4.5 Rigid Body Rotation

4.5.1 Moment of Inertia

A rigid body is a system of particles which are fixed together such that they move together, no matter the forces applied. We have seen in previous chapters that it is possible to treat rigid bodies as point masses when considering linear motion. However, when considering rotational effects as well, we need to consider the macroscopic extent of an object. Consider a rigid body in motion and look at the total kinetic energy, which is given by the sum of kinetic energy of each constituent particle:

K=\sum_{i}\frac{1}{2}m_{i}\boldsymbol{v}_{i}\cdot\boldsymbol{v}_{i}.

(4.67)

We can split the velocity of each particle into the sum of velocity around the centre of rotation and velocity along the line of motion.

$\displaystyle K$	$\displaystyle=\sum_{i}\frac{1}{2}m_{i}(\boldsymbol{v}_{i,\text{rot}}+% \boldsymbol{v}_{i,\text{lin}})^{2}$	(4.68)
	$\displaystyle=\sum_{i}\frac{1}{2}m_{i}(v_{i,\text{rot}}^{2}+v_{i,\text{lin}}^{% 2}+2\boldsymbol{v}_{i,\text{rot}}\cdot\boldsymbol{v}_{i,\text{lin}})$	(4.69)
	$\displaystyle=\underbrace{\sum_{i}\frac{1}{2}m_{i}v_{i,\text{rot}}^{2}}_{K_{% \text{rotational}}}+\underbrace{\sum_{i}\frac{1}{2}m_{i}v_{i,\text{lin}}^{2}}_% {K_{\text{linear}}}.$	(4.70)

The cross-term in the square cancels out (why?). Assuming the rigid body is rotating about a fixed axis, all particles rotate in circles with the same angular velocity $\omega$ because their distance from the axis of rotation is constant. This means the rotational kinetic energy can be written as

$\displaystyle K_{\text{rot}}$	$\displaystyle=\sum_{i}\frac{1}{2}m_{i}v_{i,\text{rot}}^{2}=\sum_{i}\frac{1}{2}% m_{i}(r_{i}\omega)^{2}$	(4.71)
	$\displaystyle=\frac{1}{2}\omega^{2}\sum_{i}m_{i}r_{i}^{2}$	(4.72)
	$\displaystyle=\frac{1}{2}I\omega^{2}.$	(4.73)

$I$ is called the moment of inertia, and it is kind of an angular equivalent of mass. Notice how it is calculated in a similar way to the centre of mass except with the square of $r$ . Also notice how in the formula for $K_{\text{rot}}$ , $I$ and $\omega$ play the role of $m$ and $v$ respectively, which shows how they are analogous to the linear quantities. However $I$ , just like the centre of mass, depends on how the mass is distributed in an object. For example, consider a solid disc and a hoop of the same mass. From the formula for $I$ , we can see that since all of the mass in the hoop is concentrated further out, the moment of inertia will be larger than the disc. This means that for the same angular velocity, the rotational kinetic energy of the hoop will be larger than the disc. $I$ also depends on the axis of rotation, since a different axis will have a different mass distribution around it.

Let’s do a few examples of calculating moments of inertia.

Example 4.7.

Consider a system of two particles with masses $m_{1}$ and $m_{2}$ separated by a distance $d$ .

What is the moment of inertia about the axis which passes through $m_{1}$ ( $A$ ) and about the axis which passes halfway between $m_{1}$ and $m_{2}$ ( $B$ )?

Lets to the axis which runs through the midpoint first. Using the formula for moment of inertia, we get

$\displaystyle I_{B}$	$\displaystyle=m_{1}r_{1}^{2}+m_{2}r_{2}^{2}$	(4.74)
	$\displaystyle=m_{1}\left(\frac{d}{2}\right)^{2}+m_{2}\left(\frac{d}{2}\right)^% {2}$	(4.75)
	$\displaystyle=\frac{m_{1}+m_{2}}{4}d^{2}.$	(4.76)

For the moment of inertia about axis $A$ , the distance of $m_{1}$ from the axis is zero, so the moment of inertia is

I_{A}=m_{2}d^{2}.

(4.77)

The key takeaway here is that any mass located at the axis of rotation contributes nothing to the moment of inertia, no matter how much it is.

