Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: rotation of rigid bodies (203-SN1-RE)

Rotation takes all the mechanics you have already seen and changes every symbol: $x$ becomes $\theta$ , $m$ becomes $I$ , $F$ becomes $\tau$ . The equations are therefore known in advance. The difficulty lies elsewhere: $I$ depends on the chosen axis, and a single object has as many moments of inertia as there are possible axes.

This sheet gathers the mistakes specific to this chapter: the axis that is not named, the lever arm confused with the distance, the pulley wrongly idealized, and the two quantities, energy and angular momentum, of which only one is conserved when the body changes shape.

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The thread of the chapter

A moment of inertia does not exist on its own: it exists ABOUT AN AXIS. As long as the axis is not named, no formula of the chapter makes sense, and that is where most of the lost marks come from.

This chapter is part of 203-SN1-RE, Mechanics

The essentials

The translation-rotation dictionary

• $x\leftrightarrow\theta$ , $v\leftrightarrow\omega$ , $a\leftrightarrow\alpha$ , $m\leftrightarrow I$ , $F\leftrightarrow\tau$ , $p=mv\leftrightarrow L=I\omega$ , $\tfrac{1}{2}mv^{2}\leftrightarrow\tfrac{1}{2}I\omega^{2}$ .
• The constant-acceleration equations carry over unchanged: $\omega=\omega_{0}+\alpha t$ , $\theta=\omega_{0}t+\tfrac{1}{2}\alpha t^{2}$ , $\omega^{2}=\omega_{0}^{2}+2\alpha\theta$ .
• Links with a point of the body at distance $r$ from the axis: $v=\omega r$ , $a_{t}=\alpha r$ , $a_{c}=\omega^{2}r$ .
• ALL angles are in RADIANS in these relations. A $\theta$ in degrees gives a wrong $v$ by a factor of $57.3$ .
• $a_{t}$ and $a_{c}$ are perpendicular: the total acceleration of a point is $\sqrt{a_{t}^{2}+a_{c}^{2}}$ , never $a_{t}$ alone.

Along the tangent,

v=\omega r

and

a_{t}=\alpha r

; toward the axis,

a_{c}=\omega^{2}r

. The two accelerations are perpendicular: the total acceleration is their vector sum.

The dictionary is no substitute for thinking: $I$ does not behave like $m$ , since it changes when the axis changes, whereas mass never does.

Torque: it is not the distance that counts

• $\tau=rF\sin\phi$ , where $\phi$ is the angle between $\vec{r}$ and $\vec{F}$ . Only the PERPENDICULAR component of the force makes the body turn.
• Another reading, often quicker: $\tau=F\times d_{\perp}$ , where $d_{\perp}$ is the distance from the axis to the LINE OF ACTION of the force; this is the lever arm.
• A force whose line of action passes through the axis has ZERO torque, whatever its magnitude.
• The sign of the torque is fixed by convention: positive counterclockwise, and this convention holds for the whole problem.
• $\Sigma\tau=I\alpha$ is Newton's second law in rotation, and it is written about the SAME axis as the $I$ used.

Three forces of the SAME magnitude at the SAME point: the perpendicular one turns at full effect, the oblique one according to its lever arm

d_{\perp}=r\sin 30°

, the one aimed at the axis not at all.

On a wrench, pulling perpendicular to the handle is optimal; pulling along the handle tightens nothing at all, however hard you pull. The sine says so.

Moment of inertia, rolling, angular momentum

• Common values, about the axis of symmetry: hoop $MR^{2}$ ; solid disc or cylinder $\tfrac{1}{2}MR^{2}$ ; solid sphere $\tfrac{2}{5}MR^{2}$ ; rod about its centre $\tfrac{1}{12}ML^{2}$ ; rod about one end $\tfrac{1}{3}ML^{2}$ .
• Parallel-axis theorem: $I=I_{cm}+Md^{2}$ , with $d$ the distance between the new axis and the centre of mass. It applies ONLY from the axis through the centre of mass.
• Rolling without slipping: $v_{cm}=\omega R$ and $a_{cm}=\alpha R$ , with $K=\tfrac{1}{2}Mv_{cm}^{2}+\tfrac{1}{2}I\omega^{2}$ .
• On a descent, the ranking depends on neither mass nor radius, only on the ratio $\dfrac{I}{MR^{2}}$ : sphere, then cylinder, then hoop, always in that order.
• $L=I\omega$ is conserved when $\Sigma\tau_{ext}=0$ , EVEN if $I$ changes along the way: this is the skater pulling her arms in.

