PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: torque, rotation and equilibrium (PHYS 101, McGill)

This sheet does not re-derive the rotational formulas: they are in your course pack and in the formulary you are given in the exam. It deals with what actually decides the mark on this chapter, which is almost never the algebra. It is knowing that the lever arm is measured to the LINE of the force and not to the point where it is applied, knowing that you may take torques about any axis you like and that choosing well deletes an unknown, and knowing that a rolling body stores part of its energy as spin.

It is written for PHYS 101, Introductory Physics, Mechanics, the algebra-based course taken at McGill University by students going into the life sciences. No derivative and no integral appears anywhere on this page, because none appears in the course: every moment of inertia is read off a formulary and every result is reached with algebra and the trigonometry of a right triangle. Once the sheet is read, the corrected exercise set for the same chapter puts each reflex to the test.

The thread of the chapter

In rotation a force no longer counts by its magnitude but by its LEVER ARM, and a mass no longer counts by its value but by its DISTANCE from the axis. What loses marks is translating the formula without translating the question, and so adding forces where the question asks for a sum of torques about an axis you were free to choose.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Angular displacement
Angular velocity
Angular acceleration
Radian measure
Torque
Lever arm
Static equilibrium
Centre of gravity
Moment of inertia
Rotational kinetic energy
Rolling without slipping
Angular momentum

The essentials

Torque: it is a geometry question before it is a physics one

• $\tau = r F \sin\phi = d\,F$ , where $\phi$ is the angle between the arm $r$ and the force, and $d = r\sin\phi$ is the LEVER ARM.
• The lever arm is the perpendicular distance from the axis to the LINE of the force, extended as far as you need. It is not the distance to the point where the force is applied.
• $\sin 90^{\circ} = 1$ gives the maximum; $\sin 0^{\circ} = 0$ says a force aimed at the axis turns nothing, however large.
• A torque is SIGNED about a fixed axis: pick one sense as positive and keep it for the whole problem. There is no such thing as adding torques as vectors in this course.
• The unit is the newton metre. It is not a joule, even though the two are dimensionally identical, and writing J for a torque costs the mark.

Same force, same point of application, two angles: on the left the lever arm is the whole bar,

0.40

m, on the right only

0.20

m, because the perpendicular is dropped onto the LINE of the force.

Before writing a single equation, draw the lines of action and drop the perpendiculars. Half the marks of this chapter are already decided on that drawing.

The two conditions of static equilibrium, and the axis you may choose

• Nothing moves means TWO conditions, not one: $\sum \vec{F} = \vec{0}$ AND $\sum \tau = 0$ .
• They are independent. A couple, two equal opposite forces on different lines, satisfies the first and not the second.
• For a body in equilibrium the torques cancel about EVERY axis, real or imaginary, inside the body or outside it.
• So choose the axis through the force you do not want: it has zero lever arm there and disappears from the equation.
• The weight of a uniform body acts at its centre of gravity, at its geometric centre. For a plank $4.0$ m long that is $2.0$ m from either end, wherever the supports happen to be.

State the axis in writing before the torque equation. A marker cannot award a torque line whose axis is not named, and two lines written about two different axes cancel each other out in the worst possible way.

Moment of inertia: mass counts once, distance counts twice

• $I = \sum m_{i} r_{i}^{2}$ . Two bodies of the same mass have quite different $I$ when the mass sits at different distances from the axis.
• From the formulary: hoop about its axis $I = mR^{2}$ ; solid disk or cylinder $I = \frac{1}{2}mR^{2}$ ; solid sphere $I = \frac{2}{5}mR^{2}$ ; thin rod about its centre $I = \frac{1}{12}mL^{2}$ , about one end $I = \frac{1}{3}mL^{2}$ ; a small body at distance $R$ , $I = mR^{2}$ .
• $I$ is defined only ONCE AN AXIS IS NAMED. The same rod has $\frac{1}{12}mL^{2}$ or $\frac{1}{3}mL^{2}$ depending on where you spin it, a factor of four.
• Parallel axis, when the formulary gives it: $I = I_{\text{cm}} + Md^{2}$ , for an axis parallel to the one through the centre of mass and a distance $d$ from it.
• Rotational Newton: $\sum \tau = I\alpha$ . The left side is the NET torque, friction and motor and weight together, exactly as $\sum F = ma$ is the net force.