4.5.2 Moment of Inertia of Continuous Objects

Just like calculating the centre of mass, when we have a continuous body we have to calculate moments of inertia using integrals. The formula is then

I=\int r^{2}\operatorname{\mathrm{d}}\!m,

(4.78)

where we have to find a way to express the mass element $\operatorname{\mathrm{d}}\!m$ in terms of length, area, or volume elements. We will now find the moments of inertia of common shapes.

Example 4.8.

Find the moment of inertia of a thin rod of uniform density rotating about its midpoint.

Let $L$ be the length of the rod and $\lambda$ be its linear mass density. Then from the diagram, we can see that $\operatorname{\mathrm{d}}\!m=\lambda\operatorname{\mathrm{d}}\!x$ . For the integral, the limits are $-L/2$ and $L/2$ and $r^{2}=x^{2}$ , so we have

I=\int_{\frac{L}{2}}^{\frac{L}{2}}x^{2}\lambda\operatorname{\mathrm{d}}\!x=% \left.\frac{\lambda x^{3}}{3}\right|_{\frac{L}{2}}^{\frac{L}{2}}=\frac{\lambda L% ^{3}}{12}.

(4.79)

If we let $M$ be the total mass then $M=\lambda L$ , and we have

I=\frac{1}{12}ML^{2}.

(4.80)

Example 4.9.

Find the moment of inertia of a uniform density thin hoop of radius $r$ in the $x$ - $y$ plane rotating around the $z$ -axis about its centre.

We can use a symmetry argument to solve this. The distance from the axis is constant, so it doesn’t change for any mass element and can be taken out of the integral. Thus the moment of inertia just the total mass multiplied by the radius squared.

I=Mr^{2}.

(4.81)

Note that this is the same result as a point mass of mass $M$ revolving around the $z$ -axis at the same radius $r$ .

Example 4.10.

Find the moment of inertia of a uniform density thin disc of radius $R$ in the $x$ - $y$ plane rotating about the $z$ -axis about its centre.

Here, we can use the result from the thin hoop. If we split the disc up into thin hoops, then we have $\operatorname{\mathrm{d}}\!I=r^{2}\operatorname{\mathrm{d}}\!m$ and we can integrate this from $0$ to $R$ . But first, we need to write the mass element in terms of the radius of the thin hoop. We can write $\operatorname{\mathrm{d}}\!m=\mu\operatorname{\mathrm{d}}\!A$ where $\mu$ is the surface density of the disc, and the area of a thin hoop is $\operatorname{\mathrm{d}}\!A=2\pi r\operatorname{\mathrm{d}}\!r$ . Putting this all together, we get

\displaystyle I=\int_{0}^{R}2\pi\mu r^{3}\operatorname{\mathrm{d}}\!r=\left.% \frac{2\pi\mu r^{4}}{4}\right|_{0}^{R}=\frac{1}{2}\pi\mu R^{4}.

(4.82)

The total mass of the disc is $M=\mu A=\mu\pi R^{2}$ , so this simplifies to

I=\frac{1}{2}MR^{2}.

(4.83)

So the moment of inertia of a thin disc is half that of a thin hoop with the same mass and radius.

Example 4.11.

Find the moment of inertia of a uniform density sphere of radius $R$ rotating about its centre.

Here we will again build upon previous results, this time using the moment of inertia of a thin disc. By aligning the axis of rotation with the $z$ -axis, we can split the sphere into thin discs of different radii and at different heights on the $z$ -axis. Then the total moment of inertia is found by integrating over $\operatorname{\mathrm{d}}\!I=\frac{1}{2}r^{2}\operatorname{\mathrm{d}}\!m$ , where $r$ is the radius of the thin disc. $\operatorname{\mathrm{d}}\!m$ is given by $\rho\operatorname{\mathrm{d}}\!V$ where $\rho$ is the volume density, and the volume element $\operatorname{\mathrm{d}}\!V$ is the area of the disc multiplied by its height, $\operatorname{\mathrm{d}}\!V=\pi r^{2}\operatorname{\mathrm{d}}\!z$ . So far, we have

\operatorname{\mathrm{d}}\!I=\frac{1}{2}\rho\pi r^{4}\operatorname{\mathrm{d}}% \!z.

