<< Chapter < Page Chapter >> Page >
  • Understand the analogy between angular momentum and linear momentum.
  • Observe the relationship between torque and angular momentum.
  • Apply the law of conservation of angular momentum.

Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum.

By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum     L size 12{L} {} as

L = . size 12{L=Iω} {}

This equation is an analog to the definition of linear momentum as p = mv size 12{p= ital "mv"} {} . Units for linear momentum are kg m /s size 12{"kg" cdot m rSup { size 8{2} } "/s"} {} while units for angular momentum are kg m 2 /s size 12{"kg" cdot m rSup { size 8{2} } "/s"} {} . As we would expect, an object that has a large moment of inertia I size 12{I} {} , such as Earth, has a very large angular momentum. An object that has a large angular velocity ω size 12{ω} {} , such as a centrifuge, also has a rather large angular momentum.

Making connections

Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation . It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles.

Calculating angular momentum of the earth


No information is given in the statement of the problem; so we must look up pertinent data before we can calculate L = size 12{L=Iω} {} . First, according to [link] , the formula for the moment of inertia of a sphere is

I = 2 MR 2 5 size 12{I= { {2 ital "MR" rSup { size 8{2} } } over {5} } } {}

so that

L = = 2 MR 2 ω 5 . size 12{L=Iω= { {2 ital "MR" rSup { size 8{2} } ω} over {5} } } {}

Earth’s mass M size 12{M} {} is 5 . 979 × 10 24 kg size 12{5 "." "979" times "10" rSup { size 8{"24"} } "kg"} {} and its radius R size 12{R} {} is 6 . 376 × 10 6 m size 12{6 "." "376" times "10" rSup { size 8{6} } m} {} . The Earth’s angular velocity ω size 12{ω} {} is, of course, exactly one revolution per day, but we must covert ω size 12{ω} {} to radians per second to do the calculation in SI units.


Substituting known information into the expression for L size 12{L} {} and converting ω size 12{ω} {} to radians per second gives

L = 0 . 4 5 . 979 × 10 24 kg 6 . 376 × 10 6 m 2 1 rev d = 9 . 72 × 10 37 kg m 2 rev/d . alignl { stack { size 12{L=0 "." 4 left (5 "." "979" times "10" rSup { size 8{"24"} } " kg" right ) left (6 "." "376" times "10" rSup { size 8{6} } " m" right ) rSup { size 8{2} } left ( { {1" rev"} over {d} } right )} {} #" "=9 "." "72" times "10" rSup { size 8{"37"} } " kg" cdot m rSup { size 8{2} } "rev/d" {} } } {}

Substituting size 12{2π} {} rad for 1 size 12{1} {} rev and 8 . 64 × 10 4 s size 12{8 "." "64" times "10" rSup { size 8{4} } s} {} for 1 day gives

L = 9 . 72 × 10 37 kg m 2 rad/rev 8 . 64 × 10 4 s/d 1 rev/d = 7 . 07 × 10 33 kg m 2 /s . alignl { stack { size 12{L= left (9 "." "72" times "10" rSup { size 8{"37"} } " kg" cdot m rSup { size 8{2} } right ) left ( { {2π" rad/rev"} over {8 "." "64" times "10" rSup { size 8{4} } " s/d"} } right ) left (1" rev/d" right )} {} #" "=7 "." "07" times "10" rSup { size 8{"33"} } " kg" cdot m rSup { size 8{2} } "/s" {} } } {}


This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia.

When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in L size 12{L} {} . The relationship between torque and angular momentum is

net τ = Δ L Δ t . size 12{"net "τ= { {ΔL} over {Δt} } } {}

This expression is exactly analogous to the relationship between force and linear momentum, F = Δ p / Δ t size 12{F=Δp/Δt} {} . The equation net τ = Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law.

Calculating the torque putting angular momentum into a lazy susan

[link] shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan’s 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk?

The given figure shows a lazy Susan on which various eatables like cake, salad grapes, and a drink are kept. A hand is shown that applies a force F, indicated by a leftward pointing horizontal arrow. This force is perpendicular to the radius r and thus tangential to the circular lazy Susan.
A partygoer exerts a torque on a lazy Susan to make it rotate. The equation net τ = Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} gives the relationship between torque and the angular momentum produced.


We can find the angular momentum by solving net τ = Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} for Δ L size 12{ΔL} {} , and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, Δ L = L size 12{ΔL=L} {} . To find the final velocity, we must calculate ω size 12{ω} {} from the definition of L size 12{L} {} in L = size 12{L=Iω} {} .

Solution for (a)

Solving net τ = Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} for Δ L size 12{ΔL} {} gives

Δ L = net τ Δt . size 12{ΔL= left ("net "τ right ) cdot Δt} {}

Because the force is perpendicular to r size 12{r} {} , we see that net τ = rF size 12{"net "τ= ital "rF"} {} , so that

L = rF Δ t = ( 0 . 260 m ) ( 2.50 N ) ( 0.150 s ) = 9 . 75 × 10 2 kg m 2 / s .

Solution for (b)

The final angular velocity can be calculated from the definition of angular momentum,

L = . size 12{L=Iω} {}

Solving for ω size 12{ω} {} and substituting the formula for the moment of inertia of a disk into the resulting equation gives

ω = L I = L 1 2 MR 2 . size 12{ω= { {L} over {I} } = { {L} over { { size 8{1} } wideslash { size 8{2} } ital "MR" rSup { size 8{2} } } } } {}

And substituting known values into the preceding equation yields

ω = 9 . 75 × 10 2 kg m 2 /s 0 . 500 4 . 00 kg 0 . 260 m = 0 . 721 rad/s . size 12{ω= { {9 "." "75" times "10" rSup { size 8{ - 2} } " kg" cdot m rSup { size 8{2} } "/s"} over { left (0 "." "500" right ) left (4 "." "00"" kg" right ) left (0 "." "260"" m" right )} } =0 "." "721"" rad/s"} {}


Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
I'm not good at math so would you help me
what is the problem that i will help you to self with?
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply
Practice Key Terms 2

Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Unit 8 - rotational motion. OpenStax CNX. Feb 22, 2016 Download for free at https://legacy.cnx.org/content/col11970/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Unit 8 - rotational motion' conversation and receive update notifications?