Energy and angular momentum in rotation

• $K_{\text{rot}} = \frac{1}{2}I\omega^{2}$ , and a body that both moves and turns has $K = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$ . The two terms are both there, always.
• Rolling without slipping: the contact point is instantaneously at rest, so $v = R\omega$ and friction does NO work.
• With $I = kmR^{2}$ , the energy balance $mgh = \frac{1}{2}mv^{2}(1+k)$ gives $v = \sqrt{\frac{2gh}{1+k}}$ : neither $m$ nor $R$ survives, only $k$ .
• $L = I\omega$ , and $L$ is conserved about an axis as long as no external torque acts about that axis.
• $L$ conserved does NOT mean $K$ conserved. With $K = \frac{L^{2}}{2I}$ , halving $I$ doubles the kinetic energy, and something had to do that work.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The translation dictionary, with a number under every line

Every formula of this chapter is a formula you already know with each symbol replaced. The table is useless as a list of symbols, which is why each line here carries the number it actually produces.

Translation	Rotation	The number it gives
$\Delta x$ , in metres	$\theta$ , in RADIANS	$\theta = 1131$ rad
Example: A wheel that makes $180$ turns has swept $180 \times 2\pi = 1131$ rad. In degrees it would be $64800$ , and $s = r\theta$ would then be false by a factor of $57.3$ .
$v = \frac{\Delta x}{\Delta t}$	$\omega = \frac{\Delta\theta}{\Delta t}$	$\omega = 188.5$ rad/s
Example: $1800$ rev/min $\times \frac{2\pi}{60} = 188.5$ rad/s. On a rim of radius $0.12$ m that is $v = r\omega = 22.6$ m/s.
$m$ , the mass	$I = \sum m r^{2}$	$I = 1.0 \times 10^{-2}$ kg m $^2$
Example: A solid disk of $2.0$ kg and radius $0.10$ m: $I = \frac{1}{2}(2.0)(0.10)^{2} = 1.0 \times 10^{-2}$ kg m $^2$ . Halve the radius and $I$ drops to a quarter.
$\sum F = ma$	$\sum \tau = I\alpha$	$\alpha = 73.5$ rad/s $^2$
Example: A net torque of $0.735$ N m on that same disk: $\alpha = \frac{0.735}{1.0 \times 10^{-2}} = 73.5$ rad/s $^2$ , and the rim accelerates at $a = R\alpha = 7.35$ m/s $^2$ .
$K = \frac{1}{2}mv^{2}$	$K = \frac{1}{2}I\omega^{2}$	$K = 474$ J
Example: A flywheel rotor, $I = 2.4 \times 10^{-3}$ kg m $^2$ at $628$ rad/s: $K = \frac{1}{2}(2.4 \times 10^{-3})(628)^{2} = 474$ J, as much as a $1$ kg mass thrown at $31$ m/s.
$p = mv$	$L = I\omega$	$\omega' = 0.415$ rad/s
Example: A carousel, $I = 291.6$ kg m $^2$ at $0.60$ rad/s, boarded by $40$ kg at $R = 1.8$ m: $\omega' = \frac{174.96}{421.2} = 0.415$ rad/s.
a heavier body has more inertia	a heavier body has a larger $I$	no number follows no such rule
Example: Two bodies of $2.0$ kg and radius $0.10$ m: as a hoop $I = 2.0 \times 10^{-2}$ , as a disk $I = 1.0 \times 10^{-2}$ kg m $^2$ . Same mass, same radius, factor two. Same form, other result: And the other way round: a $4.0$ kg disk of radius $0.05$ m has $I = 5.0 \times 10^{-3}$ kg m $^2$ , so it is TWICE as heavy as the $2.0$ kg hoop and four times easier to spin. What to do: Do not compare masses, compare $I = kmR^{2}$ line by line: read $k$ off the formulary, then compare $kmR^{2}$ .

