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Puzzled.

P5 Rotation and spin

apply torque, moment of inertia and angular momentum to spinning, rolling and balancing objects.

Before you begin: P3 · Energy and momentum, M3c · Vectors and components.

Videos coming soon

Four videos (part 1: turning motion and torque; part 2: moment of inertia and rolling; part 3: angular momentum; part 4: gyroscopes and balance) and a worked-example video (who wins the race down a slope?) are made, and will appear here once they're uploaded to YouTube.

129 steps about 114 min

Watch first

Notes available, video soon.

Notes

In this chapter

  • Describe a spin with \(\theta\), \(\omega\) and \(\alpha\), and use the constant-acceleration equations with new letters.
  • Calculate a torque, \(\tau = rF\sin\theta\), find its direction with the right-hand rule, and write it as a cross product \(\vec r \times \vec F\).
  • Use the moment of inertia, \(I = \sum mr^2\), and \(\tau = I\alpha\); know \(I\) for a ring, a disc and a sphere.
  • Find a centre of mass, and treat rolling as moving along plus spinning, with energy \(\tfrac12 Mv^2(1+k)\).
  • Use angular momentum, \(L = I\omega\), and its conservation: skaters, cats, Kepler's second law and pulsars.
  • Explain why a spinning top precesses, \(\Omega = mgr/(I\omega)\), and why gyroscopes are not what keeps a bicycle up.
  • Balance forces and torques to find the forces on beams, cables, cranes and ladders.

Describing a spin

FIRST, MAKE A PREDICTION

Two points on the same rigid wheel are at different radii. Which quantity must be the same for both during a turn?

Show the answer

Angular displacement

All points rotate through the same angle, even though their arc lengths differ with radius.

TRY IT / FOLLOW THE REASONING

Mark two points on one spoke and rotate the wheel through a quarter-turn. Label the shared angle and the two different arc lengths, then relate angle, angular speed and angular acceleration.

MAKE THE CONNECTION

Key idea

Spinning is moving in a line with new letters: distance becomes angle, speed becomes angular velocity, acceleration becomes angular acceleration.

A spinning thing is described by three quantities:

  • Angle \(\theta\), in radians (anticlockwise is positive). One full turn is \(2\pi\) rad.
  • Angular velocity \(\omega = \Delta\theta/\Delta t\), in rad/s. \(1\ \text{rev/min} = 2\pi/60\) rad/s. In calculus language, \(\omega\) is the rate of change of \(\theta\).
  • Angular acceleration \(\alpha = \Delta\omega/\Delta t\), in rad/s².

The equations of constant acceleration carry over letter for letter:

Describing a spin

moving in a line turning
\(x,\ v,\ a\) \(\theta,\ \omega,\ \alpha\)
\(v = u + at\) \(\omega = \omega_0 + \alpha t\)
\(x = ut + \tfrac12 at^2\) \(\theta = \omega_0 t + \tfrac12 \alpha t^2\)

A point at distance \(r\) from the axis moves along its circle at speed \(v = r\omega\), and speeds up along the circle at \(a = r\alpha\).

Describing a spin

Worked number: the merry-go-round and the record

On a merry-go-round everyone has the same \(\omega\), but the speed grows with \(r\): at \(\omega = 2\) rad/s, a rider 3 m out moves at 6 m/s and one 1 m out at 2 m/s.

A record at \(33\tfrac13\) rev/min has \(\omega = 3.49\) rad/s, so a point 15 cm from the centre moves at 0.52 m/s.

Checkpoint: Describing a spin (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A turntable starts from rest and speeds up at \(3.0\ \text{rad/s}^2\) for 4.0 s.

Its angular velocity after 4.0 s, in rad/s.

rad/s

The angle it has turned through in that time, in rad.

rad

Show the answer
  1. 12 rad/s
  2. 24 rad

From rest, \(\omega = \alpha t = 3.0\times4.0 = 12\) rad/s and \(\theta = \tfrac12\alpha t^2 = \tfrac12\times3.0\times16 = 24\) rad, about 3.8 turns.

Checkpoint: Describing a spin (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Two children ride a merry-go-round, one 1 m from the centre and one 3 m from it. Which statement is true?

Show the answer

They have the same \(\omega\), but the outer child moves three times as fast.

All the riders turn through the same angle in the same time, so \(\omega\) is shared. The speed is \(v = r\omega\), so the child at 3 m moves three times as fast as the one at 1 m.

Checkpoint: Describing a spin (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A bicycle wheel of radius 0.35 m turns at 20 rad/s. How fast does a point on its rim move, in metres per second?

m/s

Show the answer

7 m/s

\(v = r\omega = 0.35\times20 = 7.0\) m/s, about 25 km/h.

Torque

FIRST, MAKE A PREDICTION

You push a door with the same perpendicular force, once near the hinge and once near the handle. Which gives the larger turning effect?

Show the answer

Pushing near the handle

The torque magnitude is force times perpendicular lever arm, so a larger distance from the hinge gives more torque.

TRY IT / FOLLOW THE REASONING

Keep the force at 90° and increase the lever length. Then keep the length fixed and turn the force towards the lever, watching the torque fall.

Find the turning effect

Distance from pivot: 1 m · Angle of force: 90 °

Torque magnitude
10 N m
Perpendicular lever arm
1 m
Model assumptions

A 10 N force acts on a light lever. The angle is measured between the lever and the force.

MAKE THE CONNECTION

Key idea

A push turns something according to how hard you push and how far from the pivot you push.

The torque is

Torque

\[ \tau = r F \sin\theta \]
  • \(r\): the distance from the pivot to where the force acts.
  • \(F\): the force.
  • \(\theta\): the angle between the line from the pivot and the force.

Only the part of the force at a right angle to \(r\) turns anything. The lever arm is the shortest distance from the pivot to the line of the force, \(r\sin\theta\), so \(\tau = F \times \text{lever arm}\).

Torque

A 0.25 m spanner pivoting at its left end, pulled at its far end by 80 N at 60 degrees to the spanner. The shortest distance from the pivot to the line of the force, the lever arm, is 0.217 m, so the torque is 17.3 N m.θ = 60°pivotr = 0.25 mF = 80 Nlever arm 0.217 mτ = 17.3 N m
The same torque two ways: the part of F at a right angle to the spanner is F sin θ, or equivalently F times the lever arm r sin θ, the shortest distance from the pivot to the dashed line of the force. Here a 0.25 m spanner pulled with 80 N at 60° has a lever arm of 0.217 m, so the torque is 17.3 N m.

