Rigid body and moment of inertia: worked exercises

The Physics 1 course is about to change pace. Until October you work with the point mass: kinematics, Newton’s laws, work and energy. Between November and December comes the rigid body, that is, an object that has size, rotates and rolls. The mid-term test or the exam fall between four and ten weeks from now, between December and January, so now is the time to sort out the basics. It is for those studying Engineering, Physics, Mathematics or Chemistry who already have the first block of dynamics in hand.

The rigid body is frightening because it seems like new physics, but it isn’t: it is the same dynamics with different quantities. In my lessons I see that almost all lost marks come from three things: inconsistent units, wrong signs and texts read in a hurry. If you set up the axis and the sign well, the rest is algebra.

From translation to rotation: the dictionary

Every quantity of translation has its angular twin. Learning this correspondence is worth more than ten formulas by heart.

The bridge between the two worlds is the radius: for a point at distance $R$ from the axis, $v=\omega R$ and $a_t=\alpha R$ hold. Angles are always in radians. If you are given revolutions per minute, convert: $1\text{rev/min}=2\pi /60\text{rad/s}$.

The torque of a force about an axis is

where $\phi$ is the angle between the lever arm $r$ and the force. Only the component of the force perpendicular to the arm counts. The sign is fixed once and for all: counter-clockwise positive, clockwise negative, and it doesn’t change halfway through an exercise.

The moment of inertia

For a set of point masses it is calculated like this:

where $r_i$ is the distance of each mass from the axis of rotation. The unit of measurement is $\text{kg}\cdot {\text{m}}^2$. For continuous bodies the result is already tabulated, but you need to know about which axis:

For the same mass and radius, the further the mass is from the axis, the larger $I$ is: the ring has twice the moment of inertia of the disc. This detail decides who wins in rolling races, as we shall see.

The Huygens-Steiner theorem

If the axis doesn’t pass through the centre of mass but is parallel to one that does, at distance $d$:

Check with the rod: for one end $d=L/2$, so $I=\frac{1}{12}M{L}^2+M\frac{L^2}{4}=\frac{1}{3}M{L}^2$, which coincides with the table. With $M=3.0\text{kg}$ and $L=1.2\text{m}$ you get $0.36+1.08=1.44\text{kg}\cdot {\text{m}}^2$.

The classic mistake is starting from an $I$ calculated about an axis that is not the one through the centre of mass and adding $M{d}^2$ anyway. Steiner holds only starting from the centre of mass, never between any two axes.

The second cardinal equation and energy

For a body rotating about a fixed axis:

The kinetic energy of a body that moves and rotates at the same time is the sum of two terms:

Forgetting the second term is the most frequent mistake in energy problems with rolling bodies.

Pure rolling

A body rolls without slipping when the point of contact with the ground is at rest. The constraint is

Static friction is there, but it does no work: mechanical energy is conserved. On an inclined plane of angle $\theta$, combining $F=ma$ along the plane and $\tau =I\alpha$ about the centre, you get

The acceleration depends only on the ratio $I/(M{R}^2)$, that is on the shape, not on mass and radius. For the solid cylinder it is $\frac{2}{3}g\sin \theta$, for the solid sphere $\frac{5}{7}g\sin \theta$, for the ring $\frac{1}{2}g\sin \theta$.

Angular momentum

If the external torque about the axis is zero, $L=I\omega$ is conserved. It is the rule of the skater who pulls in her arms: $I$ decreases, $\omega$ increases. Careful, though: conservation of $L$ does not imply conservation of energy.

Exercise 1: pulley with a hanging mass

A uniform disc of mass $M=2.0\text{kg}$ and radius $R=0.10\text{m}$ rotates about its own horizontal axis. Wound around the rim is a massless rope, which doesn’t slip, from which a mass $m=0.50\text{kg}$ hangs. The system starts from rest. Calculate the acceleration of the mass, the tension in the rope and the speed after a descent of $1.0\text{m}$ (with $g=9.8{\text{m/s}}^2$).

Step 1: moment of inertia. $I=\frac{1}{2}M{R}^2=\frac{1}{2}\cdot 2.0\cdot 0.010=0.010\text{kg}\cdot {\text{m}}^2$.

Step 2: one equation for each body. For the mass, with the positive direction downwards: $mg-T=ma$. For the disc: $TR=I\alpha$. The no-slip constraint of the rope is $a=\alpha R$.

Step 3: substitute. From the disc $T=I\alpha /R=Ia/R^2$. Inserting in the first:

Here $I/R^2=0.010/0.010=1.0\text{kg}$ (it is $M/2$), so

Step 4: tension. $T=m(g-a)=0.50\cdot (9.8-3.27)\approx 3.27\text{N}$. The tension is less than $mg=4.9\text{N}$: the rope doesn’t bear the whole weight, because the mass is accelerating.