Rolling without slipping: static friction starts at the point of contact, which is at rest, points up the slope and alone gives a torque about the centre; it does no work.

When $I$ is halved at constant $L$ , $\omega$ doubles and $K=\tfrac{1}{2}I\omega^{2}$ also DOUBLES: the energy comes from the skater's muscular work, it is not conserved.

Conservation of angular momentum: the skater

• $L=I\omega$ is conserved when no EXTERNAL torque acts about the axis; internal forces can change $I$ without changing $L$ .
• Arms pulled in: the mass moves closer to the axis, $I$ decreases, and $\omega$ increases in the inverse ratio, $\omega_{2}=\dfrac{I_{1}}{I_{2}}\omega_{1}$ .
• The energy $K=\dfrac{L^{2}}{2I}$ increases in the same ratio: the work done by the arms supplies it.

View from above, same angular momentum: with arms out, the mass is far from the axis and the rotation is slow; with arms in,

I

drops and the rotation speeds up.

An object that jumps onto a merry-go-round brings its own angular momentum, $mvd$ , with $d$ its lever arm about the axis: it is the $L$ of the complete system that is conserved.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using a moment of inertia without saying about which axis

the whole question, 6 marks

What not to write

“The $2.0$ kg, $1.2$ m rod turns about a nail driven through one end: $I=\tfrac{1}{12}ML^{2}=0.24$ kg·m $^{2}$ .”

What to write

“ $\tfrac{1}{12}ML^{2}$ holds for an axis through the CENTRE. Here the axis is at the end, so $I=\tfrac{1}{12}ML^{2}+M\left(\tfrac{L}{2}\right)^{2}=\tfrac{1}{3}ML^{2}=0.96$ kg·m $^{2}$ .”

Same rod, same mass, same length: only the pivot point changes, and the moment of inertia is multiplied by four. This is why a moment-of-inertia formula is never copied without checking the axis.

Why: The moment of inertia measures how the mass is distributed ABOUT A GIVEN AXIS. Changing the axis changes the result by a factor of $4$ in this case, and all the dynamics with it.

2. Forgetting the sine, or confusing the distance with the lever arm

3 marks, and an angular acceleration almost twice too large

What not to write

“The $50$ N force is applied $0.30$ m from the axis, at $35°$ to the bar: $\tau=50\times 0.30=15$ N·m.”

What to write

“ $\tau=rF\sin 35°=0.30\times 50\times 0.574=8.6$ N·m. The lever arm is $0.30\sin 35°=0.172$ m, not $0.30$ m.”

Why: Only the component perpendicular to the arm makes the body turn. The parallel component pulls on the axis, where the support takes it up entirely, with no effect on the rotation.

3. Keeping the same tension on both sides of a pulley that has mass

the whole question, 8 marks

What not to write

“Pulley of $2.0$ kg, two masses joined: the rope has the same tension on both sides, so $T_{1}=T_{2}$ .”

What to write

“If the pulley turns and has a moment of inertia, it is precisely the DIFFERENCE in tension that makes it turn: $(T_{2}-T_{1})R=I\alpha$ , so $T_{2}\neq T_{1}$ .”

Why: Equal tensions are a consequence of $I=0$ , not a property of pulleys. A statement that gives the mass of the pulley is telling you exactly not to use them.

4. Forgetting rotational energy in the energy balance of a rolling object

the whole question, 5 marks

What not to write

“The sphere rolls without slipping from a height $h$ : $mgh=\tfrac{1}{2}mv^{2}$ , so $v=\sqrt{2gh}$ .”

What to write

“ $mgh=\tfrac{1}{2}mv^{2}+\tfrac{1}{2}I\omega^{2}$ with $\omega=\dfrac{v}{R}$ , so $v=\sqrt{\dfrac{10gh}{7}}$ , which is $85\%$ of the speed of a frictionless slide.”

Why: A rolling body stores part of its energy in rotation. Answering $\sqrt{2gh}$ amounts to saying that it slides without turning, which the statement has just ruled out.

5. Believing that mass or radius decides the race down an incline

4 marks on the reasoning questions

What not to write

“The big marble is heavier, so it arrives first.”