Torque

Torque is measured in newton-metres (N m). This looks like a joule, but torque is a turning effect, not an amount of energy, so write energy in J and torque in N m.

A force produces no torque in two cases:

  • it acts at the pivot (\(r = 0\));
  • it points along the line to the pivot (\(\sin\theta = 0\)).

Torque

Worked number: the spanner

A 0.25 m spanner pulled with 80 N at 60° gives \(0.25 \times 80 \times 0.866 = 17.3\) N m; at 90° it would be 20 N m.

Torque

Balance and the net torque. Take anticlockwise torques as positive. A non-zero net torque gives an angular acceleration, just as a net force gives an acceleration.

Torque

Worked number: the seesaw

A seesaw balances when the torques cancel: 60 kg at 1.5 m balances 45 kg at 2 m, because \(60 \times 1.5 = 45 \times 2 = 90\).

Torque

Myth

"A lever gives you energy for free." A lever trades force for distance: put the load one fifth as far from the pivot as your hand, and 100 N lifts 500 N, but your hand moves five times as far. Work in: \(100 \times 0.5 = 50\) J. Work out: \(500 \times 0.1 = 50\) J.

Checkpoint: Torque (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

You push a door with 30 N at a right angle to it, 0.80 m from the hinge. What torque do you apply about the hinge, in newton-metres?

N m

Show the answer

24 N m

\(\tau = rF\sin90^\circ = 0.80\times30 = 24\) N m.

Checkpoint: Torque (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A force acts on a nut with a spanner. In which case does it produce no torque about the nut?

Show the answer

It is pulled along the line from the nut to where it acts, straight towards the nut.

The torque is \(rF\sin\theta\), which is zero when \(r = 0\) or \(\sin\theta = 0\): a force along the line to the pivot has no lever arm.

Checkpoint: Torque (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A crowbar has its pivot 0.20 m from the load and your hand 1.0 m from the pivot. You push down with 150 N. What load can it just lift, in newtons?

N

Show the answer

750 N

Balance the torques: \(\text{load}\times0.20 = 150\times1.0\), so the load is 750 N. The hand moves five times as far as the load, so no energy is gained.

The cross product and the right-hand rule

FIRST, MAKE A PREDICTION

A force acts exactly along a lever, pointing away from its pivot. What is the torque about that pivot?

Show the answer

Zero

The force's line of action passes through the pivot, so its perpendicular lever arm and rF sin θ are zero.

TRY IT / FOLLOW THE REASONING

Use the model to compare force angles of 0°, 90° and 180°. For a nonzero torque, use curled fingers to identify the direction along the rotation axis.

Find the turning effect

Distance from pivot: 1 m · Angle of force: 90 °

Torque magnitude
10 N m
Perpendicular lever arm
1 m
Model assumptions

A 10 N force acts on a light lever. The angle is measured between the lever and the force.

MAKE THE CONNECTION

Key idea

A spin needs one arrow, along its axis. Right-hand rule: curl the fingers of your right hand the way it turns; your thumb gives the direction.

This is used for \(\vec\omega\), \(\vec\alpha\), \(\vec\tau\) and \(\vec L\). The torque is the cross product

The cross product and the right-hand rule

\[ \begin{gathered} \vec\tau = \vec r \times \vec F \\[.6em] |\vec\tau| = rF\sin\theta \end{gathered} \]
  • Direction: at a right angle to both \(\vec r\) and \(\vec F\). Point the fingers along \(\vec r\), curl them towards \(\vec F\), and the thumb gives the torque.
  • Dot versus cross: the dot product gives a number (using cosine); the cross product gives a vector (using sine).
  • Order matters: \(\vec F \times \vec r = -\,\vec r \times \vec F\).
  • On the page: a vector out of the page is drawn \(\odot\) and into the page \(\otimes\).
  • In two dimensions, with \(\vec r = (x, y)\) and \(\vec F = (F_x, F_y)\), the torque is \(\tau_z = xF_y - yF_x\), positive if anticlockwise.

Checkpoint: The cross product and the right-hand rule (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A force with components \(F_x = 5\) N and \(F_y = 10\) N acts at the point \((0.4, 0.3)\) m from the pivot. What is \(\tau_z\), in newton-metres?

N m

Show the answer

2.5 N m

\(\tau_z = 0.4\times10 - 0.3\times5 = 4.0 - 1.5 = 2.5\) N m. It is positive, so the torque is anticlockwise, out of the page.

Checkpoint: The cross product and the right-hand rule (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

If \(\vec r\times\vec F\) points out of the page, what does \(\vec F\times\vec r\) do?

Show the answer

It points into the page, with the same size.

Order matters: \(\vec F\times\vec r = -\,\vec r\times\vec F\). The size \(rF\sin\theta\) is unchanged, but the direction reverses.

Moment of inertia

FIRST, MAKE A PREDICTION

A point mass is moved to twice its distance from a fixed rotation axis. What happens to its moment of inertia?

Show the answer

It becomes four times as large

For a point mass, I = mr², so doubling radius gives a factor of four.

TRY IT / FOLLOW THE REASONING

Keep the model's mass fixed while changing the distance from the axis. Compare doubling mass with doubling radius and identify which has the larger effect.

Move a mass away from the axis

Distance from axis: 1 m · Mass: 2 kg

Moment of inertia
2 kg m²
Model assumptions

A single point mass rotating about a fixed axis. The connecting rod has negligible mass.

MAKE THE CONNECTION

Key idea

The moment of inertia \(I\) is how hard something is to spin up. Mass far from the axis counts for much more, because distance is squared.

For a small mass \(m\) at the end of a light rod of length \(r\), a force \(F\) at a right angle gives \(F = m(r\alpha)\), and the torque \(rF = mr^2\alpha\). So

Moment of inertia

\[ \begin{gathered} \tau = I\alpha \\[.6em] I = \sum m r^2 \end{gathered} \]

Moment of inertia

Worked number: twice as far

Twice as far from the axis means four times as hard to spin up. Two masses of 1 kg at 0.5 m give \(I = 0.5\) kg m²; at 1 m, 2 kg m².

