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  • 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

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Source:  OpenStax, Introduction to applied math and physics. OpenStax CNX. Oct 04, 2012 Download for free at http://cnx.org/content/col11426/1.3
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