What to write

“The arrival speed is $v=\sqrt{\dfrac{2gh}{1+I/MR^{2}}}$ : neither $M$ nor $R$ appears in it. Only the SHAPE counts, and the sphere always beats the cylinder, which always beats the hoop.”

Why: The ratio $\dfrac{I}{MR^{2}}$ is $0.4$ for a sphere, $0.5$ for a cylinder and $1$ for a hoop, whatever the mass and size. It is a pure number, specific to the geometry.

6. Conserving kinetic energy when the skater pulls her arms in

the whole reasoning question, 5 marks

What not to write

“Angular momentum is conserved, so kinetic energy is too: $\tfrac{1}{2}I\omega^{2}$ stays constant.”

What to write

“ $L=I\omega$ is conserved. If $I$ is halved, $\omega$ doubles, and $K=\dfrac{L^{2}}{2I}$ DOUBLES. The skater supplied this work with her muscles.”

Why: Pulling the arms in requires a force directed toward the axis, which does work over a radial displacement. This work is internal, so it creates no torque, but it does add energy.

7. Treating the friction of rolling without slipping as dissipative

the whole question, 5 marks

What not to write

“The sphere rolls $3.0$ m with friction, so $f\times 3.0$ must be subtracted from the energy balance.”

What to write

“In rolling WITHOUT slipping, the point of contact is at rest relative to the ground: the friction is STATIC, its point of application does not move, and it dissipates nothing.”

Why: The work done by a force is calculated at the point where it is applied. Here that point has zero velocity at every instant, hence zero work, even though the force is indispensable for making the sphere turn.

8. Working in degrees in $v=\omega r$ or in the angular equations

the whole question, and a result 57 times too large

What not to write

“The wheel turns $180°$ in $2.0$ s: $\omega=90$ deg/s, so $v=90\times 0.30=27$ m/s.”

What to write

“ $180°=\pi$ rad, so $\omega=1.57$ rad/s and $v=1.57\times 0.30=0.47$ m/s.”

Why: The radian is the only dimensionless angle that makes $v=\omega r$ homogeneous. The error factor is exactly $\dfrac{180}{\pi}=57.3$ , and it shows up in the absurdity of the result.

9. Adding moments of inertia calculated about different axes

the whole question, 6 marks

What not to write

“The system is a rod plus a mass at the end: $I=\tfrac{1}{12}ML^{2}+mR^{2}$ , with the rod about its centre and the mass about the axis.”

What to write

“All the terms must refer to the SAME axis. If the axis is the end of the rod: $I=\tfrac{1}{3}ML^{2}+mL^{2}$ .”

Why: Moments of inertia add only about a common axis. Adding two numbers calculated about two different axes amounts to adding quantities that do not describe the same motion.

Which method to choose

Torque or energy

What the question asks for, exactly as in translation

If an angular acceleration, a time or a tension is asked for → $\Sigma\tau=I\alpha$ for the rotating body, plus $\Sigma F=ma$ for the translating bodies

Example: Atwood machine with a massive pulley
If an angular velocity after a descent or a fall is asked for → energy balance with $\tfrac{1}{2}I\omega^{2}$ included

Example: rod released from the horizontal, sphere rolling down a slope
If a speed after a CHANGE OF SHAPE of the system is asked for → conservation of angular momentum, $I_{1}\omega_{1}=I_{2}\omega_{2}$

Example: skater, platform on which someone walks
If an object falls onto a rotating disc and sticks to it → conservation of angular momentum, never of energy

Example: clay dropped on a turntable
If a contact force or an axle reaction is asked for → energy first for $\omega$ , then $\Sigma F=ma_{cm}$ with $a_{c}=\omega^{2}r$

same sequence as in the energy chapter for the tension of a pendulum

The trigger phrase for conservation of angular momentum is always an internal change: “pulls her arms in”, “moves closer to the axis”, “sticks”, “comes off”. None of these events conserves kinetic energy.