Moment of inertia

\(I\) depends on the axis. For a few common shapes about their central axis:

Moment of inertia

shape (about its central axis) \(I\)
thin ring (hoop) \(MR^2\)
solid disc or cylinder \(\tfrac12 MR^2\)
solid sphere \(\tfrac25 MR^2\)
thin rod, about its middle / about its end \(\tfrac1{12}ML^2\) / \(\tfrac13 ML^2\)

Moment of inertia

Shapes of the same mass and radius, each with a bar as long as its k in I = kMR²: thin ring, k = 1; solid disc, k = 0.5; solid sphere, k = 0.4.Rthin ringI = 1 MR²Rsolid discI = ½ MR²Rsolid sphereI = ⅖ MR²
Same mass, same radius, different spread of mass. The ring has all its mass at the rim, so it is hardest to spin; the sphere keeps more of its mass near the axis. The bars show k in I = kMR²: k = 1, ½ and ⅖ for the ring, disc and sphere.

Moment of inertia

Write \(I = kMR^2\), with \(k = 1,\ \tfrac12,\ \tfrac25\). With the same mass and radius, a ring needs twice the torque of a disc for the same \(\alpha\).

Moment of inertia

Optional: where the disc's ½ comes from

Slicing a disc into thin rings of mass \(\sigma\,2\pi r\,dr\), \(I = \int_0^R 2\pi\sigma r^3\,dr = \tfrac12\pi\sigma R^4 = \tfrac12 MR^2\).

Checkpoint: Moment of inertia (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

Two 2.0 kg masses sit 0.30 m from an axis on opposite sides of a light rod. What is the moment of inertia about the axis, in kg m²?

kg m^2

Show the answer

0.36 kg m^2

\(I = \sum mr^2 = 2\times2.0\times0.30^2 = 0.36\) kg m².

Checkpoint: Moment of inertia (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A solid disc has mass 4.0 kg and radius 0.50 m. What is its moment of inertia about its central axis, in kg m²?

kg m^2

Show the answer

0.5 kg m^2

\(I = \tfrac12MR^2 = \tfrac12\times4.0\times0.25 = 0.50\) kg m².

Checkpoint: Moment of inertia (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

The same mass is moved to twice the distance from an axis. By what factor does its moment of inertia change?

Show the answer

It becomes 4 times as large.

Since \(I = mr^2\), doubling \(r\) multiplies \(I\) by \(2^2 = 4\).

Centre of mass

FIRST, MAKE A PREDICTION

A light rod joins a 1 kg mass and a 3 kg mass. Where is their centre of mass?

Show the answer

Closer to the 3 kg mass

The centre is a mass-weighted average; the larger mass pulls the average position towards itself.

TRY IT / FOLLOW THE REASONING

Place the masses at x = 0 and x = 4 m. Calculate the weighted-average position, then compare the two opposing moments about that position.

MAKE THE CONNECTION

Key idea

The centre of mass is the mass-weighted average position, \(x_{cm} = \sum mx / \sum m\).

  • A free object's centre of mass moves as if all the mass were there (a thrown spanner's centre follows a parabola) while the object spins about it.
  • Something tips over when the line down from its centre of mass leaves its base.

Centre of mass

Worked number

Masses of 2 kg at 0 and 6 kg at 1 m: \(x_{cm} = 6/8 = 0.75\) m.

Checkpoint: Centre of mass (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 3.0 kg mass sits at \(x = 0\) and a 1.0 kg mass at \(x = 2.0\) m. Where is their centre of mass, in metres?

m

Show the answer

0.5 m

\(x_{cm} = (3\times0 + 1\times2)/(3+1) = 0.50\) m, three times closer to the heavier mass.

Checkpoint: Centre of mass (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A tall box on a slope is slowly tilted. When does it tip over?

Show the answer

When the line down from its centre of mass leaves its base.

It tips when the vertical line through the centre of mass falls outside the base, so the weight then turns it over instead of back.

Rolling

FIRST, MAKE A PREDICTION

A wheel rolls without slipping on stationary ground. What is the instantaneous speed of its point of contact with the ground?

Show the answer

Zero

At contact, the centre's forward motion and the rotational backward motion cancel.

TRY IT / FOLLOW THE REASONING

Draw the centre, top and contact point of a rolling wheel. Add the forward velocity of the centre to each point's velocity relative to the centre.

MAKE THE CONNECTION

Key idea

Rolling is moving along plus spinning. The contact point is momentarily at rest, so the centre moves at \(v = \omega R\) and the top of the wheel at \(2v\).

Rolling

A wheel of radius 0.3 m rolling without slipping at 2 m/s. The point touching the ground is at rest, the centre moves at 2 m/s and the top at 4 m/s; the wheel turns at 6.7 rad/s.v = 2 m/s2v = 4 m/scontact point: at restω = v/R = 6.7 rad/s
Rolling without slipping: the point touching the ground is instantaneously still, the centre moves at v and the top at 2v. For a wheel of radius 0.30 m rolling at 2.0 m/s, the top is moving at 4.0 m/s and the wheel turns at 6.7 rad/s.

Rolling

Worked number: the wheel

A wheel of radius 0.30 m rolling at 2.0 m/s turns at \(\omega = v/R = 6.7\) rad/s. Its centre moves at 2.0 m/s, its top at 4.0 m/s, and its contact point at 0.

Rolling

A rolling object has two kinetic energies, \(\tfrac12 Mv^2\) for moving along and \(\tfrac12 I\omega^2 = \tfrac12 kMv^2\) for spinning, giving

Rolling

\[ KE = \tfrac12 Mv^2\,(1 + k) \]

Down a slope of height \(h\):

  • \(Mgh = \tfrac12 Mv^2(1+k)\), so \(v = \sqrt{2gh/(1+k)}\).
  • The acceleration along a slope at angle \(\theta\) is \(a = g\sin\theta/(1+k)\).
  • The mass and the radius cancel: only the shape matters, so a sphere beats a disc, which beats a ring, and a solid cylinder beats a hollow tube.
  • The spin takes \(k/(1+k)\) of the energy: 50% for a ring, 33% for a disc, 29% for a sphere.

Rolling

Myth

"Heavier things roll down a hill faster." The mass cancels, just as it does for falling.

Checkpoint: Rolling (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A wheel rolls without slipping with its centre moving at 5.0 m/s. How fast is the top of the wheel moving, in metres per second?

m/s

Show the answer

10 m/s

The centre moves at \(v\), and the spin adds another \(\omega R = v\) at the top, so the top moves at \(2v = 10\) m/s.

