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## A cylinder rolling down a slope

The other week I was asked to explain how a cylinder (or ball) rolling down a slope differs from e.g. a ball being dropped vertically. It is an interesting question, because it illustrates some things which are not immediately obvious. We all know that, if you drop two balls, say a tennis ball and a cannon ball, they will hit the ground at the same time. This is despite their having very different masses (weights). Galileo supposedly showed this idea by dropping objects of different weights from the tower of Pisa (although he probably never did this, see our book Ten Physicists Who Transformed Our Understanding of Reality).

With a tennis ball and a cannon ball, they clearly have very different masses (weights), but will fall to the ground at the same rate. This fact, contrary to the teachings of Aristotle, was one of the key breakthroughs which Galileo made in our understanding of motion. But, what about if we roll the two balls down a slope? If we build a track to keep them going straight, will a tennis ball roll down a slope at the same rate as a cannon ball? The answer is no, and I will explain why.

## Rolling rather than dropping

When a ball rolls down a slope, it starts off at the top of the slope with gravitational potential energy. When it starts rolling down the slope, this gravitational potential energy gets converted to kinetic energy. This is the same as when the ball drops vertically. But, in the case of the ball dropping vertically, the kinetic energy is all in the form of linear kinetic energy, given by

$\text{ linear kinetic energy } = \frac{ 1 }{ 2 } mv^{2}$

where $m$ is the mass of the ball and $v$ is its velocity (which is increasing all the time as it falls and speeds up). The gravitational potential energy is converted to linear kinetic energy as the ball drops; by the time the ball hits the bottom of its fall all of the PE has been converted to KE.

If, instead, we roll a ball down a slope, the kinetic energy is in two forms, linear kinetic energy but also rotational kinetic energy, which is given by

$\text{ rotational kinetic energy } = \frac{ 1 }{ 2 } I \omega^{2}$

where $I$ is the ball’s moment of inertia, and $\omega$ is the ball’s angular velocity, usually measured in radians per second. The key point is that the the moment of inertia for the two balls in this example (a tennis ball and a cannon ball) have a different value, because the distribution of the mass in the two balls is different. For the tennis ball it is all concentrated in the layer of the rubber near the ball’s surface, with a hollow interior. For the cannon ball, the mass is distributed throughout the ball.

## Two cylinders rolling down a slope

Let us, instead, consider the case of two cylinders rolling down a slope. One is a solid cylinder, the other is a hollow one with all of its mass concentrated near the surface. We will make the two cylinders have the same mass; this can be done by making the material from which the hollow cylinder is made denser than the material for the solid cylinder. So, even though the material of the hollow cylinder is all concentrated near the surface of the cylinder, and there is a lot less of it, if it is denser it can have equal mass.

A solid cylinder on an inclined plane. We will make the mass of this solid cylinder the same as that of the hollow cylinder, by making it of less dense material. Although it will have the same mass $m$ and the same radius $R$, it will not have the same moment of inertia $I$.

We will start both cylinders from rest near the top of the slope, and let them roll down. We will observe what happens.

A hollow rolling down an inclined plane. We will make the hollow cylinder denser than the solid one, so that they both have the same mass $m$ and the same outer radius $R$. But, they will not have the same moment of inertia $I$.

When things are dropped, the rate at which they fall is independent of the mass, but when they roll the rate at which they roll is not indpendent of the moment of inertia. In particular, it is not independent of the distribution of mass in the rolling object. As this video shows, the solid cylinder rolls down the slope faster than the hollow one!

But, why??

## Why does the solid cylinder roll down quicker?

The reason that the solid cylinder rolls down faster than the hollow cylinder has to do with the way that the potential energy (PE) is converted to kinetic energy. Because the cylinder is rolling, some of the PE is converted to rotational kinetic energy (RKE), not just to linear kinetic energy (LKE). The only way that a cylinder can roll down a slope is if there is friction between the cylinder and the slope, if the slope were perfectly smooth the cylinder would slide and not roll.

The torque (rotational force) $\tau$ is related to the angular acceleration $\alpha$ in a similar way that the linear force $F$ is related to linear accelerate $a$. From Newton’s second law we know that $F = ma$ where $m$ is the mass of the object. The rotational equivalent of this law is

$\tau = I \alpha$

where $I$ is the moment of inertia. The moment of inertia $I$ is different for a hollow cylinder and a solid cylinder. For the solid cylinder it is given by

$I_{sc} = \frac{ 1 }{ 2 } mR^{2} = 0.5mR^{2}$

where $m$ is the mass of the cylinder and $R$ is the radius of the cylinder. For the hollow cylinder, the moment of inertia is given by

$I_{hc} = \frac{ 1 }{ 2 }m(R_{2}^{2} + R_{1}^{2})$

where $R_{2} \text{ and } R_{1}$ are the outer and inner radii of the annulus of the cylinder. We are going to make the hollow cylinder such that the inner 80% is hollow, so that $R_{1} = 0.8R_{2} = 0.8R$. We will make the mass $m$ of the two cylinders the same.