Which moment of inertia to write

The geometry of the body AND the position of the axis, in that order

If the axis passes through the centre of symmetry → the formula from the table as it stands

Example: disc about its centre, $\tfrac{1}{2}MR^{2}$
If the axis is parallel to the previous one but offset by $d$ → $I=I_{cm}+Md^{2}$

Example: rod about one end, $\tfrac{1}{12}ML^{2}+M\left(\tfrac{L}{2}\right)^{2}=\tfrac{1}{3}ML^{2}$
If the body is made of several pieces → bring EACH piece to the same axis, then add

Example: rod plus point mass at the end: $\tfrac{1}{3}ML^{2}+mL^{2}$
If the body is a point mass at distance $r$ → $I=mr^{2}$ , with no coefficient at all

Example: ball on a string, satellite
If the statement gives the density and asks you to prove the formula → $I=\displaystyle\int r^{2}\,dm$ , with $dm$ expressed in terms of $dr$ or $dx$

this is the only question of the chapter where the integral calculus of 201-SN2-RE is used directly

Before writing anything, draw the axis and label it on the figure. A paper that shows the axis rarely picks the wrong formula; a paper that does not show it picks the wrong one every other time.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a system with a pulley that has mass

When to use it: The statement gives the mass or the moment of inertia of the pulley, or says it is not ideal.

1 Choose an assumed direction of motion and state it, then orient each axis in that direction, including the positive sense of rotation of the pulley.
2 Draw one free-body diagram per mass, with DIFFERENT tensions on each side of the pulley.
3 Write $\Sigma F=ma$ for each mass, then $\Sigma\tau=I\alpha$ for the pulley, that is $(T_{2}-T_{1})R=I\alpha$ .
4 Add the constraint of the non-slipping string: $a=\alpha R$ , which lets you replace $\alpha$ by $\dfrac{a}{R}$ and brings out the term $\dfrac{I}{R^{2}}$ .
5 Add the three equations to eliminate the two tensions, obtain $a$ , then work back to $T_{1}$ and $T_{2}$ separately.

Concluding sentence

“The three equations, added after substituting $\alpha=\dfrac{a}{R}$ , give $a=\dfrac{(m_{2}-m_{1})g}{m_{1}+m_{2}+I/R^{2}}=2.18$ m/s $^{2}$ , then $T_{1}=35.9$ N and $T_{2}=38.1$ N.”

The trap: The term $\dfrac{I}{R^{2}}$ has the dimensions of a MASS: it is the effective mass the pulley adds to the system. For a solid disc it is exactly $\dfrac{M}{2}$ , which gives an immediate check of the result.

Marking: Typically 2 marks for the two diagrams, 2 for the pulley equation, 1 for the string constraint, 2 for the solution, 1 for the two distinct tensions.

Handling an angular momentum conservation problem

When to use it: The system changes shape, or a body joins a body that is already rotating.

1 Justify the conservation: name the axis and say why no external torque acts about it, internal forces not counting.
2 Calculate $I_{1}$ in the initial state and $I_{2}$ in the final state, bringing each piece to the SAME axis.
3 Write $I_{1}\omega_{1}=I_{2}\omega_{2}$ and solve for the unknown.
4 If the statement asks for it, calculate $K_{1}$ and $K_{2}$ separately and comment on the difference, naming its source.

Concluding sentence

“No external torque acts about the vertical axis, so $L$ is conserved: $I_{1}\omega_{1}=I_{2}\omega_{2}$ gives $\omega_{2}=6.0$ rad/s, and the kinetic energy goes from $18$ J to $36$ J, the work of the muscles supplying the difference.”

The trap: Writing “the system is isolated” with no further detail. The weight and the axle reaction do exist: what must be said is that their TORQUE about the axis in question is zero.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The axis named on every moment of inertia

Go back over every $I$ on the paper and write beside it the axis it refers to. Two $I$ values for different axes in the same sum then jump out.

A $\tfrac{1}{12}ML^{2}$ in a problem where the nail is at the end: the axis does not match.
The term $I/R^{2}$ read as a mass

In a pulley problem, check that this term really has the dimensions of a mass and that it equals $\dfrac{M}{2}$ for a solid disc. It is a two-second check.

An $I/R^{2}=4.0$ kg for a pulley of $2.0$ kg: the moment of inertia used is not that of a disc.
The radian mode of the calculator

All the relations $v=\omega r$ , $a_{t}=\alpha r$ and $\theta=\omega t$ require radians. Check RAD on the screen and convert the degrees in the statement on the very first line.

A speed of $27$ m/s at the rim of a wheel that makes a half-turn in two seconds: factor $57$ , the angle stayed in degrees.
The comparison with a frictionless slide

A rolling body always arrives MORE SLOWLY than a body sliding without friction from the same height, since part of the energy has gone into rotation.