Checkpoint: Rolling (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A thin ring and a solid disc, with different masses and radii, are released together at the top of a slope and roll without slipping. Which arrives first?

Show the answer

The solid disc, because \(k = \tfrac12\) is smaller than the ring's \(k = 1\).

Mass and radius cancel, so only the shape counts. A disc has \(k = \tfrac12\) and a ring \(k = 1\), so the disc accelerates faster.

Checkpoint: Rolling (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

What percentage of the kinetic energy of a solid sphere rolling without slipping is in its spin?

Show the answer

28.5714285714

\(k/(1+k) = 0.4/1.4 = 0.286\), so about 29% of the energy is spin and 71% is the sphere moving along.

Angular momentum

FIRST, MAKE A PREDICTION

A skater pulls their arms inward with negligible external torque. What happens to spin rate?

Show the answer

It increases

Reducing moment of inertia while conserving angular momentum Iω requires a larger angular speed.

TRY IT / FOLLOW THE REASONING

Use the point-mass model to compare moments of inertia at radii 2 and 1 m. If angular momentum is fixed, calculate the required ratio of angular speeds; identify the work involved in pulling inward.

Move a mass away from the axis

Distance from axis: 1 m · Mass: 2 kg

Moment of inertia
2 kg m²
Model assumptions

A single point mass rotating about a fixed axis. The connecting rod has negligible mass.

MAKE THE CONNECTION

Key idea

Angular momentum is \(L = I\omega\) (kg m²/s), a vector along the axis. With no net external torque it is conserved, in size and direction.

The net torque is its rate of change, \(\tau = \Delta L/\Delta t\) (the twin of \(F = \Delta p/\Delta t\)). It is not only for spinning things: a mass moving in a straight line has \(\vec L = \vec r \times \vec p\), size \(mvd\) about a point at perpendicular distance \(d\) from its line.

Angular momentum

Worked number: the skater

A skater with \(I = 3.0\) kg m² turning once a second (\(L = 18.8\)) who pulls her arms in to \(I = 1.0\) spins three times a second.

Her kinetic energy \(\tfrac12 I\omega^2\) rises from 59 J to 178 J: \(L\) is conserved but energy is not, because her muscles do work.

Angular momentum

Where conservation of angular momentum shows up:

  • Cats and astronauts. With \(L = 0\) throughout, you can still turn over by changing shape: tuck one half (small \(I\), it turns a lot) and extend the other (large \(I\), it turns a little the other way), then swap.
  • Kepler's second law. The area swept per second is \(L/2m\); a force aimed at the Sun has no torque about it, so \(L\) and the area rate are constant. At the closest and farthest points, \(r_1v_1 = r_2v_2\).
  • Collapsing stars. \(I \propto R^2\), so shrinking a spinning core by a factor of 500 in radius speeds it up \(250\,000\)-fold: a pulsar.
  • Direction counts. A spinning wheel flipped upside down on a stool gives the person and stool the opposite change of \(\vec L\).

Checkpoint: Angular momentum (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A diver leaves the board with \(I = 15\ \text{kg m}^2\) and spins at 2.0 rad/s. She tucks, so \(I\) falls to \(6.0\ \text{kg m}^2\). What is her angular velocity now, in rad/s?

rad/s

Show the answer

5 rad/s

\(L = I\omega = 15\times2.0 = 30\) kg m²/s is conserved, so \(\omega = 30/6.0 = 5.0\) rad/s.

Checkpoint: Angular momentum (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A falling cat starts with no spin, yet it lands on its feet. Write a sentence or two on how it can turn over without breaking the conservation of angular momentum.

Show the solution

The total \(L\) stays zero, but the two halves can turn in opposite directions. The cat tucks one half (small \(I\), which turns a lot) while extending the other (large \(I\), which turns a little the other way), then swaps, and ends up facing the other way.

Checkpoint: Angular momentum (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A 2.0 kg ball moves at 3.0 m/s along a straight line that passes 4.0 m from a point. What is its angular momentum about that point, in kg m²/s?

kg m^2/s

Show the answer

24 kg m^2/s

\(L = mvd = 2.0\times3.0\times4.0 = 24\) kg m²/s. It stays 24 all along the line, because the distance \(d\) does not change.

Precession and gyroscopes

FIRST, MAKE A PREDICTION

A small torque acts perpendicular to a spinning object's angular momentum. What does it initially change?

Show the answer

Primarily the direction of angular momentum

A small perpendicular change rotates the angular-momentum vector to first order rather than increasing its magnitude along itself.

TRY IT / FOLLOW THE REASONING

Draw angular momentum and a small perpendicular change vector tip to tail. Compare the resulting direction with the original, then connect successive changes to a precessing axis.

MAKE THE CONNECTION

Key idea

A torque at a right angle to \(\vec L\) changes the direction of \(\vec L\), not its size. So a tilted top circles instead of toppling.

A tilted top has a gravity torque \(\vec r \times m\vec g\) that is horizontal, at a right angle to \(\vec L\). It changes \(\vec L\) sideways, so the axis circles: precession,

Precession and gyroscopes

\[ \begin{gathered} \Omega = \frac{mgr}{I\omega} \\[.6em] \text{(for a fast spin)} \end{gathered} \]

Precession and gyroscopes

A tilted spinning top on the ground with its angular momentum L along its axis and its weight mg pulling down at its centre of mass. The weight gives a sideways torque, so the tip of the axis circles the vertical at Ω = 4.9 rad/s while the top spins at 100 rad/s.Lweight mgΩ = 4.9 rad/sspin ω = 100 rad/s
A tilted spinning top (not to scale). Its weight mg gives a sideways torque at a right angle to L, so the axis swings round the vertical at 4.9 rad/s, the toy top's rate below.

Precession and gyroscopes

The toy top's precession rate Ω = mgr/(Iω) against its spin ω. At ω = 100 rad/s it precesses at 4.9 rad/s; at twice the spin, 200 rad/s, it precesses at half that rate, 2.45 rad/s.50100150200250246810124.9 rad/s2.45 rad/sω (rad/s)Ω (rad/s)
The toy top's precession rate against its spin: 4.9 rad/s at a spin of 100 rad/s, and twice the spin gives 2.45 rad/s, half as fast. A faster spin means a bigger L, so the same sideways torque turns it more slowly.

