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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 the moment of inertia of an annulus

Following on from my derivation of the moment of inertia of a disk, in this blog I will derive the moment of inertia of an annulus. By an annulus, I mean a disk which has the inner part missing, as shown below.

An annulus is a disk of small thickness $t$ with the inner part missing. The annulus goes from some inner radius $r_{1}$ to an outer radius $r$.

To derive its moment of inertia, we return to our definition of the moment of inertia, which for a volume element $dV$ is given by

$dI = r_{\perp}^{2} dm$

where $dm$ is the mass of the volume element $dV$. We are going to initially consider the moment of inertia about the z-axis, and so for this annulus it will be

$I_{zz} = \int _{r_{1}} ^{r} r_{\perp}^{2} dm$

where $r_{1}$ and $r$ are the inner radius and outer radius of the annulus respectively. As with the disk, the mass $dm$ of the volume element $dV$ is related to its volume and density via

$dm = \rho dV$

(assuming that the annulus has a uniform density). The volume element $dV$ can be found as before by considering a ring at a radius of $r$ which a width $dr$ and a thickness $t$. The volume of this will be

$dV = (2 \pi r dr) t$

and so we can write the mass $dm$ as

$dm = (2 \pi \rho t)rdr$

Thus we can write the moment of inertia $I_{zz}$ as

$I_{zz} = \int _{r_{1}} ^{r} r_{\perp}^{2} dm = 2 \pi \rho t \int _{r_{1}} ^{r} r_{\perp}^{3} dr$

Integrating this between $r_{1}$ and $r$ we get

$I_{zz} = 2 \pi \rho t [ \frac{ r^{4} - r_{1}^{4} }{4} ] = \frac{1}{2} \pi \rho t (r^4 - r_{1}^{4}) \text{ (Equ. 1)}$

But, we can re-write $(r^{4} - r_{1}^{4})$ as $(r^{2} + r_{1}^{2})(r^{2} - r_{1}^{2})$ (remember $x^{2} - y^{2}$ can be written as $(x+y)(x-y)$). So, wen can write Eq. (1) as

$I_{zz} = \frac{1}{2} \pi \rho t (r^{2} + r_{1}^{2})(r^{2} - r_{1}^{2}) \text{ (Equ. 2)}$

The total mass $M_{a}$ of the annulus can be found by considering the total mass of a disk of radius $r$ (which we will call $M_{2}$) and then subtracting the mass of the inner part, a disk of radius $r_{1}$ (which we will call $M_{1}$). The mass of a disk is just its density multiplied by its area multiplied by its thickness.

$M_{2} = \pi \rho t r^{2} \text{ and } M_{1} = \pi \rho t r_{1}^{2}$

so the mass $M_{a}$ of the annulus is

$M_{a} = M_{2} - M_{1} = \pi \rho t r^{2}- \pi \rho t r_{1}^{2} = \pi \rho t (r^{2} - r_{1}^{2})$

Substituting this expression for $M_{a}$ into equation (2) above, we can write that the moment of inertia for an annulus, which goes from an inner radius of $r_{1}$ to an outer radis of $r$, about the z-axis is

$\boxed{ I_{zz} = \frac{1}{2} M_{a} (r^{2} + r_{1}^{2}) }$

## Comparison to the moment of inertia of a disk

As we saw in this blog, the moment of inertia of a disk is $I_{zz} = \frac{1}{2} Mr^{2}$. It may therefore seem, at first sight, that the moment of inertia of an annulus is more than that of a disk. This would be true if they have the same mass, but if they have the same thickness and density the mass of an annulus will be much less.

Let us compare the moment of inertia of a disk and an annulus for the 4 following cases.

The same density and thickness, $r_{1} = 0.5 r$
The same density and thickness, $r_{1} = 0.9 r$
The same mass, $r_{1} = 0.5 r$
The same mass, $r_{1} = 0.9 r$

## The same density and thickness, $r_{1}=0.5r$

We are first going to compare the moment of inertia of a disk of mass $M$ with that of an annulus which goes from half the radius of the disk to the radius of the disk (i.e. $r_{1} \text{ the inner radius of the annulus, is } = 0.5 r$.