A rolling sphere arrives at $3.74$ m/s where a pure slide would give $4.43$ m/s: plausible. A value above $4.43$ is not.
The order of arrival on an incline

Solid sphere, then solid cylinder, then hoop, whatever the mass and radius. Any answer that reverses this order, or that invokes the mass, is wrong.

A line of reasoning that concludes the heavier one wins: the result contains neither $M$ nor $R$ .

The typical problem, taken apart

Atwood machine with a pulley that has mass

Two blocks of masses $m_{1}=3.00$ kg and $m_{2}=5.00$ kg are joined by an inextensible string passing over a pulley modelled as a solid disc of mass $M=2.00$ kg and radius $R=12.0$ cm.

The string does not slip on the pulley and the axle is frictionless. Determine the acceleration of the blocks, then the two tensions in the string. Take $g=9.80$ m/s $^{2}$ .

The pulley is not a mere redirection: it turns, so it absorbs part of the acceleration, and the two strands of the string do NOT pull with the same force. It is this difference that sets it spinning.

Step 1

Assumed direction: $m_{2}$ goes down, $m_{1}$ goes up, the pulley turns in the corresponding direction. Moment of inertia: $I=\tfrac{1}{2}MR^{2}=\tfrac{1}{2}(2.00)(0.120)^{2}=0.0144$ kg·m $^{2}$ .

Why

The assumed direction fixes the sign of every term, including that of the torque. Without this statement, the sign of the $I\alpha$ term cannot be justified.

Step 2

Block $m_{2}$ : $m_{2}g-T_{2}=m_{2}a$ . Block $m_{1}$ : $T_{1}-m_{1}g=m_{1}a$ . Pulley: $(T_{2}-T_{1})R=I\alpha$ .

Why

Three bodies, three equations. Writing distinct $T_{1}$ and $T_{2}$ from the very first line avoids the costliest mistake here, and makes any further justification unnecessary.

Step 3

Constraint of the non-slipping string: $a=\alpha R$ , so $\alpha=\dfrac{a}{R}$ and the pulley equation becomes $T_{2}-T_{1}=\dfrac{I}{R^{2}}a=1.00\,a$ .

Why

The term $\dfrac{I}{R^{2}}$ is $1.00$ kg here: the pulley behaves exactly like an extra mass of $1.00$ kg, half its real mass, which is the signature of the solid disc.

Step 4

Sum of the three equations: $(m_{2}-m_{1})g=\left(m_{1}+m_{2}+\dfrac{I}{R^{2}}\right)a$ , so $a=\dfrac{(5.00-3.00)(9.80)}{3.00+5.00+1.00}=\dfrac{19.6}{9.00}=2.18$ m/s $^{2}$ .

Why

The two tensions disappear by themselves because each appears twice with opposite signs. That is the only reason for orienting all the axes in the assumed direction of motion.

Step 5

$T_{1}=m_{1}(g+a)=3.00(9.80+2.18)=35.9$ N and $T_{2}=m_{2}(g-a)=5.00(9.80-2.18)=38.1$ N.

Why

Each tension is read from the equation of its own block, never from that of the pulley: the latter gives only their difference.

Step 6

Check with the pulley equation: $T_{2}-T_{1}=38.1-35.9=2.18$ N, to be compared with $\dfrac{I}{R^{2}}a=1.00\times 2.18=2.18$ N.

Why

We check with the equation that was not used to isolate the results. Had the pulley been massless, this difference would have to be zero, which also gives the limiting case to test.

Step 7

Limiting case: with $M=0$ , we would recover $a=\dfrac{19.6}{8.00}=2.45$ m/s $^{2}$ and $T_{1}=T_{2}=36.8$ N.

Why

The limiting case validates the structure of the formula: the massive pulley slows the system down and separates the tensions, exactly as the physics predicts.

The conclusion, written out

“The blocks accelerate at $2.18$ m/s $^{2}$ , the strand supporting $m_{1}$ is under $35.9$ N and the one supporting $m_{2}$ under $38.1$ N, the difference of $2.18$ N being exactly what sets the pulley spinning.”

The classic mistake on this problem: Setting $T_{1}=T_{2}$ from the start. You then get $a=2.45$ m/s $^{2}$ and a single tension, and the mass of the pulley given in the statement was never used: an unused piece of data is almost always the sign of a wrong assumption.