Precession and gyroscopes

Faster spin, slower precession.

Precession and gyroscopes

Worked number: a toy top

A top with \(m = 0.2\) kg, \(r = 0.05\) m, \(I = 2\times10^{-4}\) kg m² and \(\omega = 100\) rad/s precesses at \(4.9\) rad/s (one circuit in 1.28 s).

Precession and gyroscopes

With no torque, a gyroscope's axis stays fixed in space, which is how Foucault showed the Earth turning. The Earth itself precesses once in about 26 000 years.

Precession and gyroscopes

Myth

"Gyroscopic effects keep a bicycle upright." A bicycle with the gyroscopic effect cancelled, and with no caster effect, still balanced itself. A bicycle leaning steers into the lean; what makes that work involves trail, steering-axis tilt, mass distribution and speed together.

Checkpoint: Precession and gyroscopes (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A tilted top's spin rate is tripled. What happens to its rate of precession?

Show the answer

It falls to a third.

\(\Omega\) is proportional to \(1/\omega\): triple the spin and the precession is a third as fast.

Checkpoint: Precession and gyroscopes (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A top has mass 0.30 kg, with its centre of mass 0.040 m from the pivot, a moment of inertia of \(3.0\times10^{-4}\) kg m², and a spin of 50 rad/s. At what rate does it precess, in rad/s? Take \(g = 9.8\) m/s².

rad/s

Show the answer

7.84 rad/s

\(\Omega = (0.30\times9.8\times0.040)/(3.0\times10^{-4}\times50) = 0.1176/0.015 = 7.84\) rad/s.

Checkpoint: Precession and gyroscopes (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

What did the 2011 self-stable bicycle experiment show?

Show the answer

A bicycle can balance itself even with the gyroscopic effect cancelled.

Kooijman and colleagues built a bicycle with its gyroscopic effect cancelled and no caster effect, and it still balanced itself: the answer is in trail, steering tilt, mass distribution and speed together.

Static equilibrium and the ladder

FIRST, MAKE A PREDICTION

A body has zero net force. Does that alone guarantee that it remains in static equilibrium?

Show the answer

No; the net torque must also be zero

Opposing forces can cancel translationally while forming a couple that changes rotation.

TRY IT / FOLLOW THE REASONING

Draw a pair of equal opposite forces with separated lines of action. Then draw a ladder's force diagram and choose a pivot that eliminates one unknown from its torque equation.

MAKE THE CONNECTION

Key idea

Something at rest and staying at rest has net force zero and net torque zero (about any point). Choose the pivot where an unknown force acts and it drops out.

Static equilibrium and the ladder

Worked number: the beam

A uniform 6 m, 30 kg beam on end supports, with a 60 kg person 2 m from the left end: \(R_2 \times 6 = 30g\times3 + 60g\times2\), so \(R_2 = 35g = 343\) N and \(R_1 = 90g - 343 = 540\) N.

Static equilibrium and the ladder

A ladder of weight \(W\) leaning at angle \(\theta\) to the ground on a smooth wall:

  • the wall pushes horizontally (\(N_w\));
  • the floor pushes up (\(N_f = W\));
  • friction at the foot equals \(N_w\).

Static equilibrium and the ladder

A uniform ladder leaning on a smooth wall at 60 degrees. Four forces: its weight W at the middle, the wall pushing 0.29 W sideways at the top, the floor pushing W up at the foot, and friction 0.29 W at the foot holding it in.60°weight Wwall push N = 0.29 Wfloor push N = Wfriction f = 0.29 W
The four forces on a ladder at 60°. Take torques about the foot, where two of the forces act, and they drop out. The wall pushes 0.29 W, so friction of 0.29 W holds the foot.

Static equilibrium and the ladder

Torques about the foot: \(N_w L\sin\theta = W\,(L/2)\cos\theta\), so

Static equilibrium and the ladder

\[ \begin{gathered} f = N_w = \frac{W}{2\tan\theta} \\[.6em] \mu \ge \frac{1}{2\tan\theta} \end{gathered} \]

Steeper ladders need less friction (\(\mu \ge 0.13\) at 75°, 0.29 at 60°, 0.5 at 45°), and the higher someone climbs, the more is needed.

Checkpoint: Static equilibrium and the ladder (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A uniform 3.0 m plank of mass 12 kg rests on supports at its two ends. A 18 kg box sits 1.0 m from the left end. What force does the right-hand support exert, in newtons? Use \(g = 9.81\) m/s².

N

Show the answer

117.72 N

Torques about the left end: \(R_2\times3.0 = 12g\times1.5 + 18g\times1.0 = 36g\), so \(R_2 = 12g = 117.7\) N.

Checkpoint: Static equilibrium and the ladder (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Why is it helpful to take torques about the foot of a leaning ladder?

Show the answer

Two of the unknown forces act at the foot, so they drop out of the torque equation.

You may take torques about any point. At the foot, the floor force and friction act with no lever arm, so only the wall force and the weights are left.

Checkpoint: Static equilibrium and the ladder (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A uniform ladder leans against a smooth wall at 70° to the ground. What is the least coefficient of friction at the foot to stop it slipping, when nobody is on it?

Show the answer

0.181985117133

\(\mu \ge 1/(2\tan70^\circ) = 1/(2\times2.747) = 0.18\): a steep ladder needs little friction.

Key terms

Key terms

Angular velocity (\(\omega\))
The rate at which the angle changes, in rad/s; \(v = r\omega\) for a point at distance \(r\).
Angular acceleration (\(\alpha\))
The rate at which \(\omega\) changes, in rad/s².
Torque (\(\tau\))
The turning effect of a force, \(\tau = rF\sin\theta\), in N m.
Lever arm
The shortest distance from the pivot to the line along which a force acts, \(r\sin\theta\).

Key terms

Cross product
A product of two vectors giving a vector at a right angle to both, with size \(ab\sin\theta\): \(\vec\tau = \vec r\times\vec F\).
Right-hand rule
Curl the fingers of your right hand the way something turns; the thumb gives the direction of its \(\vec\omega\), \(\vec\tau\) or \(\vec L\).
Moment of inertia (\(I\))
The resistance to being spun up, \(I = \sum mr^2\), in kg m²; \(\tau = I\alpha\).
Centre of mass
The mass-weighted average position of an object.