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. Adding up the forces when the question asks about turning

the whole question, 8 to 10 marks

What not to write

“The plank carries $30 + 60 = 90$ kg, so each support takes $441$ N.”

What to write

“Torques about $A$ : $N_{B}(3.0) = (294)(2.0) + (588)(1.5)$ , so $N_{B} = 490$ N, and then $N_{A} = 882 - 490 = 392$ N.”

Why: The force condition alone has two unknowns and one equation: it cannot be solved. Sharing the load equally is only right when the loads are placed symmetrically, which is almost never what the problem draws. Write the torque line FIRST, because it is the one that yields a number.

2. Measuring the lever arm to the point of application

2 to 3 marks, and every later line inherits the error

What not to write

“The force acts $0.25$ m from the bolt, so $\tau = 0.25 \times 120 = 30$ N m.” Written for a force at $40^{\circ}$ to the handle.

What to write

“ $\tau = rF\sin\phi = 0.25 \times 120 \times \sin 40^{\circ} = 19.3$ N m, the lever arm being $d = 0.25\sin 40^{\circ} = 0.161$ m.”

The same trap in the other variable: three bodies of the same mass and the same radius, with

k = 1

0.5

and

0.4

. What counts is where the mass sits, not how much of it there is.

Why: The lever arm is the perpendicular distance to the LINE of the force, not to its point of application. Two forces applied at the same point can have completely different torques. The one-second test: at $\phi = 0$ the force points at the axis and must give zero, and only $\sin$ does that.

3. Writing $\tau = rF\cos\phi$

2 marks, and it is the most frequent single error on the chapter

What not to write

“ $\tau = rF\cos\phi = 0.25 \times 120 \times \cos 40^{\circ} = 23.0$ N m.”

What to write

“ $\tau = rF\sin\phi = 19.3$ N m. The component of the force that turns is the one PERPENDICULAR to the arm.”

Why: Both formulas look equally plausible on a blank page, so do not try to remember which one it is: test it. At $\phi = 90^{\circ}$ the force is perpendicular and the torque must be maximal; $\cos 90^{\circ} = 0$ would give nothing. The absurd extreme case settles it in one second, every time.

4. Believing torques must be taken about the real pivot

the question, or twenty minutes of algebra to reach the same number

What not to write

“I cannot work out the ladder problem: the two unknowns at the foot are both in the way.”

What to write

“Torques about the FOOT of the ladder: $N$ and $f$ both pass through that point, so both have zero lever arm and $N_{w}(4.0) = 117.6(1.5) + 686(2.4)$ gives $N_{w} = 456$ N directly.”

Why: For a body in equilibrium the net torque is zero about every axis, so the axis is yours to choose, and you choose the one that deletes an unknown. Ladder problems go about the foot, forearm problems about the elbow, carousel problems about the axle. One restriction only: once chosen, every torque in that equation is measured about that same axis.

5. Setting the tension equal to the weight over a pulley that has mass

4 marks, and the answer comes out four times too large

What not to write

“The block weighs $29.4$ N, so the cord pulls on the pulley with $29.4$ N and $\alpha = \frac{TR}{I} = 294$ rad/s $^2$ .”

What to write

“ $mg - T = ma$ and $TR = I\alpha$ with $a = R\alpha$ , hence $a = \frac{mg}{m + I/R^{2}} = 7.35$ m/s $^2$ and $T = m(g-a) = 7.35$ N.”

Why: If $T$ equalled $mg$ the block would not accelerate, and a block that does not accelerate does not turn the pulley. The statement contradicts itself. A pulley with mass is a third object in the problem, and it needs its own equation: that is the whole reason the chapter exists.

6. Giving a rolling body only its translational kinetic energy

3 marks, plus the physical conclusion of the question

What not to write

“ $mgh = \frac{1}{2}mv^{2}$ , so $v = \sqrt{2gh} = 4.85$ m/s for all three bodies.”

What to write

“ $mgh = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$ with $v = R\omega$ , so $v = \sqrt{\frac{2gh}{1+k}}$ : $3.43$ m/s for the hoop, $3.96$ for the disk, $4.10$ for the sphere.”