(4.84)

In order to integrate this we need a relation between $r$ , the radius of each disc, and $z$ , its height. This is given by Pythagoras’ theorem (see the diagram above) as $r^{2}=R^{2}-z^{2}$ so we get

\operatorname{\mathrm{d}}\!I=\frac{1}{2}\rho\pi(R^{2}-z^{2})^{2}\operatorname{% \mathrm{d}}\!z.

(4.85)

We can now tackle the integral:

$\displaystyle I$	$\displaystyle=\int_{-R}^{R}\frac{1}{2}\rho\pi(R^{2}-z^{2})^{2}\operatorname{% \mathrm{d}}\!z$	(4.86)
	$\displaystyle=\int_{0}^{R}\rho\pi(R^{4}+z^{4}-2R^{2}z^{2})\operatorname{% \mathrm{d}}\!z$	(4.87)
	$\displaystyle=\rho\pi\left[R^{4}z+\frac{z^{5}}{z}-\frac{2}{3}R^{2}z^{3}\right]% _{0}^{R}$	(4.88)
	$\displaystyle=\frac{8}{15}\rho\pi R^{5}.$	(4.89)

Note that in the second line we have used symmetry to ignore half of the integral and double the result. The mass of a sphere is $M=\frac{4}{3}\rho\pi R^{3}$ , so we have

I=\frac{2}{5}MR^{2}.

(4.90)

4.5.3 Helpful Results for Finding Moments of Inertia

We will now look at two useful results that help with calculating more moments of inertia.

Theorem 4.1 (Parallel Axis Theorem)

For a system of total mass $M$ and moment of inertia about an axis passing through the centre of mass $I_{\text{COM}}$ , the moment of inertia about a parallel axis a distance $d$ away is

I_{\parallel}=I_{\text{COM}}+Md^{2}.

(4.91)

Proof.

Let’s put ourselves in the centre of mass frame, align the $z$ -axis with the axis of rotation and the $x$ -axis with the along the perpendicular distance between the two axes. The moment of inertia relative to the $z$ -axis is

I_{\text{COM}}=\int(x^{2}+y^{2})\operatorname{\mathrm{d}}\!m,

(4.92)

and the moment of inertia relative to the other axis is

$\displaystyle I_{\parallel}$	$\displaystyle=\int[(x-d)^{2}+y^{2}]\operatorname{\mathrm{d}}\!m$	(4.93)
	$\displaystyle=\int(x^{2}+y^{2})\operatorname{\mathrm{d}}\!m+d^{2}\int% \operatorname{\mathrm{d}}\!m+2d\int x\operatorname{\mathrm{d}}\!m$	(4.94)
	$\displaystyle=I_{\text{COM}}+Md^{2}+2d\int x\operatorname{\mathrm{d}}\!m.$	(4.95)

The integral in the last term is $x$ coordinate of the centre of mass, which is zero in the centre of mass frame. So the last term disappears, leaving us with the result. ∎

We can use this result to calculate moments of inertia of shapes we have seen before about different axes that might be difficult to find via a straightforward approach.

Example 4.12.

Consider a uniform density thin disc of radius $R$ in the $x$ - $y$ plane, but instead of rotating around the $z$ -axis about its centre like before, it is rotating about a point on the circumference.

We know that the moment of inertia of the thin disc about the centre is $I_{\text{COM}}=\frac{1}{2}MR^{2}$ , and the distance to the parallel axis is simply the radius $R$ , so applying the parallel axis theorem we get

I_{\parallel}=\frac{1}{2}MR^{2}+MR^{2}=\frac{3}{2}MR^{2}.

(4.96)

Theorem 4.2 (Perpendicular Axis Theorem)

Suppose we have a planar lamina lying in the $x$ - $y$ plane. The moment of inertia about the $z$ axis, $I_{z}$ , is related to the moments of inertia about the $x$ and $y$ -axes, $I_{x}$ and $I_{y}$ , by

I_{z}=I_{x}+I_{y}.

(4.97)

Proof.

The proof is straightforward: manipulating the integral for moment of inertia about the $z$ -axis we get

I_{z}=\int(x^{2}+y^{2})\operatorname{\mathrm{d}}\!m=\int x^{2}\operatorname{% \mathrm{d}}\!m+\int y^{2}\operatorname{\mathrm{d}}\!m=I_{y}+I_{x}.

