen from the diagram above, the direction of the centripetal acceleration is in the $- \vec{r}$ direction, which means towards the centre of the circle.

So, in terms of the angular velocity $\omega$ we have

$\boxed{ \vec{a} = - \omega^{2} \vec{r} }$

To get the expression we had before in this blog, we need to remember the relationship between the angular velocity and the linear velocity $\vec{v}$. In terms of vectors, $\vec{ v} = \vec{ \omega } \times \vec{ r }$ (where the cross means we are talking about the vector product), but as we have $\omega^{2}$ in our expression for the centripetal acceleration, we don’t need to worry about its direction, only its magnitude, so we can simply write

$v = \omega r \rightarrow \omega = \frac{v}{r} \rightarrow \omega^{2} = \left( \frac{v}{r} \right)^{2}$

and so we can write

$\vec{a} = - \left( \frac{ v }{ r } \right)^{2} \vec{r}$

and remembering that the unit vector $\hat{r}$ can be written as

$\hat{r} = \frac{ \vec{r} }{ | \vec{r} | } \text{ so } \vec{r} = \hat{r} | \vec{r} |$

we can write

$\vec{a} = - \left( \frac{ v }{ r } \right)^{2} \cdot \hat{r} | \vec{r} | = - \frac{ v^{2} }{ | \vec{r} | } \hat{r}$

## The direction of the angular velocity vector

In this blog, I introduced the idea of angular velocity, which is the rotational equivalent of linear velocity. The angular velocity $\omega$ is usually measured in radians per second, where radians are the more natural measurement of an angle than the more familiar degrees. But, just as linear velocity is a vector and therefore has a direction, so too does angular velocity. So, what is the direction of the angular velocity vector?

## The relationship between linear velocity and angular velocity

As we saw in the blog where I introduced the concept of angular velocity, it can be defined as simply the angle $\theta$ moved per unit time, or

$\omega = \frac{ \theta }{ t }$

which of course leads to it being usually measured in radians per second. However, we can also write the angular velocity in terms of the linear velocity. To see how to do this let us remind ourselves of the definition of a radian, the more natural unit for measuring an angle. As I introduced it in this blog, measuring an angle in radians just means dividing the length of the arc $l$ by the radius of the circle $r$.

An angle measured in radians is simply the length of the arc $l$ divided by the radius $r$

We can write that the angle measured in radians is

$\theta \text{ (in radians) } = \frac{ l }{ r }$

But, the linear velocity $v$ is just defined as distance divided by time, so we can write

$v = \frac{ l }{ t }$

Re-writing $l$ in terms of $\theta \text{ and } t$ we can write

$v = \frac{ \theta r }{ t }$

and so we can write the angle $\theta$ as

$\theta = \frac{ v t }{ r }$

Using this for $\theta$ we can write the angular velocity $\omega$ as

$\omega = \frac{ v t }{ r } \cdot \frac{ 1 }{ t }, \text{ so } \boxed{ \omega = \frac{ v }{ r } }$

## The direction of the angular velocity vector $\vec{ \omega }$

Writing this in terms of vectors, and remembering that division of vectors is not defined, we instead write that

$\boxed{ \vec{ \omega } = \frac{ \vec{ r} \times \vec{ v } }{ | \vec{ r } |^{2} } }$

where $\vec{ r } \times \vec{ v }$ is the vector product (or cross-product), as I discussed in this blog here.

The direction of the radius vector $\vec{ r }$ is away from the centre of the circle, and the direction of the linear velocity $\vec{ v }$ for an object moving anti-clockwise is in the direction shown in the diagram below, tangential to the circle so at right angles to the radial vector $\vec{ r }$.

The direction of the radius vector $\vec{ r }$ is away from the centre of the circle, the direction of the linear velocity $\vec{ v }$ for an object moving anti-clockwise is as shown, at right angles to the radius vector.

To find the direction of $\vec{ \omega }$, we can use the right-hand rule, as shown in the figure below.

The right-hand rule for determining the direction of the result of the vector product

In our example here, our first-finger is in the direction of the radial vector $\vec{ r }$, and our second-finger is in the direction of the linear velocity $\vec{ v }$, leading to the angular velocity $\vec{ \omega }$ (represented by the thumb) being outwards, or towards us, as shown in the figure below.

The direction of the angular velocity vector $\vec{ \omega}$ is perpendicular to the plane of rotation of the object.

Another way to find this direction is to wrap the fingers of the right hand in the direction of the rotation, the thumb will then show the direction of the angular velocity vector.

The direction of the angular velocity can also be found as shown in this figure.