The same drop

h

pays for TWO things at once, the speed of the centre and the spin about it, which is why a rolling body is always slower than a sliding one.

Why: A rolling body stores a fixed fraction $\frac{k}{1+k}$ of its kinetic energy as spin: one half for a hoop, one third for a disk, two sevenths for a sphere. Dropping the term makes every rolling body arrive together, and at the speed of a frictionless slide, which the experiment flatly contradicts.

7. Reading the net torque as the applied torque

2 marks, and it hides the very quantity the question asks for

What not to write

“The motor supplies $\tau = I\alpha = 0.188$ N m.”

What to write

“ $\sum \tau = I\alpha = 0.188$ N m is the NET torque, so the motor supplies $\tau_{\text{motor}} = 0.188 + 0.050 = 0.238$ N m against the friction torque.”

Why: This is the rotational twin of confusing the tension with the weight, and it has the same cure: $I\alpha$ is the SUM of the torques, exactly as $ma$ is the sum of the forces. Whenever a problem names a friction torque, the applied torque and $I\alpha$ are two different numbers, and the question almost always asks for the first.

8. Conserving kinetic energy where only angular momentum survives

the whole question, and the equation looks perfectly correct

What not to write

“The carousel keeps its $52.5$ J, so $\frac{1}{2}(421.2)\omega'^{2} = 52.5$ and $\omega' = 0.50$ rad/s.”

What to write

“No external torque, so $L$ is conserved: $(291.6)(0.60) = (421.2)\omega'$ gives $\omega' = 0.415$ rad/s. The kinetic energy falls from $52.5$ J to $36.3$ J, lost under the child's shoes.”

Why: A body joining a rotating one is the rotational version of a perfectly inelastic collision: momentum survives, energy does not. And the reverse trap sits next to it, when the skater pulls her arms in: there $L$ is conserved and $K$ RISES, because her muscles did work. The rule to carry: $L$ is conserved when no external torque acts, $K$ only when nothing rubs and nothing pushes.

9. Reading a negative support force as an arithmetic slip

2 marks, and the conclusion the examiner was actually testing

What not to write

“I get $N_{A} = -19.6$ N, which is impossible, so I must have made a sign error somewhere.”

What to write

“ $N_{A} = -19.6$ N: the support would have to PULL down, which a plank simply resting on it cannot do. The plank has therefore already tipped about $B$ , and the model no longer applies.”

Why: A negative answer in statics is information, not an error. It says the direction you assumed is wrong, and for a one-sided contact, a support, a floor, a cable, that reversal means contact is lost. The tipping condition is exactly $N = 0$ , which is how the limiting distance is found.

10. Feeding revolutions per minute into a formula that expects radians

1 to 2 marks, and an answer that is absurd by a factor of nearly ten

What not to write

“ $v = r\omega = 0.12 \times 1800 = 216$ m/s.”

What to write

“ $\omega = 1800 \times \frac{2\pi}{60} = 188.5$ rad/s, so $v = 0.12 \times 188.5 = 22.6$ m/s.”

Why: $s = r\theta$ , $v = r\omega$ and $a_{t} = r\alpha$ are true only in radians, because the radian is defined as the arc divided by the radius. Every one of them needs the conversion, and the factor is $\frac{2\pi}{60} = 0.1047$ from rev/min to rad/s. The order-of-magnitude reflex catches it too: a bench grinder rim does not move at $780$ km/h.

Which method to choose

Which equation, from the SHAPE of the question

Do not choose by topic, choose by what the statement gives you and what it asks. Five shapes cover every question of this chapter.