For the disk, its mass will be

$M = \rho t (\pi r^{2}) = \pi \rho t r^{2}$

The mass of the annulus, $M_{a}$, will be this mass less the mass of the missing part $M_{1}$, so

$M_{a} = M - M_{1} = M - \pi \rho t (r_{1})^{2} = \pi \rho t (r^{2} - (0.5r)^{2})= \pi \rho t (1-0.25)r^{2}$

$M_{a} = \pi \rho t (0.75)r^{2} = 0.75 M$

The moment of inertia of the disk will be

$I_{d} = \frac{1}{2} M r^{2}$
The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M_{a} (r^{2} + r_{1}^{2}) = \frac{1}{2} (0.75M)(r^{2} + (0.5r)^{2}) = \frac{1}{2} (0.75M)(1.25r^{2}) = \frac{1}{2} (0.9375) M r^{2}$

So, for this case, $I_{a} = 0.9375 I_{d}$, i.e. slightly less than the disk.

## The same density and thickness, $r_{1}=0.9r$

Let us now consider the second case, with an annulus of the same density and thickness as the disk, and its inner radius being 90% of the outer radius, $r_{1} = 0.9r$. Now, the mass of the missing part of the disk, $M_{1}$ will be

$M_{1} = \rho t (\pi r_{1}^{2}) = \rho t \pi (0.9r)^{2} = 0.81 \rho t \pi r^{2} = 0.81M$

which means that the mass of the annulus, $M_{a}$ is

$M_{a} = M - M_{1} = M-0.81M=0.19M$

The moment of inertia of the annulus will then be

$I_{a} = \frac{1}{2}M_{a}(r^{2}+r_{1}^{2}) = \frac{1}{2}(0.19M)(r^{2}+(0.9r)^{2})=\frac{1}{2}(0.19M)((1.81)r^{2} = \frac{1}{2}(0.1539)Mr^{2}$

and so in this case

$I_{a} = 0.1539 I_{d}$

which is much less than the moment of inertia of the disk.

## The same mass, $r_{1}=0.5r$

In this third case, the mass of the annulus is the same as the mass of the disk, and its inner radius is 50% of the radius of the disk. This would, of course, require the annulus to either have a greater density than the disk, or to be thicker (or both). So, $M_{a} = M$. The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M(r^{2} + r_{1}^{2}) = \frac{1}{2} M(r^{2} + (0.5r)^{2}) = \frac{1}{2} M(r^{2} + 0.25r^{2}) = \frac{1}{2} M(1.25)r^{2}$

$I_{a}= 1.25 I_{d}$

## The same mass, $r_{1}=0.9r$

The last case we will consider is an annulus with its inner radius being 90% of the outer radius, but its mass the same. So, $M_{a} = M$. The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M(r^{2} + r_{1}^{2}) = \frac{1}{2} M(r^{2} + (0.9r)^{2}) = \frac{1}{2} M(r^{2} + 0.81r^{2}) = \frac{1}{2} M(1.81)r^{2}$

$I_{a}= 1.81 I_{d}$

## Summary

To summarise, we have

The same density and thickness, $r_{1} = 0.5 r, \; \; I_{a}=0.9375 I_{d}$
The same density and thickness, $r_{1} = 0.9 r, \; \; I_{a}=0.1539 I_{d}$
The same mass, $r_{1} = 0.5 r, \; \; I_{a}=1.25 I_{d}$
The same mass, $r_{1} = 0.9 r, \; \; I_{a}=1.81 I_{d}$

So, as these calculations show, if keeping the mass of a flywheel down is important, then a larger moment of inertia will be achieved by concentrating most of that mass in the outer parts of the flywheel, as this photograph below shows.

If keeping mass down is important, a flywheel’s moment of inertia can be increased by concentrating most of the mass in its outer parts

In the next blogpost in this series I will calculate the moment of inertia of a solid sphere.