Thus, for the hollow cylinder, we can now write

$I_{hc} = \frac{ 1 }{ 2 }m(R^{2} + (0.8R)^{2}) = \frac{ 1 }{ 2 }mR^{2}(1+0.64) = \frac{ 1 }{ 2 }mR^{2}(1.64) = 0.82 mR^{2}$

The cylinder accelerates down the slope due to the component of its weight which acts down the slope. This component is $mg sin(\theta)$ where $g$ is the acceleration due to gravity and $\theta$ is the angle of the slope from the horizontal. To make the maths easier, we are going to set $\theta = 30^{\circ}$, as $sin(30) =0.5$.

Friction always acts in the opposite direction to the direction of motion, and in this case the friction $F_{f}$ is related to the torque $\tau$ via the equation

$\tau = F_{f}R \text{ (1)}$

so we can write

$F_{f}R = \tau = I \alpha \text{ (2)}$

where $\alpha$ is the rotational acceleration. Re-arranging this to give $F_{f}$, we have

$F_{f} = \frac{I \alpha}{ R }$

The force down the slope, $F (=ma)$ is just the component of the weight down the slope minus the frictional force $F_{f}$ acting up the slope.

$ma = mg\sin(30) - F_{f} = 0.5mg - \frac{ I \alpha }{ R } \text{ (3)}$

The angular acceleration $\alpha$ is given by $\alpha = a/R$ where $a$ is the linear acceleration. So, we can re-write Eq. (3) as

$ma = 0.5mg - \frac{ Ia }{ R^{2} } \text{ (4)}$

Now we will put in the moments of inertia for the solid cylinder and the hollow cylinder. For the solid cylinder, we can write

$ma = 0.5mg - \frac{ 0.5mR^{2}a }{ R^{2} } = 0.5mg - 0.5ma$

The mass $m$ can be cancelled out, and assuming $g=9.8 \text{ m/s/s}$, we have

$a = 0.5g - 0.5a \rightarrow 1.5a = 4.9 \rightarrow \boxed{ a = 3.27 \text{ m/s/s (5)} }$

Notice that Equation (5) does not have the mass $m$ in it, as this cancels out. It also does not have the radius $R$ of the cylinder in it; the acceleration of the cylinder as it rolls down the slope is independent of both the mass and the radius of the cylinder.

For the hollow cylinder, again using Eq. (4), we have

$ma = 0.5mg - \frac{ 0.82maR^{2} }{ R^{2} } = 0.5mg - 0.82ma$

This simplifies to

$a = 4.9 - 0.82a \rightarrow 1.82a = 4.9 \rightarrow \boxed{ a = 2.69 \text{ m/s/s} (6)}$

As with Equation (5), Equation (6) is independent of both mass and radius.

So, as we can see, the linear acceleration $a$ for the hollow cylinder is 2.69 m/s/s, less than the linear acceleration for the solid cylinder, which was 3.27 m/s/s. This is why the solid cylinder rolls down the slope quicker than the hollow cylinder! And, the result is independent of both the mass and the radius of either cylinder. Therefore, a less massive solid cylinder will roll down a slope faster than a more massive hollow one, which may seem contradictory.

## Summary

All objects falling vertically fall at the same rate, but this is not true for objects which roll down a slope. We have shown above that a solid cylinder will roll down a slope quicker than a hollow one. This is because their moments of inertia are different, it requires a greater force to get the hollow cylinder turning than it does the solid cylinder. Remember, the meaning of the word ‘inertia’ is a reluctance to change velocity, so in this case a reluctance to start rolling from being stationary. A larger moment of inertia means a greater reluctance to start rolling.

The solid cylinder will start turning more quickly from being stationary than the hollow cylinder, and this means that it will roll down the slope quicker. This result is independent of the masses (and radii) of the two cylinders; even a less massive solid cylinder will roll down a slope quicker than a more massive hollow one, which may be counter-intuitive.

## 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.

## Why are there $360^{\circ}$ in a circle?

I would imagine all of you reading this know that when we learn about angles, we learn that we measure angles in degrees, and that there are $360^{\circ}$ in a full circle. Why this strange number?

It was the Babylionians who chose to have $360^{\circ}$ in a full circle. When you think about it, they could have chosen any figure, say 400 or 10 or 1,000. Why did they choose 360? Well, it turns out that 360 was a special number to the Babylonians, or to be more correct 12 and 30 were, and 360 is 12 multiplied by 30.

12 was special to the Babylonians because there are 12 and a bit cycles of the Moon in a year. It is from the time between e.g. each new Moon (or full Moon) that we get our idea of months. 30 was special to the Babylonians because this was, roughly speaking, the number of days in each lunar cycle.

So, if 12 and 30 were special, then the number produced by multiplying them together, 360, was even more special. That’s why we are stuck with the strange number for the degrees in a circle.