If nothing moves, and a force or a reaction is asked → the two equilibrium conditions, $\sum \tau = 0$ FIRST

Example: ladder, plank on two supports, forearm, hinged sign

take the torques about the point where the unknown you do NOT want is applied
If something accelerates and a rope, a pulley or an axle turns → $\sum F = ma$ and $\sum \tau = I\alpha$ together, linked by $a = R\alpha$

Example: block on a cord over a massive pulley: $a = \frac{mg}{m + I/R^{2}} = 7.35$ m/s $^2$

one equation per object plus the rolling or unwinding constraint; three equations, three unknowns
If a speed is asked after a drop or a distance, and no time appears → energy, $mgh = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$

Example: a sphere rolling down $1.20$ m arrives at $4.10$ m/s

rolling without slipping means friction does no work, so mechanical energy is conserved
If two bodies join, separate, or change shape, with nothing twisting them from outside → $I_{1}\omega_{1} = I_{2}\omega_{2}$ , angular momentum conserved

Example: skater, rotating stool, child stepping onto a carousel

never conserve the kinetic energy here as well, it is the trap the question is built on
If the angular acceleration is constant and a time or a number of turns appears → the four rotational kinematics equations, twins of the linear ones

Example: $6000$ rev/min reached in $8.0$ s means $400$ revolutions

the fastest route is usually the average rate: $50$ rev/s for $8.0$ s is $400$ turns, with no formula at all

One question can need two branches in sequence, typically dynamics then energy. What never happens is two branches at once on the same unknown: if you find yourself writing an energy balance and a torque equation for the same number, one of them is wrong.

Which axis to take in a statics problem

The axis is free, so it is a tool. Run down this list and stop at the first line that applies.

If a hinge, a joint or an axle is drawn → take the axis there

Example: the elbow, because the joint force then has zero lever arm

the force at a hinge is unknown in size AND direction, so deleting it removes two unknowns at once
If two unknown forces act at the same point → take the axis at that point

Example: the foot of a ladder carries both $N$ and $f$ : one axis kills both
If the question asks when contact is lost or when the body tips → set the reaction to zero and take the axis at the OTHER support

Example: $294(1.0) = 588(d-3.0)$ gives $d = 3.5$ m
If you already have an answer and want to check it → redo the torque equation about a different axis

Example: the plank gives $N_{A} = 392$ N about $A$ and about $B$ alike

this is the only check in the chapter that catches a wrong lever arm, and it costs two lines

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 static equilibrium problem

When to use it: The statement says a body is at rest, in balance, or on the point of slipping, and asks for a force, a tension or a position.

1 Draw the body alone and put EVERY force on it, at its point of application, with a name.
2 Name the axis in writing, and say in half a line why: “I take torques about the elbow, where the joint force has zero lever arm.”
3 Write the torque equation with each lever arm shown as a product, $F \times d$ , before any arithmetic.
4 Solve for the single unknown that is left, then use $\sum F = 0$ for whatever else is asked.
5 Conclude with a sentence giving magnitude, unit and DIRECTION, and read the sign rather than deleting it.

Concluding sentence

“Taking torques about the hinge, where the hinge force has zero lever arm: $T(2.0\sin 36.9^{\circ}) = (78.4)(1.0) + (117.6)(1.6)$ , so $T = 222$ N.”

The trap: The step that gets skipped is the second, and it is the one that carries the marks. A torque equation whose axis is not named cannot be credited, because the marker has no way to tell a right lever arm from a wrong one.

Marking: typically 1 mark for the labelled diagram, 1 for naming the axis, 3 for the torque equation with correct lever arms, 2 for the force equations, 1 for a conclusion with a direction.

Solving a rolling or an angular momentum problem

When to use it: The statement asks for a speed after a drop, or for the new rate after something joins, leaves or changes shape.

1 Say which quantity is conserved and WHY, in one sentence naming what is absent: no friction that slides, or no external torque.
2 Write $I$ for each body from the formulary, with the axis named, and show the $kmR^{2}$ form.
3 Write the conservation line in full symbols before substituting a single number.
4 Substitute, keeping $v = R\omega$ visible wherever a body rolls.
5 Check the kinetic energy separately, and say where any difference went.

Concluding sentence

“No external torque acts about the axle, so $L$ is conserved: $(291.6)(0.60) = (291.6 + 40 \times 1.8^{2})\,\omega'$ , giving $\omega' = 0.415$ rad/s.”