## Derivation of the centripetal force

I have mentioned a few times in previous blogs that an object moving in a circle at a constant speed does so because of a force acting towards the centre. We call this force the centripetal force. The force is given by the equation

$F = \frac{ mv^{2} }{ r }$

where $m$ is the mass of the object moving in the circle, $v$ is its speed, and $r$ is the radius of the circle. More correctly, remembering that force is a vector, it should be written

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where $|\,\vec{r}\,|$ is the magnitude (size) of the radial vector $\vec{r}$, and $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

But, from where does this formula come?

## The acceleration of an object moving in a circle

The acceleration of any object is defined at the change in its velocity divided by the change in time. Both acceleration and velocity are vectors, so mathematically we can write this as

$\vec{a} = \frac{ \Delta \vec{v} }{ \Delta t }$

We are going to consider an object moving at a constant speed in a circle, as illustrated in the diagram below. In a time $\Delta t$ the object has moved through an angle $\theta$, and its initial velocity $v_{1}$ has changed to $v_{2}$, where the only change in the velocity is its direction. Remember, the velocity is tangential to the radius, so makes a right angle with the radius vector $\vec{r}$. The direction of the radius vector is, by definition, radially outwards.

We will consider an object moving in a circle with a constant speed. It moves through an angle $\theta$ in time $t$.

In my blog about vectors I mentioned that, to combine vectors which have different directions, we need to split the vectors into components, add the components and then recombine the resultant components. The components need to be at right-angles to each other, and usually (but not always) we choose the x and y-directions when the vectors are in two dimensions.

To find the acceleration of our object in this example, we want to find the change or difference in the velocity, that is $\Delta \vec{ v } = \vec{ v_{2} } - \vec{ v_{1} }$. We start by splitting the two vectors into their x and y-components.

$\vec{ v_{1} } = v_{1}\hat{x} + v_{1}\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v_{2}\hat{x} + v_{2}\hat{y}$

Looking at our diagram, we can write

$\vec{ v_{1} } = 0\hat{x} + v\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v\sin(\theta)\hat{x} + v\cos(\theta)\hat{y}$

where $v$ is the speed, the size of the vectors $\vec{ v_{1} } \text{ and } \vec{ v_{2} }$.

The change in the velocity in the x-direction, which we will call $(\Delta v)\hat{x}$ is then just

$(\Delta v)\hat{x} = v\sin(\theta) - 0 = v\sin(\theta)$

Similarly, the change in the velocity in the y-direction, which we will call $(\Delta v)\hat{y}$ is given by

$(\Delta v)\hat{y} = v\cos(\theta) - v = v(\cos(\theta) - 1)$

To correctly calculate the acceleration, we need to find the change in velocity with time as the time interval tends to zero. This means the angle $\theta$ tends towards zero also, and when $\theta$ is very small (and expressed in radians) we can write

$\cos(\theta) \rightarrow 1 \; \; \text{ and } \sin(\theta) \rightarrow \theta$

So we can then write

$(\Delta v)\hat{x} \rightarrow v\theta \; \; \text{ and } (\Delta v)\hat{y} \rightarrow v(1 - 1) =0$

The overall change in velocity, $\Delta \vec{v}$ is then just the change in velocity in the x-direction, $\Delta \vec{v} = v\theta$. The direction of the change in velocity is in the positive x-direction, which as $\theta \rightarrow 0$ is along the radial vector towards the centre of the circle (that is, in the $- \vec{r}$ direction).

However, we can do a substation for the angle $\theta$. Remember, the arc-length, which we shall call $l$ is related to the angle $\theta$ via our definition of the radian (see this blog here), $\theta = l/r$, so we can write

$\Delta \vec{v} = \frac{ vl }{ |\,\vec{r}\,| }\hat{r} \; \text{ (Equation 1)}$

The acceleration $\vec{a} = \Delta \vec{v} / \Delta t$. But the speed $v$ and time $t$ are related via $v = l/t$, so we can write that $l = vt$. Substituting this into equation 1 above gives

$\frac{ vl }{ r } = \frac{ v^{2}t }{ r } \; \text{ (Equation 2)}$

and so the acceleration $\vec{a}$ is given by

$\vec{a} = \frac{ \Delta \vec{v} }{ t } = \frac{ v^{2}t }{ |\,\vec{r}\,| t } = \boxed{ - \frac{ v^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where we have added the minus sign to remind us that the change in velocity, and hence the acceleration, is directed towards the centre of the circle.

If you prefer to think of vectors pictorially, then the direction of $\Delta\vec{v} = \vec{v_{2}} - \vec{v_{1}}$ can be seen from this diagram:

This shows the direction of $\Delta \vec{v} = \vec{v_{2}} - \vec{v_{1}}$, and it is along the radius vector towards the centre of the circle, in the opposite direction to the radius vector $\vec{r}$.

The centripetal force is found by remembering that $\vec{F} = m\vec{a}$ (Newton’s 2nd law), so finally we can write that the centripetal force is

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$