Learn by heart

• A moment of inertia is ALWAYS given with its axis: $\tfrac{1}{12}ML^{2}$ at the centre, $\tfrac{1}{3}ML^{2}$ at the end.
• Parallel-axis theorem: $I=I_{cm}+Md^{2}$ , and only from the axis through the centre of mass.
• $\tau=rF\sin\phi$ : a force directed toward the axis has zero torque.
• Radians are compulsory in $v=\omega r$ , $a_{t}=\alpha r$ and $a_{c}=\omega^{2}r$ .
• Rolling without slipping: $v=\omega R$ , $K=\tfrac{1}{2}Mv^{2}+\tfrac{1}{2}I\omega^{2}$ , and static friction dissipates nothing.
• On a slope, the ranking depends only on $\dfrac{I}{MR^{2}}$ : sphere $0.4$ , cylinder $0.5$ , hoop $1$ .
• Pulley with nonzero mass: $T_{1}\neq T_{2}$ , and $\dfrac{I}{R^{2}}$ adds like a mass in the denominator.
• $L=I\omega$ is conserved when $\Sigma\tau_{ext}=0$ ; $K=\dfrac{L^{2}}{2I}$ , on the other hand, changes as soon as $I$ changes.

Frequently asked questions

Why does the moment of inertia change with the chosen axis?

Because it measures how far from the axis the mass is distributed, and that distance changes with the axis. A rod turning about its centre has four times less inertia than the same rod turning about one end. The parallel-axis theorem lets you go from one to the other by adding the mass times the square of the offset.

When are the two tensions on a pulley different?

As soon as the pulley has mass and turns. It is precisely the gap between the two tensions that supplies the torque needed to set it spinning. An ideal massless pulley has zero moment of inertia and the two tensions are equal. If the statement gives the mass of the pulley, it is telling you not to confuse them.

Which object reaches the bottom of an incline first?

The solid sphere, ahead of the solid cylinder, ahead of the hoop, always in that order. Neither mass nor radius matters: all that counts is the ratio of the moment of inertia to the product of the mass and the square of the radius, which is two fifths, one half and one for these three shapes. The smaller this number, the less energy goes into rotation.

Why does a skater spin faster when she pulls her arms in?

Because angular momentum, the product of the moment of inertia and the angular velocity, is conserved when there is no external torque. By bringing her mass closer to the axis, she decreases her moment of inertia, so her angular velocity increases by the same factor. Her kinetic energy increases too: her muscles supply this work.

Does the friction on a rolling object dissipate energy?

No, as long as the rolling is without slipping. The point of contact is at rest relative to the ground at every instant, so the friction force does not move and does no work. It is still indispensable, since it creates the torque that makes the object turn.

Should rotation be done in degrees or radians?

In radians, without exception, in every relation that connects an angular quantity to a linear one. An angle in degrees gives a speed or an acceleration that is wrong by a factor of about fifty-seven. The first line of the paper should therefore convert the degrees given in the statement.

Practise it

Corrected exercises: Rotation of rigid bodies, 203-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys1-rigid-body-rotation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: rotation of rigid bodies (203-SN1-RE)

The essentials

The translation-rotation dictionary

Torque: it is not the distance that counts

Moment of inertia, rolling, angular momentum

Conservation of angular momentum: the skater

The mistakes that cost marks

1. Using a moment of inertia without saying about which axis

2. Forgetting the sine, or confusing the distance with the lever arm

3. Keeping the same tension on both sides of a pulley that has mass

4. Forgetting rotational energy in the energy balance of a rolling object

5. Believing that mass or radius decides the race down an incline

6. Conserving kinetic energy when the skater pulls her arms in

7. Treating the friction of rolling without slipping as dissipative

8. Working in degrees in v=ωrv=\omega rv=ωr or in the angular equations

9. Adding moments of inertia calculated about different axes

Which method to choose

Torque or energy

Which moment of inertia to write

How the answer is expected to be written

Solving a system with a pulley that has mass

Handling an angular momentum conservation problem

Check before you hand in

The typical problem, taken apart

Atwood machine with a pulley that has mass

Learn by heart

Frequently asked questions

Why does the moment of inertia change with the chosen axis?

When are the two tensions on a pulley different?

Which object reaches the bottom of an incline first?

Why does a skater spin faster when she pulls her arms in?

Does the friction on a rolling object dissipate energy?

Should rotation be done in degrees or radians?

Corrected exercises: Rotation of rigid bodies, 203-SN1-RE

See also

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8. Working in degrees in $v=\omega r$ or in the angular equations