The trap: Writing the conservation line without the justification loses the mark even when the number is right, and it is a one-sentence mark. The second trap is to conserve energy as well, in a situation where the question is built on it not being conserved.

Check before you hand in

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

Redo the torques about a second axis

Take any other point, redo the torque equation, and see whether the same number comes out. For a body in equilibrium it must, and this is the only check that catches a lever arm measured to the wrong place.

The plank gives $N_{A} = 392$ N about $A$ , through $N_{B}$ ; about $B$ it gives $N_{A}(3.0) = 294(1.0) + 588(1.5) = 1176$ , hence $392$ N again. Two lines, and the whole question is secured.
Push the angle to its extremes

Before trusting a torque formula, set the angle to $0^{\circ}$ and to $90^{\circ}$ . A force aimed at the axis must give zero, a force perpendicular to the arm must give the maximum. Only $\sin$ does both.

$rF\cos\phi$ would give $30$ N m for a pull straight along the wrench handle, which no one has ever managed to loosen a bolt with.
Compare a rolling speed with the free slide

Any body that rolls must be SLOWER than the same body sliding without friction down the same drop. If your answer matches $\sqrt{2gh}$ or beats it, the $\frac{1}{2}I\omega^{2}$ term has gone missing.

From $h = 1.20$ m the ceiling is $\sqrt{2(9.8)(1.20)} = 4.85$ m/s. A sphere gives $4.10$ , a disk $3.96$ , a hoop $3.43$ : all below, and in the right order.
Check the units, and never write a joule for a torque

A torque is in newton metres, an energy in joules, and the two have the same dimensions without being the same thing. Check as well that every $\omega$ in a formula is in rad/s and not in rev/min, and that $I$ is in kg m $^2$ .

$\tau = 0.238$ N m and $K = 474$ J in the same problem, and writing $0.238$ J would cost the mark even though the arithmetic was right.
Read the sign before you erase it

A negative torque means the other sense of rotation. A negative support force means contact has been lost and the body has tipped. Neither is an arithmetic mistake, and both are usually the point of the question.

$N_{A} = -19.6$ N on the plank says the left end has lifted. Bolt it down and the same $19.6$ N becomes a genuine tension in the bolt.

The typical problem, taken apart

A shop sign on a hinged beam

A uniform beam $2.0$ m long and of mass $8.0$ kg is hinged to a wall at $O$ and held horizontal by a cable running from its far end to a point on the wall $1.5$ m above the hinge.

A sign of mass $12$ kg hangs from the beam $1.6$ m from the wall. Take $g = 9.8$ m/s $^2$ .

Find the tension in the cable, then the force the hinge exerts on the beam, in magnitude and direction.

The cable meets the beam at

36.9^{\circ}

, so its lever arm about the hinge is

2.0\sin 36.9^{\circ} = 1.2

m and not

2.0

m: that factor

0.6

is the whole difficulty of the problem.

Step 1

Axis at the hinge $O$ . The hinge force is unknown in magnitude AND direction, but its line passes through $O$ , so its lever arm there is zero.

Why

Two unknowns are deleted by one choice. Any other axis would leave three unknowns in one equation, which no amount of algebra recovers. This single line is worth a mark on its own on most marking schemes.

Step 2

Geometry: $\tan\phi = \frac{1.5}{2.0}$ , so $\phi = 36.9^{\circ}$ , $\sin\phi = 0.6$ and $\cos\phi = 0.8$ . The lever arm of the cable is $d = 2.0 \times 0.6 = 1.2$ m.

Why

The lever arm is computed before the physics, because it is a geometry question. Note the $3$ , $4$ , $5$ triangle hiding in $1.5$ and $2.0$ : the sine is exactly $0.6$ , so no decimals are lost in the next line.

Step 3

Torques about $O$ : $T(1.2) = (78.4)(1.0) + (117.6)(1.6) = 78.4 + 188.2 = 266.6$ , so $T = \frac{266.6}{1.2} = 222$ N.

Why

The beam weight acts at its centre of gravity, $1.0$ m from the hinge, because the beam is uniform. The sign acts where it hangs. Both weights are vertical, so their lever arms are plain horizontal distances, which is why the figure is drawn horizontal in the first place.