Key terms

Angular momentum (\(L\))
\(L = I\omega\) for a spinning body, \(\vec r\times\vec p\) for a particle; conserved when there is no net torque.
Precession
The slow circling of a spinning axis under a sideways torque, \(\Omega = mgr/(I\omega)\).
Equilibrium
Net force zero and net torque zero; stable, unstable or neutral.

History and sources

History

Where a primary text wasn't read, the sources are encyclopedic or institutional pages, and the wording below is cautious.

Archimedes and the lever (3rd century BC)

Archimedes and the lever (3rd century BC). In On the Equilibrium of Planes, Book I, Archimedes proved that magnitudes balance at distances inversely proportional to their weights, and went on to the centres of gravity of a triangle and a trapezium. The lever itself was far older. The line "give me a place to stand and I will move the Earth" is reported by Pappus more than five hundred years later: legend. Sources: On the Equilibrium of Planes; Archimedes (Wikiquote); Archimedes and the Law of the Lever.

The word "torque" (1884)

The word "torque" (1884). Said to have been suggested by James Thomson, brother of Lord Kelvin, and first in print in 1884 in Silvanus P. Thompson's Dynamo-Electric Machinery; from Latin torquere, "to twist". Source: Torque.

Hamilton, Gibbs and Heaviside (1843, 1880s)

Hamilton, Gibbs and Heaviside (1843, 1880s). On 16 October 1843, walking beside the Royal Canal in Dublin, William Rowan Hamilton saw how to multiply quaternions and cut the rule \(i^2 = j^2 = k^2 = ijk = -1\) into a stone of Brougham (Broome) Bridge, as he told his son Archibald in 1865. The modern dot and cross products were built from such ideas in the 1880s by J. Willard Gibbs at Yale (notes printed 1881–84 for his students) and, independently, Oliver Heaviside. Sources: Hamilton's letters on the discovery; Broom Bridge; Quaternions: a history (arXiv).

Euler and the moment of inertia (1765)

Euler and the moment of inertia (1765). Leonhard Euler's Theoria motus corporum solidorum seu rigidorum named and systematised the momentum inertiae; others, such as Johann Segner, had met the idea earlier. Sources: Euler's Theoria motus (17centurymaths); A historical discussion of angular momentum (arXiv).

Galileo's ramps (1638)

Galileo's ramps (1638). In Two New Sciences (Third Day) Galileo described rolling a bronze ball down a smooth groove and timing it with a water clock; distances in equal times went as 1, 3, 5, 7. (A solid ball rolls at 5/7 of the sliding acceleration, a fact he did not have.) Sources: Reconstructing Galileo's experiment (IMEKO); OU Libraries: inclined plane.

Marey's falling cat (1894) and Kane and Scher (1969)

Marey's falling cat (1894) and Kane and Scher (1969). Étienne-Jules Marey photographed a falling cat at 12 frames a second (Comptes Rendus 119, 1894): it starts with no spin. Kane and Scher (International Journal of Solids and Structures 5, 1969) modelled it with two bendable halves. Sources: Falling cat problem; Photographs of a Falling Cat.

Newton's area theorem (1687)

Newton's area theorem (1687). Principia, Book I, Section II, Proposition 1: for any force aimed at a fixed centre, a body sweeps equal areas in equal times. Source: Newton's Principia, Book I, §2 (Ian Bruce).

The first pulsar (1967–68)

The first pulsar (1967–68). Jocelyn Bell, a Cambridge research student, noticed a signal pulsing every 1.337 s (recorded 28 November 1967). The discovery paper by Hewish, Bell, Pilkington, Scott and Collins is in Nature 217 (1968) 709. Thomas Gold explained it as a rotating neutron star (Nature 218, 1968). The 1974 Nobel Prize in Physics went to Hewish and Martin Ryle, not to Bell. Sources: Physics Today; 50 Years of Pulsars (arXiv).

Foucault's gyroscope (1851–52)

Foucault's gyroscope (1851–52). After his pendulum (1851), Léon Foucault built a spinning wheel in gimbals (1852) whose axis stays fixed in space while the Earth turns, and named it the gyroscope. Sources: Foucault's gyroscope; Foucault pendulum.

Hipparchus and precession (about 130 BC)

Hipparchus and precession (about 130 BC). Hipparchus compared the position of Spica with older records and found the sky's pattern slides by about 1° per century; Newton identified the cause as the Sun's and Moon's pull on the Earth's equatorial bulge. The wobble takes about 26 000 years. Sources: Hipparchus and the discovery of precession.

The self-stable bicycle (2011)

The self-stable bicycle (2011). J. D. G. Kooijman, J. P. Meijaard, J. M. Papadopoulos, A. Ruina and A. L. Schwab, "A bicycle can be self-stable without gyroscopic or caster effects", Science 332 (2011) 339–342. Sources: SWOV record; ScienceDaily.

Worked example: who wins the race?

The problem

Problem. A ring, a solid disc and a solid sphere, all with the same mass and radius, are released together from rest at the top of a slope 2.0 m long at 30°. They roll without slipping. Who arrives first, how long does each take, and how much friction does the slope need?

Solution

  1. Forces. Weight \(mg\), the slope's push \(N = mg\cos\theta\), and friction \(f\) up the slope at the contact point.
  2. Moving. \(ma = mg\sin\theta - f\).
  3. Turning about the centre. \(fR = I\alpha = (kmR^2)(a/R)\), so \(f = kma\).
  4. Solve. \(ma = mg\sin\theta - kma\), so \(a = g\sin\theta/(1+k)\). The mass and radius cancel.

Solution

  1. Numbers. \(g\sin30^\circ = 4.905\). Ring (\(k=1\)): \(a = 2.45\) m/s². Disc (\(k=\tfrac12\)): \(3.27\). Sphere (\(k=\tfrac25\)): \(3.50\).
  2. Times. \(t = \sqrt{2L/a} = \sqrt{4/a}\): ring 1.28 s, disc 1.11 s, sphere 1.07 s. Speeds \(v = at\): 3.13, 3.62, 3.74 m/s.