Step 5: speed. $v=\sqrt{2ah}=\sqrt{2\cdot 3.27\cdot 1.0}\approx 2.56\text{m/s}$.

Check with energy. The work done by gravity is $mgh=4.9\text{J}$. With $\omega =v/R=25.6\text{rad/s}$: $\frac{1}{2}m{v}^2=1.63\text{J}$ and $\frac{1}{2}I{\omega }^2=3.27\text{J}$, sum $4.9\text{J}$. The numbers add up.

To train on closed-answer questions there is the Physics 1 practice test: the correct answer is free, the guided step-by-step solution unlocks with your first lesson. There is no practice test dedicated to the rigid body alone: that is the reference page for the course.

Exercise 2: solid cylinder against ring

A solid cylinder and a ring, with the same mass and the same radius, start together from rest at the top of an inclined plane of $30^{\circ }$ and length $L=2.0\text{m}$ and roll without slipping. Which arrives first, and at what speed?

Solid cylinder, $I/(M{R}^2)=\frac{1}{2}$:

Ring, $I/(M{R}^2)=1$:

The descent time is $t=\sqrt{2L/a}$:

• cylinder: $t=\sqrt{4.0/3.27}\approx 1.11\text{s}$;
• ring: $t=\sqrt{4.0/2.45}\approx 1.28\text{s}$.

The final speeds are $v=\sqrt{2aL}$: about $3.6\text{m/s}$ for the cylinder and $3.1\text{m/s}$ for the ring. The cylinder wins, whatever the mass and radius, because the race depends only on shape. The ring keeps all its mass far from the axis and spends a larger share of the available energy on rotation.

A check that closes the problem well: a body sliding without friction would have $a=g\sin \theta =4.9{\text{m/s}}^2$, greater than both. If your result exceeds this value, you have got something wrong.

Exercise 3: the disc that couples on

A disc with $I_1=0.50\text{kg}\cdot {\text{m}}^2$ rotates at ${\omega }_1=12\text{rad/s}$ about a fixed vertical axis. From above a second disc at rest, with $I_2=0.30\text{kg}\cdot {\text{m}}^2$, falls on top of it, coaxially, and the two drag each other by friction until they rotate together. Calculate the final angular velocity and the energy dissipated.

There are no external torques about the axis, so angular momentum is conserved:

For the energy:

• initial: $\frac{1}{2}\cdot 0.50\cdot 12^2=36\text{J}$;
• final: $\frac{1}{2}\cdot 0.80\cdot 7.5^2=22.5\text{J}$;
• dissipated: $36-22.5=13.5\text{J}$, which is $37.5\%$ of the initial.

It is the rotational counterpart of the perfectly inelastic collision. Conserving $L$ and not conserving $E_k$ is the point the exam wants to see, and many students invent a conservation of energy that isn’t there.

The mistakes that cost marks

1. Wrong axis in Steiner. Always start from $I_{CM}$. Write the starting axis on the sheet before adding $M{d}^2$.
2. Rotational term forgotten. If a body rolls, the kinetic energy has two addends. The check is simple: if the body rotates, $\frac{1}{2}I{\omega }^2$ must be there.
3. Constraint $v=\omega R$ ignored or used with the wrong radius. In rolling the radius is that of the rolling body; in a pulley it is that of the groove on which the rope is wound.
4. Units and dimensional check. Before handing in, check that $I$ is in $\text{kg}\cdot {\text{m}}^2$, $\omega$ in $\text{rad/s}$, $\tau$ in $\text{N}\cdot \text{m}$. A centimetre left in place of the metre gets $I$ wrong by a factor of 10,000. It is the first filter I apply when marking.
5. Signs. Choose the positive direction of rotation and stick to it for all torques. A mass that goes down and a pulley that turns must have directions consistent with each other.
6. Text read halfway. “Rolls without slipping”, “the rope doesn’t slip”, “fixed axis”: every phrase of the text is a condition. Underline them before writing formulas.

How to prepare in the weeks that remain

The rigid body is learned only by solving. Redo the three exercises above by hand changing the data, then move on to mixed problems with pulleys, rotating rods and rolling uphill. If the previous block is still fragile, revise kinematics and dynamics for the Physics 1 mid-term, where you will find the method for setting up forces. For energy balances you need work, energy and collisions, which tackles the same ideas at a simpler level. If the course continues with the analytical formulation, the next step is Lagrangian dynamics in Rational Mechanics. And to revisit an inclined plane without rotation from scratch there is dynamics and the inclined plane.

Rotation exercises follow the same scheme every time: drawing with axis and directions, equations for each body, constraints, result and energy check. The general picture of the subject is on the physics page.

If you want a check on your method

If every rotation problem leaves you with a different doubt, there is usually a single cause, upstream: axis, signs or constraint. Write to me from the contact page and we’ll do a free introductory call: you send me one or two exercises that don’t work out for you and I tell you where the method breaks.