Step 4

Forces, horizontal: the cable pulls the beam toward the wall with $T\cos\phi = 222.1 \times 0.8 = 178$ N, so the hinge must push outward with $H = 178$ N.

Why

The torque equation is exhausted, so the force condition takes over. Note that the tension exceeds the total weight of $196$ N: a shallow cable is always under more tension than the load it carries, which is why sign brackets are anchored high.

Step 5

Forces, vertical: $V + T\sin\phi = 78.4 + 117.6$ , so $V = 196 - 222.1 \times 0.6 = 196 - 133.3 = 62.7$ N upward.

Why

The cable already carries $133$ N of the $196$ N of weight, so the hinge only supplies the rest. Had the cable been steeper, $V$ would have turned negative, meaning the hinge pulls DOWN, exactly as the plank's support did.

Step 6

Hinge force: $\sqrt{178^{2} + 62.7^{2}} = 188$ N, at $\arctan\frac{62.7}{178} = 19.4^{\circ}$ above the horizontal.

Why

A hinge force is a vector and the question asks for a direction, so it gets one. Answering with $H$ and $V$ separately is half an answer and usually half the marks.

Step 7

Check, torques about the FAR END of the beam: the cable and the horizontal hinge force both have zero lever arm there, so $V(2.0) = (78.4)(1.0) + (117.6)(0.4) = 125.4$ , hence $V = 62.7$ N.

Why

A second axis, an independent equation, the same number. This is the two-line check that protects the whole problem, and it is the reason the freedom of choosing the axis matters twice over.

The conclusion, written out

“The cable carries a tension of $222$ N, and the hinge pushes on the beam with $188$ N directed outward and $19.4^{\circ}$ above the horizontal.”

The classic mistake on this problem: The classic loss here is taking the lever arm of the cable as $2.0$ m instead of $1.2$ m, which returns $T = 133$ N, comfortably less than the weight it is holding. The absurdity is visible without any further calculation, and reading the answer against the weight of $196$ N catches it in a second.

Learn by heart

• $\tau = rF\sin\phi = d\,F$ , where $d$ is the perpendicular distance from the axis to the LINE of the force. In newton metres, never joules.
• Equilibrium is TWO conditions: $\sum \vec{F} = \vec{0}$ and $\sum \tau = 0$ . The second holds about any axis you choose, so choose the one that deletes an unknown.
• Formulary: hoop $mR^{2}$ , disk $\frac{1}{2}mR^{2}$ , sphere $\frac{2}{5}mR^{2}$ , rod about its centre $\frac{1}{12}mL^{2}$ , rod about its end $\frac{1}{3}mL^{2}$ , point mass $mR^{2}$ . Parallel axis: $I = I_{\text{cm}} + Md^{2}$ .
• $\sum \tau = I\alpha$ , with the NET torque on the left, motor minus friction.
• $K = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$ , and rolling without slipping means $v = R\omega$ with friction doing no work.
• Rolling down a drop: $v = \sqrt{\frac{2gh}{1+k}}$ with $I = kmR^{2}$ . Neither $m$ nor $R$ appears, and a rolling body is always slower than a sliding one.
• $L = I\omega$ is conserved when no external torque acts. The kinetic energy is not: $K = \frac{L^{2}}{2I}$ , so halving $I$ doubles $K$ .
• $s = r\theta$ , $v = r\omega$ and $a_{t} = r\alpha$ need RADIANS. From rev/min to rad/s, multiply by $\frac{2\pi}{60} = 0.1047$ .
• A point on a spinning body has $a_{t} = r\alpha$ AND $a_{c} = r\omega^{2}$ , perpendicular to each other. At constant speed $a_{t} = 0$ and $a_{c}$ is still there.

Frequently asked questions

How do you work out the torque of a force in PHYS 101?

Multiply the size of the force by its lever arm, which is the perpendicular distance from the axis to the line along which the force acts, extended as far as you need. If you know the distance from the axis to the point where the force is applied, multiply that distance by the force and then by the sine of the angle between them. A force pointing straight at the axis has no lever arm at all, so it produces no torque however large it is.