## Radians – more natural units

Mathematicians, physicists and a lot of other scientists most often use a different unit to measure angles, a radian. It is a more natural unit than degrees. A radian is very simply defined. If we have a circle with a radius $r$, and we draw an angle $\theta$ as shown in the diagram below, then we will call the length of the arc produced $l$.

The definition of the radian is quite simply that if $\theta$ is measured in radians then

$\boxed { \theta = \frac{l}{r} }$

## Converting between degrees and radians

Converting between these two ways of measuring an angle is quite simple. Remember that the circumference of a circle, the distance all the way around, is $2\pi r$. But this is just the length $l$ of our arc when $\theta = 360^{\circ}$. So when $\theta=360^{\circ}$ this is equivalent to $\theta = \frac{2\pi r}{r} = 2 \pi$ radians.

This table shows the conversion from degrees to radians for some commonly used angles.

 Degrees Radians 0 0 30 $\frac{\pi}{6}$ 45 $\frac{\pi}{4}$ 60 $\frac{\pi}{3}$ 90 $\frac{\pi}{2}$ 180 $\pi$ 270 $\frac{3\pi}{2}$ 360 $2 \pi$

## The sine function

The sine function, $\sin(\theta)$, can be multiplied by a factor to produce a regularly varying function of any amplitude. For example, $4 \sin(\theta)$ would look like this (remember we are expressing $\theta$ in radians, not degrees).

The sine function is very useful in many areas of mathematics and physics.

## Time varying waves

Suppose we have a wave travelling out from a stone which has been dropped into the water in a smooth pond. The ripples (waves) will travel outwards from where the stone was dropped. If we look at the height of the water at some point on the surface of the pond, then we find that the height $y$ of the water will vary with time $t$ as $y=A + R\sin (\omega t + \phi)$. In this expression $A$ is the mean (average) height of the water (the level of the water before we dropped the stone), $R$ is the size of the wave (the peak height), $\omega$ is something called the angular velocity, $t$ is time (in seconds), and $\phi$ is something called the phase angle.

### Angular velocity

Linear velocity, the velocity of something moving in a straight line in, say, the x-direction is just defined as $v=\frac{x}{t}$. If an object is moving in a circle, we can define a similar concept called the angular velocity, which is not the linear distance divided by time, but rather the angular distance divided by time. But, the angular distance is just $\theta$ (when expressed in radians), so the angular velocity $\omega = \frac{ \theta }{t}$ and is measured in radians per second. Re-arranging this, we can see that $\theta = \omega t$, so in other words $\omega t$ has the same units as radians, they multiply together to measure an angle. This means that $\sin{\theta}$ is equivalent to $\sin(\omega t)$, enabling us to consider the variation of a wave as a function of time.

When expressed in radians, the angle for one complete cycle of a sine curve is just given by $\theta = 2\pi$. Using our relationship between $\theta$ and $\omega t$ we can say that the time taken for one complete cycle, which we called the period $T$ is just given by $T = \frac{2\pi}{\omega}$. So, depending on the value of the angular velocity $\omega$, this determines how many seconds it takes for the sine function to complete one cycle.

### The phase angle

The phase angle, $\phi$ above, is needed as our zero time may not be when the expression $R \sin( \omega t)$ is zero. So, it just allows us to correct for when we start timing, just in case this is not at the moment when our sine function has zero value.

### Putting this together

Below are plots of two sine curves, the first (the blue curve) is given by $y = 4 \sin ( \frac{\pi}{4} t)$. The second one (the green curve) is of $y = 4 \sin ( \frac{\pi}{4} t + 0.5 )$ (remember the phase angle $\phi$ is expressed in radians, and one radian is approximately $57^{\circ}$.

Because $\omega=\frac{\pi}{4}$, this means the period of the wave will be $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/4} = 8s$. You can see that the blue curve does indeed complete one cycle in 8 seconds. The green curve has the same value of $\omega$, so it has the same period. However, it is shifted in time due to the phase angle $\phi$. So, at a time of $t=0$, the green curve is not zero, and we say that it is ahead or leads the blue curve as it reaches its first peak before the blue curve does. The number of seconds it leads the blue curve can be found by considering that $\phi = \omega t$ so $t = \frac{\phi}{\omega}$, which in this example is $t = \frac{1/2}{\pi/4} = \frac{2}{\pi} \approx 0.63 s$.

In tomorrow’s blog I will discuss an exam question I have been going over with my students, where the height of water in metres in a harbour is given by e.g. $y = 6\sin(\frac{\pi}{4}t) + 8\cos(\frac{\pi}{4}t) + 11$ (where $t$ is expressed in hours). I will discuss how we can work out the maximum and minimum heights of the water, when e.g. the first maximum occurs, and how many hours a day we can use the harbour if we need a minimum of e.g. 2 metres of water in the harbour.