## Derivation of the moment of inertia of a disk

In physics, the rotational equivalent of mass is something called the moment of inertia. The definition of the moment of inertia of a volume element $dV$ which has a mass $dm$ is given by

$dI = r_{\perp}^{2} dm$

where $r_{\perp}$ is the perpendicular distance from the axis of rotation to the volume element. To find the total moment of inertia of an object, we need to sum the moment of inertia of all the volume elements in the object over all values of distance from the axis of rotation. Normally we consider the moment of inertia about the vertical (z-axis), and we tend to denote this by $I_{zz}$. We can write

$I_{zz} = \int _{r_{1}} ^{r_{2}} r_{\perp}^{2} dm$

The moment of inertia about the other two cardinal axes are denoted by $I_{xx}$ and $I_{yy}$, but we can consider the moment of inertia about any convenient axis.

## Derivation of the moment of inertia of a disk

In this blog, I will derive the moment of inertia of a disk. In upcoming blogs I will derive other moments of inertia, e.g. for an annulus, a solid sphere, a spherical shell and a hollow sphere with a very thin shell.

For our purposes, a disk is a solid circle with a small thickness $t$ ($t \ll r$, small in comparison to the radius of the disk). If it has a thickness which is comparable to its radius, it becomes a cylinder, which we will discuss in a future blog. So, our disk looks something like this.

A disk of small thickness $t$, with a radius of $r$

To calculate the moment of inertia of this disk about the z-axis, we sum the moment of inertia of a volume element $dV$ from the centre (where $r=0$) to the outer radius $r$.

$I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} dm \text{ (Equ. 1)}$

The mass element $dm$ is related to the volume element $dV$ via the equation
$dm = \rho dV$ (where $\rho$ is the density of the volume element). We will assume in this example that the density $\rho(r)$ of the disk is uniform; but in principle if we know its dependence on $r, \; \rho (r) = f(r)$, this would not be a problem.

The volume element $dV$ can be calculated by considering a ring at a radius $r$ with a width $dr$ and a thickness $t$. The volume of this ring is just this rings circumference multiplied by its width multiplied by its thickness.

$dV = (2 \pi r dr) t$

so we can write

$dm = \rho (2 \pi r dr) t$

and hence we can write equation (1) as

$I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} \rho (2 \pi r dr) t = 2 \pi \rho t \int_{r=0} ^{r=r} r_{\perp} ^{3} dr$

Integrating between a radius of $r=0$ and $r$, we get

$I_{zz} = 2 \pi \rho t [ \frac{ r^{4} }{ 4 } -0 ] = \frac{1}{2} \pi \rho t r^{4} \text{ (Equ. 2)}$

If we now define the total mass of the disk as $M$, where

$M = \rho V$

and $V$ is the total volume of the disk. The total volume of the disk is just its area multiplied by its thickness,

$V = \pi r^{2} t$

and so the total mass is

$M = \rho \pi r^{2} t$

Using this, we can re-write equation (2) as

$\boxed{ I_{zz} = \frac{1}{2} \pi \rho t r^{4} = \frac{1}{2} Mr^{2} }$

## What are the moments of inertia about the x and y-axes?

To find the moment of inertia about the x or the y-axis we use the perpendicular axis theorem. This states that, for objects which lie within a plane, the moment of inertia about the axis parallel to this plane is given by

$I_{zz} = I_{xx} + I_{yy}$

where $I_{xx}$ and $I_{yy}$ are the two moments of inertia in the plane and perpendicular to each other.

We can see from the symmetry of the disk that the moment of inertia about the x and y-axes will be the same, so $I_{zz} = 2I_{xx}$. Therefore we can write

$\boxed{ I_{xx} = I_{yy} = \frac{1}{2}I_{zz} = \frac{1}{4} Mr^{2} }$

## Flywheels

Flywheels are used to store rotational energy. This is useful when the source of energy is not continuous, as they can help provide a continuous source of energy. They are used in many types of motors including modern cars.

It is because of an disk’s moment of inertia that it can store rotational energy in this way. Just as with mass in the linear case, it requires a force to change the rotational speed (angular velocity) of an object. The larger the moment of inertia, the larger the force required to change its angular velocity. As we can see above from the equation for the moment of inertia of a disk, for two flywheels of the same mass a thinner larger one will store more energy than a thicker smaller one because its moment of inertia increases as the square of the radius of the disk.

Sometimes mass is a critical factor, and next time I will consider the case of an annulus, where the inner part of the disk is removed.