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What we have here is, in fact, another conservation law. If the net torque is zero , then angular momentum is constant or conserved . We can see this rigorously by considering net τ = Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} for the situation in which the net torque is zero. In that case,

net τ = 0 size 12{"net "τ=0} {}

implying that

Δ L Δ t = 0 . size 12{ { {ΔL} over {Δt} } =0} {}

If the change in angular momentum Δ L size 12{ΔL} {} is zero, then the angular momentum is constant; thus,

L = constant net τ = 0 size 12{L="constant " left ("net "τ=0 right )} {}

or

L = L net τ = 0 . size 12{L=L'" " left ("net "τ=0 right )} {}

These expressions are the law of conservation of angular momentum    . Conservation laws are as scarce as they are important.

An example of conservation of angular momentum is seen in [link] , in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both F size 12{F} {} and r size 12{r} {} are small, and so τ size 12{τ} {} is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that

L = L . size 12{L=L'} {}

Expressing this equation in terms of the moment of inertia,

= I ω , size 12{Iω=I'ω'} {}

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I size 12{I'} {} is smaller, the angular velocity ω size 12{ω'} {} must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

The image a shows an ice skater spinning on the tip of her skate with both her arms and one leg extended. The image b shows the ice skater spinning on the tip of one skate, with her arms crossed and one leg supported on another.
(a) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy.

Calculating the angular momentum of a spinning skater

Suppose an ice skater, such as the one in [link] , is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2 . 34 kg m 2 size 12{2 "." "34"`"kg" cdot m rSup { size 8{2} } } {} with her arms extended and of 0 . 363 kg m 2 size 12{0 "." "363"`"kg" cdot m rSup { size 8{2} } } {} with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this?

Strategy

In the first part of the problem, we are looking for the skater’s angular velocity ω size 12{ { {ω}} sup { ' }} {} after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by

KE rot = 1 2 2 . size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}

Solution for (a)

Because torque is negligible (as discussed above), the conservation of angular momentum given in = I ω size 12{Iω= { {I}} sup { ' } { {ω}} sup { ' }} {} is applicable. Thus,

L = L size 12{L=L'} {}

or

= I ω size 12{Iω=I'ω'} {}

Solving for ω and substituting known values into the resulting equation gives

ω = I I ω = 2.34 kg m 2 0 .363 kg m 2 0.800 rev/s = 5.16 rev/s.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
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ninjadapaul
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
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ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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Abhi
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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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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