Which point should I take as the axis in a statics problem?

Any point you like, because for a body that is not moving the torques cancel about every possible axis. Use that freedom: take the axis at the point where an unknown force acts, since its lever arm there is zero and it drops out of the equation. That is why ladder problems are worked about the foot, forearm problems about the elbow, and hinged beam problems about the hinge. Once you have chosen, measure every torque in that equation about the same point.

Why does a solid disk beat a hoop down a ramp?

Both start with the same energy from the same drop, and both have to spend part of it on spinning rather than on moving forward. The hoop carries all of its mass at the rim, as far from the axis as possible, so at a given speed it needs more energy to spin, and half of everything it has goes into rotation. The disk keeps much of its mass near the axis and only pays a third. The mass and the radius of each body cancel out completely, so a large hoop and a small one finish together.

Do I need calculus for the rotation chapter of PHYS 101?

No, and that is the whole point of this course as opposed to the calculus-based one. Every moment of inertia you need is given to you in a formulary rather than worked out by integration, and every problem is solved with algebra, ratios and the trigonometry of a right triangle. If a solution you find online starts by differentiating a position or integrating a force, it is written for another course and you cannot hand it in, even if its final number happens to be right.

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

Nothing twists her from outside, so her angular momentum, the product of her moment of inertia and her rate of turning, cannot change. Pulling her arms in brings mass closer to the axis, which cuts her moment of inertia, so her rate of turning has to rise in exactly the same proportion to keep the product fixed. Her kinetic energy rises as well, and that is not free: her muscles do the work of pulling her arms inward against the outward pull she feels while spinning.

What is the difference between a torque in newton metres and an energy in joules?

They have the same units written out, a newton multiplied by a metre, and they are still completely different quantities. A torque is a force multiplied by a distance measured at right angles to it, and it makes things turn. An energy is a force multiplied by a distance measured along it, and it makes things change speed. Because of that, a torque is always written in newton metres and never in joules, and a marker will take the mark off if you swap them.

Practise it

Corrected exercises: Torque, rotation and equilibrium, PHYS 101 at McGill

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

Previous sheet Momentum, impulse and collisions Next sheet Oscillations and simple harmonic motion

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Get in touch for a first session. On rotation we start by drawing the lever arms and choosing the axis, because that is where the marks are won or lost, well before the calculator comes out.

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Revision sheet: torque, rotation and equilibrium (PHYS 101, McGill)

The essentials

Torque: it is a geometry question before it is a physics one

The two conditions of static equilibrium, and the axis you may choose

Moment of inertia: mass counts once, distance counts twice

Energy and angular momentum in rotation

The rules in table form

The translation dictionary, with a number under every line

The mistakes that cost marks

1. Adding up the forces when the question asks about turning

2. Measuring the lever arm to the point of application

3. Writing τ=rFcos⁡ϕ\tau = rF\cos\phiτ=rFcosϕ

4. Believing torques must be taken about the real pivot

5. Setting the tension equal to the weight over a pulley that has mass

6. Giving a rolling body only its translational kinetic energy

7. Reading the net torque as the applied torque

8. Conserving kinetic energy where only angular momentum survives

9. Reading a negative support force as an arithmetic slip

10. Feeding revolutions per minute into a formula that expects radians

Which method to choose

Which equation, from the SHAPE of the question

Which axis to take in a statics problem

How the answer is expected to be written

Solving a static equilibrium problem

Solving a rolling or an angular momentum problem

Check before you hand in

The typical problem, taken apart

A shop sign on a hinged beam

Learn by heart

Frequently asked questions

How do you work out the torque of a force in PHYS 101?

Which point should I take as the axis in a statics problem?

Why does a solid disk beat a hoop down a ramp?

Do I need calculus for the rotation chapter of PHYS 101?

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

What is the difference between a torque in newton metres and an energy in joules?

Corrected exercises: Torque, rotation and equilibrium, PHYS 101 at McGill

See also

Looking for a PHYS 101 tutor in Montreal?

3. Writing $\tau = rF\cos\phi$