(4.98)

Note that $\int x^{2}\operatorname{\mathrm{d}}\!m=I_{y}$ , not $I_{x}$ since $x^{2}$ in this case denotes the perpendicular distance from the axis of rotation, which must be around the $y$ -axis. ∎

Example 4.13.

Consider the uniform density thin disc again. What is its moment of inertia about the $x$ and $y$ axes?

We know that $I_{z}=\frac{1}{2}MR^{2}$ for the thin disc, and by the perpendicular axis theorem we have $I_{z}=I_{x}+I_{y}$ . In this case we can invoke symmetry of the disc to argue that $I_{x}=I_{y}$ , and therefore $I_{z}=2I_{x}=2I_{y}$ . Therefore we must have

I_{x}=I_{y}=\frac{I_{z}}{2}=\frac{1}{4}MR^{2}.

(4.99)

Theorem 4.3 (Stretch Rule)

The moment of inertia of a rigid object is unchanged when the object is stretched parallel to the axis of rotation and the distribution of mass is kept the same (except along the axis of rotation). This allows us to take a planar object and extrude it into 3D along the $z$ -axis while keeping the same mass and moment of inertia.

Proof.

Have an object of cross-sectional area in the $x$ - $y$ plane $A$ and height $L$ . Then the moment of inertia around the $z$ -axis can be expressed as

I_{z}=\int(x^{2}+y^{2})\rho(x,y,z)\operatorname{\mathrm{d}}\!V=\int_{0}^{L}% \operatorname{\mathrm{d}}\!z\int_{A}(x^{2}+y^{2})\rho(x,y,z)\operatorname{% \mathrm{d}}\!A.

(4.100)

If we stretch this object along $z$ by a factor of $a$ , then we have to divide the mass density by $a$ and change the upper limit on the $z$ integral to the new height $a L$ . This means the total mass remains unchanged. Then the new moment of inertia is

I_{z}^{\prime}=\int_{0}^{aL}\operatorname{\mathrm{d}}\!z\int_{A}(x^{2}+y^{2})% \frac{\rho(x,y,z/a)}{a}\operatorname{\mathrm{d}}\!A.

(4.101)

Making the substitution $z^{\prime}=z/a$ , we get

$\displaystyle I_{z}^{\prime}$	$\displaystyle=\int_{0}^{L}a\operatorname{\mathrm{d}}\!z^{\prime}\int_{A}(x^{2}% +y^{2})\frac{\rho(x,y,z^{\prime})}{a}\operatorname{\mathrm{d}}\!A$	(4.102)
	$\displaystyle=\int_{0}^{L}\operatorname{\mathrm{d}}\!z^{\prime}\int_{A}(x^{2}+% y^{2})\rho(x,y,z^{\prime})\operatorname{\mathrm{d}}\!A$	(4.103)
	$\displaystyle=I_{z}.$	(4.104)

∎

Using the stretch rule, we can say that the moment of inertia of a cylinder rotating about its main axis is the same as a thin disc.

Let’s now look at some problems which combine all the ideas of this chapter.

Example 4.14.

A light inextensible cable is being pulled out from a reel with a force of $9\text{\,}\mathrm{N}$ . The reel weighs $50\text{\,}\mathrm{kg}$ and has a radius of $0.6\text{\,}\mathrm{m}$ . After $2\text{\,}\mathrm{m}$ of the cable have been pulled out, what it its velocity?

To solve this, we will look at the rotational kinetic energy of the pulley before and after and then invoke the work-energy theorem. Its initial kinetic energy is zero, and its final kinetic energy is its rotational kinetic energy $K_{\text{rot}}=\frac{1}{2}I\omega^{2}$ . By the work-energy theorem, the final kinetic energy is equal to the work done, which is the length of rope pulled out multiplied by the tension $W=Ts$ . Putting these two equations together and rearranging for $\omega$ , we get

\omega=\sqrt{\frac{2Ts}{I}}.

(4.105)

By the stretch rule, the moment of inertia of the reel is the same as a thin disc, namely $I=\frac{1}{2}MR^{2}$ , so we get

\omega=\sqrt{\frac{4Ts}{MR^{2}}}=\sqrt{\frac{4\cdot$9\text{\,}\mathrm{N}$\cdot% $2\text{\,}\mathrm{m}$}{$50\text{\,}\mathrm{kg}$\cdot($0.6\text{\,}\mathrm{m}$% )^{2}}}=$2\text{\,}\mathrm{rad}\text{\,}{\mathrm{s}}^{-1}$.