Solution

  1. Order. The sphere wins; the disc is 0.04 s behind it and the ring 0.21 s behind. A frictionless sliding block would take \(\sqrt{4/4.905} = 0.90\) s.
  2. Check by energy. \(h = 2.0\sin30^\circ = 1.0\) m, \(v = \sqrt{2gh/(1+k)}\): \(\sqrt{19.62/2} = 3.13\), \(\sqrt{19.62/1.5} = 3.62\), \(\sqrt{19.62/1.4} = 3.74\) ✓.
  3. Friction. \(\mu \ge f/N = k\tan\theta/(1+k)\), with \(\tan30^\circ = 0.577\): ring 0.29, disc 0.19, sphere 0.16. On a slipperier slope they would slide as well as roll.

Practice

Practice

Use \(g = 9.81\) m/s² unless told otherwise.

★ Torques

1.1 A 0.30 m spanner is pulled at its end with 45 N, at 40° to the spanner. What is the torque on the nut, in newton-metres? Then compare it with a pull at 90°.

N m

Solution

\(\tau = rF\sin\theta = 0.30 \times 45 \times \sin40^\circ = 13.5 \times 0.643 \approx 8.7\) N m. At 90° it would be \(0.30 \times 45 \times 1 = 13.5\) N m. Check: the slanted pull gives less than the square-on pull, as it must ✓.

Problem 1.2 (★ Torques)

1.2 A 35 kg child sits 1.6 m from the pivot of a seesaw. Where must a 28 kg child sit on the other side to balance it, in metres from the pivot?

m

Solution

The torques are equal: \(35g \times 1.6 = 28g \times d\). The \(g\) cancels: \(d = 35 \times 1.6/28 = 2.0\) m. Check: the lighter child sits further out ✓.

Problem 1.3 (★ Torques)

1.3 A wheel of radius 0.40 m starts from rest and speeds up steadily at 1.5 rad/s² for 6.0 s. Find its final \(\omega\) and the angle turned (in turns). Then find the speed of a point on the rim, in metres per second.

m/s

Solution

\(\omega = \alpha t = 1.5 \times 6.0 = 9.0\) rad/s. \(\theta = \tfrac12\alpha t^2 = \tfrac12 \times 1.5 \times 36 = 27\) rad, which is \(27/2\pi \approx 4.3\) turns. Rim speed \(v = r\omega = 0.40 \times 9.0 = 3.6\) m/s. Check: 9 rad/s is about 86 rev/min, and 3.6 m/s is a brisk walk ✓.

Problem 1.4 (★ Torques)

1.4 A force of 20 N points straight up the page at a point 0.5 m to the right and 0.2 m above the pivot. Find the torque about the pivot, in newton-metres, and say which way it points.

N m

Solution

With \(\vec r = (0.5, 0.2)\) and \(\vec F = (0, 20)\): \(\tau_z = xF_y - yF_x = 0.5\times20 - 0.2\times0 = 10\) N m. The sign is positive, so the torque points out of the page (anticlockwise). Check with the right-hand rule: fingers along \(\vec r\) (right), curl towards \(\vec F\) (up): thumb out of the page ✓.

Problem 1.5 (★ Torques)

1.5 A record turns at 45 rev/min. What is its angular velocity, in radians per second?

rad/s

Solution

One turn is \(2\pi\) rad and a minute is 60 s: \(\omega = 45 \times 2\pi/60 = 4.71\) rad/s. Check: 45 is a little more than \(33\tfrac13\), so \(\omega\) should be a little more than that record's 3.49 rad/s ✓.

★★ Rolling motion

2.1 A solid disc has mass 2.0 kg and radius 0.15 m. What torque gives it an angular acceleration of 8.0 rad/s², in newton-metres?

N m

Solution

\(I = \tfrac12 MR^2 = \tfrac12 \times 2.0 \times 0.15^2 = 0.0225\) kg m². \(\tau = I\alpha = 0.0225 \times 8.0 = 0.18\) N m.

Problem 2.2 (★★ Rolling motion)

2.2 A solid cylinder of mass 0.50 kg rolls without slipping at 2.0 m/s. Find its total kinetic energy, in joules, and the part that is spin.

J

Solution

Moving along: \(\tfrac12 \times 0.50 \times 2.0^2 = 1.0\) J. Spin: \(\tfrac12 kMv^2 = \tfrac12 \times \tfrac12 \times 0.50 \times 4 = 0.5\) J. Total \(= 1.5\) J (\(= \tfrac12Mv^2(1+k)\)). The spin is a third of the total ✓ (\(k/(1+k) = \tfrac13\)).

Problem 2.3 (★★ Rolling motion)

2.3 A solid sphere is released from rest at the top of a ramp 1.5 m high and rolls to the bottom without slipping. How fast is it moving at the bottom, in metres per second? How fast would a frictionless sliding block be?

m/s

Solution

Sphere: \(v = \sqrt{2gh/(1+k)} = \sqrt{2\times9.81\times1.5/1.4} = \sqrt{21.0} \approx 4.58\) m/s. Block: \(v = \sqrt{2gh} = \sqrt{29.4} \approx 5.42\) m/s. Check: the roller is slower, because some energy went into spin ✓.

Problem 2.4 (★★ Rolling motion)

2.4 Masses of 2 kg at \(x = 0\), 3 kg at \(x = 2\) m and 5 kg at \(x = 5\) m lie on a light rod. Where is the centre of mass, in metres?

m

Solution

\(x_{cm} = (2\times0 + 3\times2 + 5\times5)/(2+3+5) = 31/10 = 3.1\) m. Check: it lies between the extremes, nearer the heavy mass ✓.

Problem 2.5 (★★ Rolling motion)

2.5 A solid disc is released from rest at the top of a slope 2.0 m long, tilted at 30° to the level. It rolls without slipping. How long does it take to reach the bottom, in seconds?

s

Solution

For a disc \(k = \tfrac12\), so \(a = g\sin\theta/(1+k) = 9.81\times0.5/1.5 = 3.27\) m/s². From rest, \(L = \tfrac12at^2\), so \(t = \sqrt{2L/a} = \sqrt{4.0/3.27} = 1.11\) s. Check: this is the disc's time in the Worked example, and a little longer than the sphere's 1.07 s ✓.

Problem 2.6 (★★ Rolling motion)

2.6 A skater with \(I = 3.0\) kg m² turns at 6.28 rad/s (once a second). She pulls in her arms and \(I\) becomes 1.0 kg m². What is her new angular velocity, in radians per second?

rad/s

Solution

\(L = I\omega\) is conserved: \(3.0\times6.28 = 1.0\times\omega_2\), so \(\omega_2 = 18.8\) rad/s, three turns a second. Check: a third of the moment of inertia gives three times the spin ✓.