(4.106)

Now that we have the final angular velocity of the reel, the speed of the cable is just the tangential speed of the reel at its edge, which is

v_{\text{cable}}=v_{\text{reel}}=R\omega=$0.6\text{\,}\mathrm{m}$\cdot$2\text{% \,}\mathrm{rad}\text{\,}{\mathrm{s}}^{-1}$=$1.2\text{\,}\mathrm{m}\text{\,}{% \mathrm{s}}^{-1}$.

(4.107)

Example 4.15.

Suppose a dumbbell-shaped pendulum is allowed to fall from horizontal. The weights at each end are spherical with radius $r=$10\text{\,}\mathrm{cm}$$ and they weigh $m_{\text{weight}}=$10\text{\,}\mathrm{kg}$$ . The bar is $l=$60\text{\,}\mathrm{cm}$$ long and weighs $m_{\text{bar}}=$1\text{\,}\mathrm{kg}$$ . The pivot point is at the left end of the bar. What is angular velocity of the dumbbell when it is vertical?

The first step to solving this problem is working out the moment of inertia of the dumbbell about the pivot point. If we split the dumbbell into the two weights ( $A$ for the one above the pivot and $B$ for the one below) and the bar, the total moment of inertia can be written as

I=I_{A}+I_{\text{bar}}+I_{B}.

(4.108)

Using the parallel axis theorem, the moment of inertia of the bar is $I_{\text{bar}}=\frac{1}{12}m_{\text{bar}}l^{2}+m_{\text{bar}}\left(\frac{1}{2}% l\right)^{2}=\frac{1}{3}m_{\text{bar}}l^{2}$ . For the weights, we can again use the parallel axis theorem since we know that the moment of inertia of a sphere is $\frac{2}{5}mr^{2}$ . Hence we get

$\displaystyle I$	$\displaystyle=\left(\frac{2}{5}m_{\text{weight}}r^{2}+m_{\text{weight}}r^{2}% \right)+\frac{1}{3}m_{\text{bar}}l^{2}+\left(\frac{2}{5}m_{\text{weight}}r^{2}% +m_{\text{weight}}(l+r)^{2}\right)$	(4.109)
	$\displaystyle=\frac{9}{5}\cdot$10\text{\,}\mathrm{kg}$\cdot($0.1\text{\,}% \mathrm{m}$)^{2}+\frac{1}{3}\cdot$1\text{\,}\mathrm{kg}$\cdot($0.6\text{\,}% \mathrm{m}$)^{2}+$10\text{\,}\mathrm{kg}$\cdot($0.7\text{\,}\mathrm{m}$)^{2}$	(4.110)
	$\displaystyle=$5.2\text{\,}\mathrm{kg}\text{\,}{\mathrm{m}}^{2}$.$	(4.111)

Now, to use energy conservation we also need to know the potential energy difference. To do this we need to know where the centre of mass is, since the potential energy difference is just $-Mgh$ , where $M$ is the total mass and $h$ is the distance that the centre of mass dropped. By symmetry, the centre of mass is clearly just the centre of the dumbbell which is a distance of $30\text{\,}\mathrm{cm}$ from the pivot. Thus the height by which the centre of mass has dropped when it reaches the bottom is $30\text{\,}\mathrm{cm}$ . Setting the zero point of the potential energy to be the initial height, the total energy before is zero since the dumbbell is not moving. The final energy is

E_{f}=\frac{1}{2}I\omega^{2}-Mgh=0.

(4.112)

Rearranging for $\omega$ , we get

$\displaystyle\omega$	$\displaystyle=\sqrt{\frac{2Mgh}{I}}$	(4.113)
	$\displaystyle=\sqrt{\frac{2\cdot$21\text{\,}\mathrm{kg}$\cdot$9.8\text{\,}% \mathrm{m}\text{\,}{\mathrm{s}}^{-2}$\cdot$0.3\text{\,}\mathrm{m}$}{$5.2\text{% \,}\mathrm{kg}\text{\,}{\mathrm{m}}^{2}$}}$	(4.114)
	$\displaystyle=$4.87\text{\,}\mathrm{rad}\text{\,}{\mathrm{s}}^{-1}$.$	(4.115)