★★★ The ladder, and a flipped wheel

3.1 A uniform 6.0 m ladder of mass 20 kg leans against a smooth wall at 65° to the ground. A 70 kg person stands 3.0 m up the ladder (measured along it). Find the friction force at the foot, in newtons, and the least coefficient of friction.

Hint

Take torques about the foot. The ladder's weight acts at its middle, the person's at 3.0 m along, and the wall's push at height \(L\sin\theta\).

N

Solution

Torques about the foot: \(N_w \times 6.0\sin65^\circ = 20g \times 3.0\cos65^\circ + 70g \times 3.0\cos65^\circ = 270g\cos65^\circ\). So \(N_w = 270g/(6.0\tan65^\circ) = 45g/2.145 \approx 206\) N. Friction \(f = N_w \approx 206\) N. The floor's push is \(N_f = 90g \approx 883\) N, so \(\mu \ge 206/883 \approx 0.23\). Check: with no person it would be \(f = 20g/(2\tan65^\circ) = 45.7\) N, so the climber raises the requirement a lot ✓.

Problem 3.2 (★★★ The ladder, and a flipped wheel)

3.2 A person sits on a freely turning stool (the person and stool have \(I = 1.5\) kg m²), holding a bicycle wheel (a thin ring of mass 1.2 kg and radius 0.34 m) with its axle vertical, spinning at 12 rad/s. They flip the wheel upside down. How fast do they turn, in radians per second?

Hint

The wheel's angular momentum changes from \(+L\) to \(-L\), a change of \(2L\). The person and stool must gain \(2L\).

rad/s

Solution

\(I_{wheel} = MR^2 = 1.2 \times 0.34^2 = 0.1387\) kg m², so \(L = I\omega = 0.1387 \times 12 = 1.665\) kg m²/s. After the flip the wheel has \(-1.665\), so the person and stool have \(+3.329\): \(\omega = L/I = 3.329/1.5 \approx 2.2\) rad/s. Check: about a third of a turn per second ✓.

Problem 3.3 (★★★ The ladder, and a flipped wheel)

3.3 Describe, using angular momentum, why a spinning top at a slant circles instead of toppling, and what happens to its rate of circling if its spin is doubled.

Solution

Gravity's torque \(\vec r\times m\vec g\) is horizontal, at a right angle to \(\vec L\). It changes \(\vec L\) sideways without changing its length, so the axis swings round: precession, \(\Omega = mgr/(I\omega)\). Doubling \(\omega\) halves \(\Omega\) ✓: it circles half as fast, though it still does not topple.

Problem 3.4 (★★★ The ladder, and a flipped wheel)

3.4 A top with \(m = 0.15\) kg has its centre of mass 0.04 m from the pivot. Its moment of inertia is \(1.5\times10^{-4}\) kg m² and it spins at 120 rad/s. At what rate does it precess, in radians per second? Take \(g = 9.8\) m/s².

rad/s

Solution

\(\Omega = \dfrac{mgr}{I\omega} = \dfrac{0.15\times9.8\times0.04}{1.5\times10^{-4}\times120} = \dfrac{0.0588}{0.018} \approx 3.27\) rad/s, one circuit in about 1.9 s. Check: the toy top in the Notes spins at 100 rad/s and precesses at 4.9 rad/s; this one is lighter and spins faster, so it is slower ✓.

Problem 3.5 (★★★ The ladder, and a flipped wheel)

3.5 A uniform 4.0 m beam of mass 20 kg rests on supports at its two ends. A 50 kg person stands 1.0 m from the left end. What force does the right-hand support exert, in newtons?

Hint

Take torques about the left-hand support, so its unknown force drops out.

N

Solution

Torques about the left end: \(R_2\times4.0 = 20g\times2.0 + 50g\times1.0 = 90g\), so \(R_2 = 22.5g \approx 221\) N. The left support carries the rest: \(R_1 = 70g - 221 = 466\) N. Check: the person is nearer the left end, so it carries more ✓.

Check yourself

Answer these out loud, from memory. Then try again in a few days.

Check yourself

  • How are \(\theta\), \(\omega\) and \(\alpha\) related, and how do the constant-acceleration equations carry over?

Check yourself

  • How are \(v = r\omega\) and \(a = r\alpha\) found for a point at distance \(r\)?

Check yourself

  • What is torque, and why does only the perpendicular part of a force count?

Check yourself

  • What is the lever arm, and how does it give the same torque?

Check yourself

  • Why is a newton-metre of torque not a joule?

Check yourself

  • What does the right-hand rule give, and how is the cross product \(\vec r\times\vec F\) found?

Check yourself

  • When is the net torque on an object zero, and what happens when it is not?

Check yourself

  • What is the moment of inertia, and why does distance from the axis count squared?

Check yourself

  • What are \(I\) for a ring, a disc and a sphere, and why do they differ?

Check yourself

  • What is the centre of mass, and when does an object tip over?

Check yourself

  • What does rolling without slipping mean for the contact point and the top?

Check yourself

  • Why does the mass cancel in a rolling race, and why does the sphere win?

Check yourself

  • What is angular momentum, and when is it conserved?

Check yourself

  • Why does a skater speed up when she pulls her arms in, and where does the extra energy come from?

Check yourself

  • How can a cat or an astronaut turn over with zero angular momentum?

Check yourself

  • How does conservation of angular momentum explain Kepler's second law, and pulsars?

Check yourself

  • Why does a spinning top precess, and what does the rate depend on?

Check yourself

  • What did the self-stable bicycle experiment show?

Check yourself

  • What are the two conditions for equilibrium, and how does choosing the pivot help?

Check yourself

  • Why does a ladder need friction at the foot, and how does the angle change how much?

Summary

Summary

KEEP THE THREAD GOING

Next: Matter in bulk: solids and fluids

Continue when you’re ready, or return to any question above.

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Further learning

Free textbook: OpenStax, University Physics Volume 1 (not yet checked): Chapter 10: Fixed-Axis Rotation · Chapter 11: Angular Momentum · Chapter 12: Static Equilibrium and Elasticity.

Reading: Photographs of a Falling Cat (1894), The Public Domain Review · Falling cat problem · Foucault's gyroscope.

Next: P6 · Matter in bulk: solids and fluids · Progress